% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
%                                                                         %
% The Project Gutenberg EBook of Plane Geometry, by George Albert Wentworth
%                                                                         %
% This eBook is for the use of anyone anywhere at no cost and with        %
% almost no restrictions whatsoever.  You may copy it, give it away or    %
% re-use it under the terms of the Project Gutenberg License included     %
% with this eBook or online at www.gutenberg.org                          %
%                                                                         %
%                                                                         %
% Title: Plane Geometry                                                   %
%                                                                         %
% Author: George Albert Wentworth                                         %
%                                                                         %
% Release Date: July 3, 2010 [EBook #33063]                               %
%                                                                         %
% Language: English                                                       %
%                                                                         %
% Character set encoding: ISO-8859-1                                      %
%                                                                         %
% *** START OF THIS PROJECT GUTENBERG EBOOK PLANE GEOMETRY ***            %
%                                                                         %
% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %

\def\ebook{33063}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%                                                                  %%
%% Packages and substitutions:                                      %%
%%                                                                  %%
%% book:     Required.                                              %%
%% inputenc: Standard DP encoding. Required.                        %%
%%                                                                  %%
%% ifthen:   Logical conditionals. Required.                        %%
%%                                                                  %%
%% amsmath:  AMS mathematics enhancements. Required.                %%
%% amssymb:  Additional mathematical symbols. Required.             %%
%%                                                                  %%
%% alltt:    Fixed-width font environment. Required.                %%
%%                                                                  %%
%% footmisc: Extended footnote capabilities. Optional.              %%
%%                                                                  %%
%% indentfirst: Indent first word of each sectional unit. Optional. %%
%%                                                                  %%
%% fancyhdr: Enhanced running headers and footers. Required.        %%
%% graphicx: Standard interface for graphics inclusion. Required.   %%
%%                                                                  %%
%% geometry: Enhanced page layout package. Required.                %%
%% hyperref: Hypertext embellishments for pdf output. Required.     %%
%%                                                                  %%
%% enumerate: Configurable enumeration labels. Required.            %%
%% supertabular: The supertabular environment. Required.            %%
%%                                                                  %%
%%                                                                  %%
%% Producer's Comments:                                             %%
%%                                                                  %%
%% lacheck: ..................................... OK                %%
%%   Numerous false positives                                       %%
%%                                                                  %%
%% PDF pages: 326 (if ForPrinting set to false)                     %%
%% PDF page size: 6 x 8.5in (non-standard)                          %%
%% PDF bookmarks: created, point to ToC entries                     %%
%% PDF document info: filled in                                     %%
%% Images: 313 pdf diagrams                                         %%
%%                                                                  %%
%% Summary of log file:                                             %%
%% * One harmless underfull hbox on title page.                     %%
%%                                                                  %%
%% Compile History:                                                 %%
%%                                                                  %%
%% Jeremy Weatherford <jweather -at- xidus -dot- net> (PG xidus)    %%
%%     ca. 2005                                                     %%
%%     Fedora Core 3 Linux                                          %%
%%     tetex-2.0.2-21 RPM                                           %%
%%     tetex-latex-2.0.2-21 RPM                                     %%
%%                                                                  %%
%% Andrew D. Hwang (DP:adhere)                                      %%
%%     July 2010                                                    %%
%%     texlive2007, GNU/Linux                                       %%
%%                                                                  %%
%% Majority of figures contributed by:                              %%
%% Owen Whitby <owhitby -at- aol -dot- com> (PG nilrem)             %%
%%                                                                  %%
%% Command block:                                                   %%
%%                                                                  %%
%%     pdflatex x2                                                  %%
%%                                                                  %%
%%                                                                  %%
%% July 2010: pglatex.                                              %%
%%   Compile this project with:                                     %%
%%   pdflatex 33063-t.tex ..... TWO times                           %%
%%                                                                  %%
%%   pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6)                 %%
%%                                                                  %%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\listfiles
\documentclass[12pt,titlepage]{book}[2005/09/16]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%% PACKAGES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage[latin1]{inputenc}[2006/05/05]

\usepackage{ifthen}[2001/05/26]  %% Logical conditionals

\usepackage{amsmath}[2000/07/18] %% Displayed equations
\usepackage{amssymb}[2002/01/22] %% and additional symbols

\usepackage{alltt}[1997/06/16]   %% boilerplate, credits, license

                                 %% extended footnote capabilities
\IfFileExists{footmisc.sty}{%
  \usepackage[symbol]{footmisc}[2005/03/17]
}{}

\IfFileExists{indentfirst.sty}{%
  \usepackage{indentfirst}[1995/11/23]
}{}

\usepackage{fancyhdr}
\usepackage{graphicx}[1999/02/16]%% For diagrams

\usepackage{enumerate}[1999/03/05]
\usepackage{supertabular}[2004/02/20]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%% Interlude:  Set up PRINTING (default) or SCREEN VIEWING %%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% ForPrinting=true (default)           false
% Asymmetric margins                   Symmetric margins
% Black hyperlinks                     Blue hyperlinks
% Start Preface, ToC, etc. recto       No blank verso pages
%
% Chapter-like ``Sections'' start both recto and verso in the scanned
% book. This behavior has been retained.
\newboolean{ForPrinting}

%% UNCOMMENT the next line for a PRINT-OPTIMIZED VERSION of the text %%
%\setboolean{ForPrinting}{true}

%% Initialize values to ForPrinting=false
\newcommand{\Margins}{hmarginratio=1:1}     % Symmetric margins
\newcommand{\HLinkColor}{blue}              % Hyperlink color
\newcommand{\PDFPageLayout}{SinglePage}
\newcommand{\TransNote}{Transcriber's Note---}
\newcommand{\TransNoteCommon}{%
  Minor typographical corrections and presentational changes have
  been made without comment.
  \bigskip
}

\newcommand{\TransNoteText}{%
  \TransNoteCommon

  This PDF file is optimized for screen viewing, but may easily be
  recompiled for printing. Please see the preamble of the \LaTeX\
  source file for instructions.
}
%% Re-set if ForPrinting=true
\ifthenelse{\boolean{ForPrinting}}{%
  \renewcommand{\Margins}{hmarginratio=2:3} % Asymmetric margins
  \renewcommand{\HLinkColor}{black}         % Hyperlink color
  \renewcommand{\PDFPageLayout}{TwoPageRight}
  \renewcommand{\TransNoteText}{%
    \TransNoteCommon

    This PDF file is optimized for printing, but may easily be
    recompiled for screen viewing. Please see the preamble of the
    \LaTeX\ source file for instructions.
  }
}{% If ForPrinting=false, don't skip to recto
  \renewcommand{\cleardoublepage}{\clearpage}
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%  End of PRINTING/SCREEN VIEWING code; back to packages  %%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\ifthenelse{\boolean{ForPrinting}}{%
  \setlength{\paperwidth}{8.5in}%
  \setlength{\paperheight}{11in}%
  \usepackage[body={5.5in,7.33in},\Margins]{geometry}[2002/07/08]
}{%
  \setlength{\paperwidth}{6in}%
  \setlength{\paperheight}{8.5in}%
  \usepackage[body={5.5in,7.33in},\Margins,includeheadfoot]{geometry}[2002/07/08]
}

\providecommand{\ebook}{00000}    % Overridden during white-washing
\usepackage[pdftex,
  hyperref,
  hyperfootnotes=false,
  pdftitle={The Project Gutenberg eBook \#\ebook: Plane Geometry},
  pdfauthor={George Albert Wentworth},
  pdfkeywords={Jeremy Weatherford, Stan Goodman, Kevin Handy,
               Project Gutenberg Online Distributed Proofreading Team},
  pdfstartview=Fit,    % default value
  pdfstartpage=1,      % default value
  pdfpagemode=UseNone, % default value
  bookmarks=true,      % default value
  linktocpage=false,   % default value
  pdfpagelayout=\PDFPageLayout,
  pdfdisplaydoctitle,
  pdfpagelabels=true,
  bookmarksopen=true,
  bookmarksopenlevel=0,
  colorlinks=true,
  linkcolor=\HLinkColor]{hyperref}[2007/02/07]

% Re-crop screen-formatted version, accommodating wide displays
\ifthenelse{\boolean{ForPrinting}}
  {}
  {\hypersetup{pdfpagescrop= 0 30 612 765}}


%%%% Fixed-width environment to format PG boilerplate %%%%
\newenvironment{PGtext}{%
\begin{alltt}
\fontsize{9.2}{10.5}\ttfamily\selectfont}%
{\end{alltt}}
\newlength{\TmpLen}

% [** TN: The Note to Teachers makes verbal reference to a portion of the book]
\newcommand{\PageName}[1]{%
  \ifthenelse{\equal{\pageref{page:PageName}}{17}}{%
    seventeen % Value at current text block size
  }{%
    \ifthenelse{\equal{\pageref{page:PageName}}{14}}{%
      fourteen
    }{%
      \ifthenelse{\equal{\pageref{page:PageName}}{15}}{%
        fifteen
      }{%
        \ifthenelse{\equal{\pageref{page:PageName}}{16}}{%
          sixteen
        }{%
          \ifthenelse{\equal{\pageref{page:PageName}}{18}}{%
            eighteen
          }{%
            \ifthenelse{\equal{\pageref{page:PageName}}{19}}{%
              nineteen
            }{%
              \pageref{page:PageName} % Punt
            }%
          }%
        }%
      }%
    }%
  }%
}


\newcommand{\DittoMark}{``}%''
\newcommand{\Ditto}{\quad\DittoMark\quad}
\DeclareInputText{176}{\ifmmode{{}^\circ}\else\textdegree\fi}

\newcommand{\Headings}[1]{\textbf{\small#1}}

%[** TN: Make all parentheses upright.
%        Code due to Will Robertson in comp.text.tex, Sept 3, 2008]
\makeatletter
\DeclareRobustCommand\IfMath{%
  \ifmmode
    \expandafter\@firstoftwo
  \else
    \expandafter\@secondoftwo
  \fi}
\begingroup
\global\let\origlparen=(
\global\let\origrparen=)
\catcode`\(=\active
\catcode`\)=\active
\AtBeginDocument{%
  \catcode`\(=\active
  \catcode`\)=\active
  \def({\IfMath{\origlparen}{\textup\origlparen}}%
  \def){\IfMath{\origrparen}{\textup\origrparen}}%
}

\endgroup
\makeatother

% tweak page layout
\emergencystretch=11pt
\raggedbottom
\setlength{\parskip}{1mm}

\makeatletter

% convenience
\newcommand{\hsp}{\hspace{\stretch{1}}}

% Page separators
\newcommand{\scanpage}[1]{}

\newcommand{\indexbf}[1]{\textbf{#1}\label{#1}}
\newcommand{\indexemph}[1]{\emph{#1}\label{#1}}


% section heading styles in the text
\renewcommand{\part}[2][]{%
  \def\mypartname{#2}%
  \clearpage
  \thispagestyle{plain}%
  \addcontentsline{toc}{part}{#2}%
  \def\firstchap{yes}%
  {\centering \normalfont \Huge #2\par}%
}

% chapter
\renewcommand{\chapter}{
  \ifthenelse{\equal{\firstchap}{yes}}{\vspace*{30\p@}}{\clearpage\thispagestyle{plain}}%
  \def\firstchap{no}%
  \global\@topnum\z@
  \@afterindentfalse
  \secdef\@chapter\@schapter}

\def\firstchap{yes}
% modify chapter headings
\renewcommand{\@makechapterhead}[1]{%
  {\centering \large \scshape #1\par\vskip 40\p@}}

% section
\def\section{\@startsection{section}{3}{\z@}
  {\baselineskip}{\parskip}%
  {\bfseries\centering}}

% subsection
\def\subsection{\@startsection{subsection}{4}{\z@}
  {\baselineskip}{\parskip}%
  {\bfseries\centering}}

\newcommand{\proposition}[1]{\filbreak\bigskip\centerline{\textsc{\theproposition{} #1}}\medskip}

% numbering depth and TOC depth
\setcounter{tocdepth}{1}
\setcounter{secnumdepth}{6}

% section naming and numbering styles
\def\partname{}
\def\chaptername{}
\def\sectionname{}
\def\subsectionname{Proposition}

\renewcommand{\thepart}{}
\renewcommand{\thechapter}{}
\renewcommand{\thesection}{}
\renewcommand{\thesubsection}{}

\newcounter{proposition}[chapter]
\renewcommand{\theproposition}{\stepcounter{proposition}Proposition \Roman{proposition}.}

% running headers
\pagestyle{fancy}
\renewcommand{\chaptermark}[1]{\markboth{BOOK \Roman{chapter}. \mypartname}{}}
\renewcommand{\sectionmark}[1]{\markright{#1}}

\newcommand{\SetRunningHeads}{%
  \fancyhf{}
  \ifthenelse{\boolean{ForPrinting}}{%
    \fancyhead[LE,RO]{\Headings{\thepage}}
  }{%
    \fancyhead[R]{\Headings{\thepage}}
  }
  \fancyhead[CO]{\Headings{\rightmark}}
  \fancyhead[CE]{\Headings{\leftmark}}
}
\renewcommand{\headrulewidth}{0.5pt}
\renewcommand{\footrulewidth}{0.0pt}
\addtolength{\headheight}{2.5pt}
\fancypagestyle{plain}{
  \fancyhead{}
  \renewcommand{\headrulewidth}{0pt}
}

\def\mypartname{}

% numbered points
\newcounter{point}
\newcounter{exnum}[part]

\newcommand{\pointno}{\stepcounter{point}\textbf{\thepoint{}.}}
\newcommand{\exno}{\stepcounter{exnum}\textbf{\theexnum{}.}}
\newenvironment{point}{\medskip\pointno~}{\filbreak}
\newenvironment{proof}{\begin{point}}{\end{point}}
\newenvironment{proofex}{\medskip\textbf{Ex.~\exno~}}{\filbreak}

\newcommand{\pp}[1]{\medskip\pointno~#1\filbreak}
\newcommand{\ex}[1]{\medskip\textbf{Ex.~\exno~}\ #1\filbreak}
\newcommand{\defn}[1]{\textsc{Def.~}#1}
\newcommand{\ax}[1]{\textbf{Axiom.~}\textit{#1}}
\newcommand{\thm}[1]{\textbf{Theorem.~}\textit{#1}}

% an outdented line heading several exercises
% eg, ``To construct a triangle equivalent to:''
\newcommand{\exheader}[1]{\bigskip#1}

\newcommand{\note}[2][foo]{\medskip\begin{small}\textsc{Note%
  \ifthenelse{\equal{#1}{foo}}{.~}{~#1.~}}%
  #2\end{small}\filbreak}

\newcommand{\cor}[2][foo]{\textsc{Cor.~%
  \ifthenelse{\equal{#1}{foo}}{}{#1.~}}%
  \begin{itshape}#2\end{itshape}}


% proof contents
\newlength{\dentwidth}\setlength{\dentwidth}{\textwidth}
\addtolength{\dentwidth}{-\parindent}

\newcommand{\pnote}[1]{%
  \noindent\parbox{\linewidth}{\spaceskip0.5em plus 1em minus 0.25em\centering\small\textit{#1}}%
}
\newcommand{\obs}[1]{\textit{#1}}
\newcommand{\fig}[1]{\fbox{figure here}\par\lett{#1}}
\newcommand{\lett}[1]{\boldmath\textbf{#1}\unboldmath}
\newcommand{\prove}[2][To prove that ]{\textit{#1#2}}
\newcommand{\proveq}[3][\indent To prove that ]{\eq[\textit{#1}]{#2}{#3}{}}

\newcommand{\qed}{\textsc{q.e.d.}}
\newcommand{\qef}{\textsc{q.e.f.}}

\newcommand{\step}[3][]{%
  \setlength{\TmpLen}{\parindent}%
  \noindent\parbox{\textwidth}{\setlength{\parindent}{\TmpLen}\rlap{#1}\hfill#2\hfill\llap{#3}}%
}

\newlength{\eqalign}\setlength{\eqalign}{.5\dentwidth}

\newcommand{\eq}[4][]{%
  \setlength{\TmpLen}{\parindent}%
  \noindent\parbox{\textwidth}{\setlength{\parindent}{\TmpLen}\rlap{#1}\hfill~#2~#3~\hfill\llap{#4}}%
}

% math symbols
\DeclareMathOperator{\arc}{arc}
\DeclareMathOperator{\base}{base}
\DeclareMathOperator{\rect}{rect.}

% ugly parallelogram
 \def\Par{%
   \mathchoice
      {\@parallelogram\scriptsize}%
      {\@parallelogram\scriptsize}%
      {\@parallelogram\tiny}%
      {\@parallelogram\tiny}}
\def\@parallelogram#1{%
  \textnormal{\setbox\z@\hbox{#1/}\dimen@\wd\z@
   \@tempdima 2.45\dimen@
   \vbox{\offinterlineskip
      \hbox{\kern.8\dimen@\vrule\@width\@tempdima\@height.4\p@}%
      \kern-.0\p@
      \hbox to\@tempdima{#1/\hfil\rlap/}%
      \kern-.5\p@
      \hbox{\kern.1\dimen@\vrule\@width\@tempdima\@height.4\p@}}}}


\newcommand{\smallrule}{\medskip\centerline{\rule{10em}{.2ex}}\medskip}

\newcommand{\figc}[2]{\centerline{\includegraphics{./images/#1.pdf}}%
  \ifthenelse{\equal{#2}{}}{}{\par\small\boldmath\textbf{#2}\unboldmath}}

\newcommand{\figcc}[2]{\centerline{\includegraphics{./images/#1.pdf}\qquad\includegraphics{./images/#2.pdf}}}

\newcommand{\figccc}[3]{\centerline{\includegraphics{./images/#1.pdf}\qquad%
  \includegraphics{./images/#2.pdf}\qquad%
  \includegraphics{./images/#3.pdf}}}

\newcommand{\figcccc}[4]{\centerline{\includegraphics{./images/#1.pdf}\quad%
  \includegraphics{./images/#2.pdf}\quad%
  \includegraphics{./images/#3.pdf}\quad%
  \includegraphics{./images/#4.pdf}}}

% lists
\newcounter{Lcount}
\newenvironment{myenum}
 {\setcounter{Lcount}{0}%
  \begin{list}{\arabic{Lcount}.}
   {\usecounter{Lcount}%
    \setlength{\itemsep}{0pt}%
    \setlength{\leftmargin}{2\parindent}}}
 {\end{list}}

\makeatother

%%%%%%%%%%%%%%%%%%%%%%%% START OF DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{document}

\pagestyle{empty}
\pagenumbering{Alph}

\phantomsection
\pdfbookmark[-1]{FRONT MATTER.}{FRONT MATTER}

%%%% PG BOILERPLATE %%%%
\phantomsection
\pdfbookmark[0]{PG BOILERPLATE.}{BOILERPLATE}

\begin{center}
\begin{minipage}{\textwidth}
\small
\begin{PGtext}
The Project Gutenberg EBook of Plane Geometry, by George Albert Wentworth

This eBook is for the use of anyone anywhere at no cost and with
almost no restrictions whatsoever.  You may copy it, give it away or
re-use it under the terms of the Project Gutenberg License included
with this eBook or online at www.gutenberg.org


Title: Plane Geometry

Author: George Albert Wentworth

Release Date: July 3, 2010 [EBook #33063]

Language: English

Character set encoding: ISO-8859-1

*** START OF THIS PROJECT GUTENBERG EBOOK PLANE GEOMETRY ***
\end{PGtext}
\end{minipage}
\end{center}

\clearpage


%%%% Credits and transcriber's note %%%%
\begin{center}
\begin{minipage}{\textwidth}
\begin{PGtext}
Produced by Jeremy Weatherford, Stan Goodman, Kevin Handy
and the Online Distributed Proofreading Team at
http://www.pgdp.net
\end{PGtext}
\end{minipage}
\end{center}
\vfill

\begin{minipage}{0.85\textwidth}
\small
\phantomsection
\pdfbookmark[0]{TRANSCRIBER'S NOTE.}{TRANSCRIBER'S NOTE}
\subsection*{\centering\normalfont\scshape%
\normalsize\MakeLowercase{\TransNote}}%

\raggedright
\TransNoteText
\end{minipage}


%%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%%

\frontmatter
\pagestyle{fancy}
\thispagestyle{empty}

\begin{titlepage}
\null\vfil
\begin{center}
{\LARGE PLANE GEOMETRY \par}
\vspace{3em}
{BY \par}
{\large G.A. WENTWORTH \par}
\textsc{Author of a Series of Text-Books in Mathematics}
\vspace{3em}

\emph{REVISED EDITION}

\vspace{3em}

\vfil\null

{\large GINN \& COMPANY}

BOSTON · NEW YORK · CHICAGO · LONDON

\end{center}
\end{titlepage}

\newpage
\thispagestyle{empty}

\begin{center}
\null\vfil
Entered, according to Act of Congress, in the year 1888, by

G.A. WENTWORTH

in the Office Of the Librarian of Congress, at Washington

\smallrule

\textsc{Copyright, 1899}

\textsc{By G.A. WENTWORTH}

\smallrule

{\small ALL RIGHTS RESERVED}

{\small 67 10}

\null\vfil

The Athenæum Press
\smallrule

\settowidth{\TmpLen}{PRIETORS · BOSTON · U.S.A.}%
\parbox{\TmpLen}{GINN \& COMPANY · PRO\-PRIETORS ·
BOSTON · U.S.A.}
\end{center}

\cleardoublepage
\SetRunningHeads

\section*{PREFACE.}
\pdfbookmark[0]{PREFACE.}{PREFACE}

Most persons do not possess, and do not easily acquire, the power of
abstraction requisite for apprehending geometrical conceptions, and for
keeping in mind the successive steps of a continuous argument. Hence,
with a very large proportion of beginners in Geometry, it depends mainly
upon the \emph{form} in which the subject is presented whether they pursue the
study with indifference, not to say aversion, or with increasing interest
and pleasure.

Great care, therefore, has been taken to make the pages attractive.
The figures have been carefully drawn and placed in the middle of
the page, so that they fall directly under the eye in immediate connection
with the text; and in no case is it necessary to turn the page in
reading a demonstration. Full, long-dashed, and short-dashed lines of
the figures indicate given, resulting, and auxiliary lines, respectively.
Bold-faced, italic, and roman type has been skilfully used to distinguish
the hypothesis, the conclusion to be proved, and the proof.

As a further concession to the beginner, the reason for each statement
in the early proofs is printed in small italics, immediately following the
statement. This prevents the necessity of interrupting the logical train
of thought by turning to a previous section, and compels the learner to
become familiar with a large number of geometrical truths by constantly
seeing and repeating them. This help is gradually discarded, and the
pupil is left to depend upon the knowledge already acquired, or to find
the reason for a step by turning to the given reference.

It must not be inferred, because this is not a geometry of interrogation
points, that the author has lost sight of the real object of the study.
The training to be obtained from carefully following the logical steps
of a complete proof has been provided for by the Propositions of the
\scanpage{005.png}%
Geometry, and the development of the power to grasp and prove new
truths has been provided for by original exercises. The chief value of
any Geometry consists in the happy combination of these two kinds
of training. The exercises have been arranged according to the test
of experience, and are so abundant that it is not expected that any
one class will work them all out. The methods of attacking and proving
original theorems are fully explained in the first Book, and illustrated
by sufficient examples; and the methods of attacking and solving
original problems are explained in the second Book, and illustrated by
examples worked out in full. None but the very simplest exercises are
inserted until the student has become familiar with geometrical methods,
and is furnished with elementary but much needed instruction in the
art of handling original propositions; and he is assisted by diagrams
and hints as long as these helps are necessary to develop his mental
powers sufficiently to enable him to carry on the work by himself.

The law of converse theorems, the distinction between positive and
negative quantities, and the principles of reciprocity and continuity
have been briefly explained; but the application of these principles is
left mainly to the discretion of teachers.

The author desires to express his appreciation of the valuable suggestions
and assistance which he has received from distinguished educators
in all parts of the country. He also desires to acknowledge his obligation
to Mr.~Charles Hamilton, the Superintendent of the composition
room of the Athen\ae{}um Press, and to Mr.~I.~F. White, the compositor,
for the excellent typography of the book.

Criticisms and corrections will be thankfully received.

\hfill G.~A. WENTWORTH.

\textsc{Exeter}, N.H., June, 1899.
\scanpage{006.png}%


\clearpage
\section*{NOTE TO TEACHERS.}
\pdfbookmark[0]{NOTE TO TEACHERS.}{NOTE TO TEACHERS}

It is intended to have the first \PageName\ pages of this book simply read
in the class, with such running comment and discussion as may be useful
to help the beginner catch the spirit of the subject-matter, and not
leave him to the mere letter of dry definitions. In like manner, the
definitions at the beginning of each Book should be read and discussed
in the recitation room. There is a decided advantage in having the
definitions for each Book in a single group so that they can be included
in one survey and discussion.

For a similar reason the theorems of limits are considered together.
The subject of limits is exceedingly interesting in itself, and it was
thought best to include in the theory of limits in the second Book every
principle required for Plane and Solid Geometry.

When the pupil is reading each Book for the first time, it will be
well to let him write his proofs on the blackboard in his own language,
care being taken that his language be the simplest possible,
that the arrangement of work be vertical, and that the figures be
accurately constructed.

This method will furnish a valuable exercise as a language lesson,
will cultivate the habit of neat and orderly arrangement of work, and
will allow a brief interval for deliberating on each step.

After a Book has been read in this way, the pupil should review
the Book, and should be required to draw the figures free-hand. He
should state and prove the propositions orally, using a pointer to indicate
on the figure every line and angle named. He should be encouraged
in reviewing each Book, to do the original exercises; to state
the converse propositions, and determine whether they are true or
false; and also to give well-considered answers to questions which
may be asked him on many propositions.
\scanpage{007.png}%

The Teacher is strongly advised to illustrate, geometrically and
arithmetically, the principles of limits. Thus, a rectangle with a constant
base $b$, and a variable altitude $x$, will afford an obvious illustration
of the truth that the product of a constant and a variable is
also a variable; and that the limit of the product of a constant and a
variable is the product of the constant by the limit of the variable.
If $x$ increases and approaches the altitude $a$ as a limit, the area of the
rectangle increases and approaches the area of the rectangle $ab$ as a
limit; if, however, $x$ decreases and approaches zero as a limit, the area
of the rectangle decreases and approaches zero as a limit.

An arithmetical illustration of this truth may be given by multiplying
the approximate values of any repetend by a constant. If, for example,
we take the repetend $0.3333$ etc., the approximate values of the repetend
will be $\frac{3}{10}$, $\frac{33}{100}$, $\frac{333}{1000}$,
$\frac{3333}{10000}$, etc., and these values multiplied by $60$
give the series $18$, $19.8$, $19.98$, $19.998$, etc., which evidently approaches
$20$ as a limit; but the product of $60$ into $\frac{1}{3}$ (the limit of the repetend
$0.333$ etc.) is also $20$.

Again, if we multiply $60$ into the different values of the decreasing
series  $\frac{1}{30}$, $\frac{1}{300}$, $\frac{1}{3000}$, $\frac{1}{30000}$,
etc., which approaches zero as a limit, we
shall get the decreasing series $2$, $\frac{1}{5}$, $\frac{1}{50}$,
$\frac{1}{500}$, etc.; and this series evidently
approaches zero as a limit.

The Teacher is likewise advised to give frequent written examinations.
These should not be too difficult, and sufficient time should be
allowed for accurately constructing the figures, for choosing the best
language, and for determining the best arrangement.

The time necessary for the reading of examination books will be
diminished by more than one half, if the use of symbols is allowed.

\textsc{Exeter}, N.H., 1899.
\scanpage{008.png}%


\clearpage
\phantomsection\pdfbookmark[0]{TABLE OF CONTENTS.}{CONTENTS}
\tableofcontents

\mainmatter
\scanpage{009.png}%
\scanpage{010.png}%


\phantomsection%
\part{GEOMETRY.}
\markboth{GEOMETRY.}{} % fake a chapter heading, no chapters yet

\section{INTRODUCTION.}

\begin{point}%
If a block of wood or stone is cut in the shape represented
in Fig.~1, it will have \emph{six flat
faces}.

Each face of the block is called a
surface\label{surface}; and if the faces are made
smooth by polishing, so that, when a
straight edge is applied to any one
of them, the straight edge in every
part will touch the surface, the faces
are called \textbf{plane surfaces}, or \indexbf{planes}.

% [Illustration: Fig. 1.]
\centering{\includegraphics{./images/woodcutsmall.pdf}}

\centering{\small\textsc{Fig. 1.}}
\end{point}

\pp{The intersection\label{intersection} of any two of these surfaces is called
a \indexbf{line}.}

\pp{The intersection of any three of these lines is called
a \indexbf{point}.}

\begin{point}%
The block extends in three principal directions:

\begin{list}{}{\setlength{\itemsep}{0pt}}
\item From left to right, $A$ to $B$.

\item From front to back, $A$ to $C$.

\item From top to bottom, $A$ to $D$.
\end{list}

These are called the \textbf{dimensions}\label{dimensions} of the block, and are named
in the order given, \textbf{length}, \textbf{breadth} (or \textit{width}), and \textbf{thickness}
(\textit{height} or \textit{depth}).
\end{point}
\scanpage{011.png}%

\begin{point}%
A \textbf{solid}, in common language, is a limited portion of
space \textit{filled with matter}; but in Geometry we have nothing
to do with \textit{the matter} of which a body is composed; we study
simply its \textit{shape} and \textit{size}; that is, we regard a solid as a
limited portion of space which may be occupied by a physical
body, or marked out in some other way. Hence,

\textit{A geometrical solid\label{geometrical1} is a limited portion of space.}
\end{point}

\begin{point}%
The surface\label{surface2} of a solid is simply the boundary of the
solid, that which separates it from surrounding space. The
surface is no part of a solid and has no thickness. Hence,

\textit{A surface has only two dimensions, length and breadth.}
\end{point}

\begin{point}%
A line\label{line2} is simply a boundary of a surface, or the intersection
of two surfaces. Since the surfaces have no thickness,
a line has no thickness. Moreover, a line is no part of a
surface and has no width. Hence,

\textit{A line has only one dimension, length.}
\end{point}

\begin{point}%
A point\label{point2} is simply the extremity of a line, or the intersection
of two lines. A point, therefore, has no thickness,
width, or length; therefore, no magnitude. Hence,

\textit{A point has no dimension, but denotes position simply.}
\end{point}

\begin{point}%
It must be distinctly understood at the outset that the
points, lines, surfaces, and solids of Geometry are \textit{purely ideal},
though they are represented to the eye in a material way.
Lines, for example, drawn on paper or on the blackboard, will
have some width and some thickness, and will so far fail of
being \textit{true lines}; yet, when they are used to help the mind
in reasoning, it is assumed that they represent true lines,
without breadth and without thickness.
\end{point}
\scanpage{012.png}%

\figc{012aaZ10}{}
\begin{point}%
A point is \textit{represented} to the eye by a fine dot, and
named by a letter, as $A$ (Fig.~2). A line is named by two
letters, placed one at each end, as $BF$.
A surface\label{surface3} is represented and named by
the lines which bound it, as $BCDF$. A
solid\label{geometrical2} is represented by the faces which
bound it.
\end{point}

\pp{A point in space may be considered by itself, without
reference to a line\label{line3}.}

\pp{If a point moves in space, its path is a line. This line
may be considered apart from the idea of a surface.}

\pp{If a line moves in space, it generates, in general, a surface.
A surface can then be considered apart from the idea of a solid.}

\begin{point}%
If a surface moves in space, it generates, in general, a
solid.

\filbreak
\figc{012bbZ14}{}
Thus, let the upright surface $ABCD$
(Fig.~3) move to the right to the position
$EFGH$, the points $A$, $B$, $C$, and $D$ generating
the lines $AE$, $BF$, $CG$, and $DH$,
respectively. The lines $AB$, $BC$, $CD$,
and $DA$ will generate the surfaces $AF$.
$BG$, $CH$, and $DE$, respectively. The surface $ABCD$ will generate the
solid $AG$.
\end{point}

\pp{\indexbf{Geometry} is the science which treats of \textit{position, form},
and \textit{magnitude}.}

\pp{A \indexbf{geometrical figure} is a combination of points, lines,
surfaces, or solids.}

\begin{point}%
{\indexbf{Plane Geometry} treats of figures all points of which are
in the same plane.}

\indexbf{Solid Geometry} treats of figures all points of which are not
in the same plane.
\end{point}
\scanpage{013.png}%


\section{GENERAL TERMS.}

\begin{point}%
A \indexbf{proof} is a course of reasoning by which the truth or
falsity of any statement is logically established.
\end{point}

\begin{point}%
An \textbf{axiom}\label{axiom} is a statement admitted to be true without
proof.
\end{point}

\begin{point}%
A \indexbf{theorem} is a statement to be proved.
\end{point}

\begin{point}%
A \textbf{construction}\label{construction} is the representation of a required figure
by means of points and lines.
\end{point}

\begin{point}%
A \indexbf{postulate} is a construction admitted to be possible.
\end{point}

\begin{point}%
A \textbf{problem} is a construction to be made so that it shall
satisfy certain given conditions.
\end{point}

\begin{point}%
A \indexbf{proposition} is an axiom, a theorem, a postulate, or a
problem.
\end{point}

\begin{point}%
A \textbf{corollary} is a truth that is easily deduced from known
truths.
\end{point}

\begin{point}%
A \indexbf{scholium} is a remark upon some particular feature of
a proposition.
\end{point}

\begin{point}%
The \textbf{solution of a problem} consists of four parts:

1. The \emph{analysis}\label{analysis}, or course of thought by which the construction
of the required figure is discovered.

2. The \emph{construction} of the figure with the aid of ruler and
compasses.

3. The \emph{proof} that the figure satisfies all the conditions.

4. The \emph{discussion} of the limitations, if any, within which
the solution is possible.
\end{point}
\scanpage{014.png}%

\begin{point}%
A theorem consists of two parts: the \indexbf{hypothesis}, or
that which is assumed; and the \indexbf{conclusion}, or that which is
asserted to follow from the hypothesis.
\end{point}

\settowidth{\TmpLen}{\qquad\qquad Its contradictory:\quad}%
\begin{point}%
The \textbf{contradictory} of a theorem\label{contradictory} is a theorem which must
be true if the given theorem is false, and must be false if the
given theorem is true. Thus, \\[0.5\baselineskip]
\parbox{\TmpLen}{\qquad\qquad A theorem:} If $A$ is $B$, then $C$ is $D$.\\
\parbox{\TmpLen}{\qquad\qquad Its contradictory:} If $A$ is $B$, then $C$ is not $D$.
\end{point}

\begin{point}%
The \indexbf{opposite} of a theorem is obtained by making \emph{both
the hypothesis and the conclusion negative}. Thus, \\[0.5\baselineskip]
\parbox{\TmpLen}{\qquad\qquad A theorem:} If $A$ is $B$, then $C$ is $D$.\\
\parbox{\TmpLen}{\qquad\qquad Its opposite:} If $A$ is not $B$, then $C$ is not $D$.
\end{point}

\begin{point}%
The \textbf{converse}\label{converse1} of a theorem is obtained by \emph{interchanging
the hypothesis and conclusion}. Thus, \\[0.5\baselineskip]
\parbox{\TmpLen}{\qquad\qquad A theorem:} If $A$ is $B$, then $C$ is $D$.\\
\parbox{\TmpLen}{\qquad\qquad Its converse:} If $C$ is $D$, then $A$ is $B$.
\end{point}

\begin{point}%
The converse of a truth is not \emph{necessarily} true.

Thus, Every horse is a quadruped is true, but the converse, Every
quad\-ru\-ped is a horse, is not true.
\end{point}

\begin{point}%
\textit{If a direct proposition and its opposite are true, the
converse proposition is true; and if a direct proposition and
its converse are true, the opposite proposition is true}.

Thus, if it were true that

1. If an animal is a horse, the animal is a quadruped;

2. If an animal is not a horse, the animal is not a quadruped;\\
it would follow that

3. If an animal is a quadruped, the animal is a horse.

Moreover, if 1 and 3 were true, then 2 would be true.
\end{point}
\scanpage{015.png}%


\section[GENERAL AXIOMS.]{\pointno\hsp GENERAL AXIOMS.\hsp\phantom{XX.}}
\label{generalaxioms}

1. Magnitudes which are equal to the same magnitude, or
equal magnitudes, are equal to each other.

2. If equals are added to equals, the sums are equal.

3. If equals are taken from equals, the remainders are equal.

4. If equals are added to unequals, the sums are unequal
in the same order; if unequals are added to unequals in the
same order, the sums are unequal in that order.

5. If equals are taken from unequals, the remainders are
unequal in the same order; if unequals are taken from equals,
the remainders are unequal in the reverse order.

6. The doubles of the same magnitude, or of equal magnitudes
are equal; and the doubles of unequals are unequal.

7. The halves of the same magnitude, or of equal magnitudes
are equal; and the halves of unequals are unequal.

8. The whole is greater than any of its parts.

9. The whole is equal to the sum of all its parts.


\section[SYMBOLS AND ABBREVIATIONS.]
  {\pointno\hsp SYMBOLS AND ABBREVIATIONS. \hsp\phantom{XX.}}

\label{abbr}\label{symbols}%
\noindent\small\enlargethispage{8pt}%
\begin{tabular*}{\dentwidth}{rl@{\extracolsep{\fill}}l@{\extracolsep{0pt}}@{\dots}l}

$>$ & is (or are) greater than. & Def. & definition. \\
$<$ & is (or are) less than. & Ax. & axiom. \\
$\Bumpeq$ & is (or are) equivalent to. & Hyp. & hypothesis. \\
$\therefore$ & therefore. & Cor. & corollary. \\
$\perp$ & perpendicular. & Scho. & scholium. \\
$\perp_s$ & perpendiculars. & Ex. & exercise. \\
$\parallel$ & parallel.\qquad $\parallel_s$ parallels. & Adj. & adjacent. \\
$\angle$ & angle.\qquad $\angle_s$ angles. & Iden. & identical. \\
$\triangle$ & triangle.\qquad $\triangle_s$ triangles. & Const. & construction. \\
$\Par$ & parallelogram. & Sup. & supplementary. \\
$\Par_s$ & parallelograms. & Ext. & exterior. \\
$\odot$ & circle.\qquad $\odot_s$ circles. & Int. & interior. \\
rt. & right.\qquad  st.\ straight. & Alt. & alternate. \\
\end{tabular*}

\qed\ stands for quod erat demonstrandum, \emph{which was to be proved}.

\qef\ stands for quod erat faciendum, \emph{which was to be done.}

The signs $+$, $-$, $×$, $\div$, $=$, have the same meaning as in Algebra.
\scanpage{016.png}%

\normalsize

\phantomsection%
\part{PLANE GEOMETRY.}

\chapter{BOOK I\@. RECTILINEAR FIGURES.}

\section{DEFINITIONS.}

\vspace{2ex}
\figc{016aaZ37}{}
\begin{point}%
A \indexbf{straight line} is a line such that any part of it, however
placed on any other part, will lie wholly in that part if
its extremities lie in that part, as~$AB$.
\end{point}

\begin{point}%
A \indexbf{curved line} is a line no part of
which is straight, as~$CD$.
\end{point}

\begin{point}%
A \textbf{broken line} is made up of different
straight lines, as~$EF$.
\end{point}

\note{A straight line is often called simply a \emph{line}.}

\begin{point}%
A \textbf{plane surface}, or a \indexbf{plane}, is a surface in which, if any
two points are taken, the straight line joining these points
lies wholly in the surface.
\end{point}

\begin{point}%
A \textbf{curved surface}\label{curvedsurf} is a surface no part of which is plane.
\end{point}

\begin{point}%
A \indexbf{plane figure} is a figure all points of which are in the
same plane.
\end{point}

\begin{point}%
Plane figures which are bounded by straight lines are
called \indexbf{rectilinear} figures; by curved lines, \indexbf{curvilinear} figures.
\end{point}

\begin{point}%
Figures that have the \emph{same shape} are called \indexbf{similar}.
Figures that have the \textit{same size but not the same shape} are
called \textbf{equivalent}\label{equivalent1}. Figures that have the \textit{same shape and the
same size} are called \textbf{equal}\label{equal} or \textbf{congruent}\label{congruent}.
\end{point}
\scanpage{017.png}%


\section{THE STRAIGHT LINE.}

\pp{\textbf{Postulate.} A straight line can be drawn from one point
to another.}

\pp{\textbf{Postulate.} A straight line can be produced indefinitely.}

% footnote keeps this from being a normal \ax
\pp{\textbf{Axiom.}\footnote{The general axioms on page \pageref{generalaxioms} apply to all magnitudes. Special
geometrical axioms will be given when required.}
\textit{Only one straight line can be drawn from one
point to another.} Hence, two points \textit{determine} a straight line.}

\pp{\cor[1]{Two straight lines which have two points in
common coincide and form but one line.}}

\pp{\cor[2]{Two straight lines can intersect in only one
point.}}

For if they had two points common, they would coincide
and not intersect.

Hence, two intersecting lines \textit{determine} a point.

\pp{\ax{A straight line is the shortest line that can be
drawn from one point to another.}\label{axiomstraight}}

\pp{\defn{The \textbf{distance}\label{distance1} between two points is the length of
the straight line that joins them.}}

\pp{A straight line determined by two points may be considered
as prolonged indefinitely.}

\pp{If only the part of the line between two fixed points is
considered, this part is called a \textbf{segment} of the line\label{lineseg}.}

\pp{For brevity, we say ``the line $AB$,'' to designate a segment
of a line limited by the points $A$ and $B$.}

\pp{If a line is considered as extending from a fixed point,
this point is called the \indexbf{origin} of the line.}
\scanpage{018.png}%

\figc{018aaZ55}{}
\begin{point}%
If any point, $C$, is taken in a given straight line, $AB$,
the two parts $CA$ and $CB$ are
said to have \emph{opposite directions}
from the point $C$ (Fig.~5).

Every straight line, as $AB$, may be considered as extending
in either of two opposite directions, namely, from $A$ towards
$B$, which is expressed by $AB$, and read segment $AB$; and
from $B$ towards $A$, which is expressed by $BA$, and read segment
$BA$.
\end{point}

\begin{point}%
If the magnitude of a given line is changed, it becomes
longer or shorter.

Thus (Fig.~5), by prolonging $AC$ to $B$ we add $CB$ to $AC$, and $AB = AC+CB$.
By diminishing $AB$ to $C$, we subtract $CB$ from $AB$, and
$AC = AB-CB$.

If a given line increases so that it is prolonged by its own
magnitude several times in succession, the line is \emph{multiplied},
and the resulting line is called a \emph{multiple} of the given line.

\figc{018bbZ56}{}
Thus (Fig.~6), if $AB = BC = CD = DE$,
then $AC = 2AB$, $AD = 3AB$,
and $AE = 4AB$. Hence,

\textit{Lines of given length may be added and subtracted; they
may also be multiplied by a number.}
\end{point}


\section{THE PLANE ANGLE.}

\label{angle}
\figc{018ccZ57}{}
\begin{point}%
The \emph{opening} between two straight lines drawn from the
same point is called a \textbf{plane angle}. The two
lines, $ED$ and $EF$, are called the \textbf{sides}\label{anglesides}, and $E$,
the point of meeting, is called the \indexbf{vertex} of
the angle.

The size of an angle depends upon the \emph{extent
of opening} of its sides, and not upon the length of its sides.
\end{point}
\scanpage{019.png}%

\begin{point}%
If there is but one angle at a given vertex, the angle is
designated by a capital letter placed at the vertex, and is read
by simply naming the letter.

\figcc{019aaZ58}{019bbZ58}

If two or more angles have the same vertex, each
angle is designated by three letters, and is read by
naming the three letters, the one at the vertex
between the others. Thus, $DAC$ (Fig.~8) is the
angle formed by the sides $AD$ and $AC$.

An angle is often designated by placing a
small \emph{italic} letter between the sides and near
the vertex, as in Fig.~9.
\end{point}

\begin{point}%
\textbf{Postulate of Superposition.}\label{superposition} Any figure may be moved
from one place to another without altering its size or
shape.
\end{point}

\begin{point}%
The \textbf{test of equality} of two geometrical magnitudes is
that they may be made to coincide throughout their whole
extent. Thus,

Two straight lines are equal, if they can be placed one upon
the other so that the points at their extremities coincide.

Two angles are equal, if they can be placed one upon the
other so that their vertices coincide and their sides coincide,
each with each.
\end{point}

\begin{point}%
A line or plane that divides a geometric magnitude into
\emph{two equal parts} is called the \textbf{bisector}\label{bisector} of the magnitude.

If the angles $BAD$ and $CAD$ (Fig.~8) are equal, $AD$ \emph{bisects}
the angle $BAC$.
\end{point}

\begin{point}%
Two angles are called \textbf{adjacent angles}\label{adjacent1}
when they have the same vertex and a common
side between them; as the angles $BOD$
and $AOD$ (Fig.~10).
\end{point}
\scanpage{020.png}%

\figcc{019ccZ62}{020aaZ63}
\begin{point}%
When one straight line meets another
straight line and makes the \emph{adjacent angles
equal}, each of these angles is called a \textbf{right
angle}\label{right}; as angles $DCA$ and $DCB$ (Fig.~11).
\end{point}

\begin{point}%
A \indexbf{perpendicular} to a straight line is a straight line that
makes a right angle with it.

Thus, if the angle $DCA$ (Fig.~11) is a right angle, $DC$ is perpendicular
to $AB$, and $AB$ is perpendicular to $DC$.
\end{point}

\begin{point}%
The point (as $C$, Fig.~11) where a perpendicular meets
another line is called the \indexbf{foot} of the perpendicular.
\end{point}

\begin{point}%
When the sides of an angle extend in opposite directions,
so as to be in the same straight line, the angle is called a
\textbf{straight angle}.\label{straight}

\figc{020bbZ66}{}

Thus, the angle formed at $C$ (Fig.~12) with its sides $CA$ and $CB$ extending
in opposite directions from $C$ is a straight angle.
\end{point}

\pp{\cor{A right angle is half a straight angle.}}

\figcc{020ccZ68}{020ddZ69}
\begin{point}%
An angle less than a right angle is called
an \textbf{acute angle}\label{acute}; as, angle $A$ (Fig.~13).
\end{point}

\begin{point}%
An angle greater than a right angle and
less than a straight angle is called an \textbf{obtuse
angle}\label{obtuse}; as, angle $AOD$ (Fig.~14).
\end{point}

\begin{point}%
An angle greater than a straight angle and less than
two straight angles is called a \textbf{reflex angle}\label{reflex}; as, angle $DOA$,
indicated by the dotted line (Fig.~14).
\end{point}
\scanpage{021.png}%

\begin{point}%
Angles that are neither right nor straight angles are
called \textbf{oblique angles}\label{oblique}; and intersecting lines that are not perpendicular
to each other are called \indexbf{oblique lines}.
\end{point}

\subsection{EXTENSION OF THE MEANING OF ANGLES.}

\figc{021aaZ72}{}
\begin{point}%
Suppose the straight line $OC$ (Fig.~15) to move in
the plane of the paper from coincidence with $OA$, about the
point $O$ as a pivot, to the position $OC$; then the line $OC$
describes or generates \emph{the angle $AOC$}, and
the magnitude of the angle $AOC$ depends
upon the \emph{amount of rotation} of the line
from the position $OA$ to the position $OC$.

If the rotating line moves from the position
$OA$ to the position $OB$, \emph{perpendicular}
to $OA$, it generates the right angle $AOB$;
if it moves to the position $OD$, it generates
the obtuse angle $AOD$; if it moves to
the position $OA'$, it generates the straight angle $AOA'$; if it
moves to the position $OB'$ it generates the reflex angle $AOB'$,
indicated by the dotted line; and if it moves to the position
$OA$ again, it generates two straight angles. Hence,
\end{point}

\begin{point}%
\textit{The angular magnitude about a point in a plane is equal
to two straight angles, or four right angles; and the angular
magnitude about a point on one side of a straight line drawn
through the point is equal to a straight angle, or two right
angles.}
\end{point}

\begin{point}%
The whole angular magnitude about a point in a plane
is called a \indexbf{perigon}; and two angles whose sum is a perigon are
called \indexbf{conjugate angles}.
\end{point}

\note{This \emph{extension of the meaning of angles} is necessary in the
applications of Geometry, as in Trigonometry, Mechanics, etc.}
\scanpage{022.png}%

\figccc{022aaZ75}{022bbZ76}{022ccZ77}
\begin{point}%
When two angles have the same vertex, and
the sides of the one are prolongations of the sides of
the other, they are called \indexbf{vertical angles}; as, angles
$a$ and $b$, $c$ and $d$ (Fig.~16).
\end{point}

\begin{point}%
Two angles are called \textbf{complementary}\label{complementary} when
their sum is equal to a right angle; and each is
called the \emph{complement}\label{complement} of the other; as, angles $DOB$
and $DOC$ (Fig.~17).
\end{point}

\begin{point}%
Two angles are called \textbf{supplementary}\label{supplementary} when
their sum is equal to a straight angle; and each
is called the \indexemph{supplement} of the other; as, angles
$DOB$ and $DOA$ (Fig.~18).
\end{point}


\subsection{UNIT OF ANGLES.}

\begin{point}%
By adopting a suitable unit for measuring angles we
are able to express the magnitudes of angles by numbers.

If we suppose $OC$ (Fig.~15) to turn about $O$ from coincidence
with $OA$ until it makes \emph{one three hundred sixtieth} of a
revolution, it generates an angle at $O$, which is taken as the
unit for measuring angles. This unit is called a \emph{degree}.

The degree is subdivided into sixty equal parts, called
\emph{minutes}, and the minute into sixty equal parts, called \emph{seconds}.

Degrees, minutes, and seconds are denoted by symbols.
Thus, $5$~degrees $13$~minutes $12$~seconds is written $5°\ 13'\ 12''$.

A right angle is generated when $OC$ has made \emph{one fourth} of
a revolution and contains~$90°$; a straight angle, when $OC$ has
made \emph{half} of a revolution and contains~$180°$; and a perigon,
when $OC$ has made a complete revolution and contains~$360°$.
\end{point}

\note{The natural angular unit is one complete revolution. But
this unit would require us to express the values of most angles by fractions.
The advantage of using the degree as the unit consists in its convenient
size, and in the fact that $360$~is divisible by so many different
integral numbers.}
\scanpage{023.png}%

\figc{023aaZ79}{}

\begin{point}%
By the method of superposition we are able to compare
magnitudes of the same kind. Suppose we have two angles,
$ABC$ and $DEF$ (Fig.~19). Let
the side $ED$ be placed on the
side $BA$, so that the vertex $E$
shall fall on $B$; then, if the
side $EF$ falls on $BC$, the angle
$DEF$ equals the angle $ABC$;
if the side $EF$ falls between
$BC$ and $BA$ in the position shown by the dotted line $BG$, the
angle $DEF$ is less than the angle $ABC$; but if the side $EF$
falls in the position shown by the dotted line $BH$, the angle
$DEF$ is greater than the angle $ABC$.
\end{point}

\figc{023bcZ80}{}

\begin{point}%
If we have the angles $ABC$ and $DEF$ (Fig.~20), and
place the vertex $E$ on $B$ and the side $ED$ on $BC$, so that the
angle $DEF$ takes the position $CBH$, the angles $DEF$ and $ABC$
will together be equal to the angle $ABH$.

If the vertex $E$ is placed on $B$, and the side $ED$ on $BA$, so
that the angle $DEF$ takes the position $ABF$, the angle $FBC$
will be the difference between the angles $ABC$ and $DEF$.

If an angle is increased by its own magnitude two or more
times in succession, the angle is \emph{multiplied} by a number.

Thus, if the angles $ABM$, $MBC$, $CBP$, $PBD$ (Fig.~21) are all equal,
the angle $ABD$ is $4$~times the angle $ABM$. Therefore,

\textit{Angles may be added and subtracted; they may also be multiplied
by a number.}
\label{page:PageName}% [** TN: For verbal ref. in Note to Teachers]
\end{point}
\scanpage{024.png}%


\pagebreak
\section{PERPENDICULAR AND OBLIQUE LINES.}

\proposition{Theorem.}

\begin{proof}%
\obs{All straight angles are equal.}

\figc{024aaZ81}{Let the angles $ACB$ and $DEF$ be any two straight angles.}

\prove{$\angle ACB = \angle DEF$}.

\textbf{Proof.} Place the $\angle ACB$ on the $\angle DEF$, so that
the vertex $C$ shall fall on the vertex $E$, and the side $CB$ on the
side $EF$.

\step{Then $CA$ will fall on $ED$,}{§~47}

\pnote{(because $ACB$ and $DEF$ are straight lines).}

\step{$\therefore \angle ACB = \angle DEF$.}{§~60}

\hfill\qed

\end{proof}

\pp{\cor[1]{All right angles are equal.}\hfill~Ax.~7}

\begin{point}%
\cor[2]{At a given point in a given line
  there can be but one perpendicular to the line.}

\figc{024bbZ83}{}
For, if there could be two $\perp_s$, we should have rt.~$\angle_s$ of
different magnitudes; but this is impossible, §~82.
\end{point}

\pp{\cor[3]{The complements of the same angle
  or of equal angles are equal.}\hfill~Ax.~3}

\pp{\cor[4]{The supplements of the same angle
  or of equal angles are equal.}\hfill~Ax.~3}

\note{The beginner must not forget that in Plane Geometry all
the points of a figure are in the same plane.  Without this
restriction in Cor.~2, an indefinite number of perpendiculars can be
erected at a given point in a given line.}
\scanpage{025.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two adjacent angles have their exterior sides
in a straight line, these angles are supplementary.}

\figc{025aaZ86}{Let the exterior sides $OA$ and $OB$ of the adjacent angles $AOD$ and
$BOD$ be in the straight line $AB$.}

\prove{$\angle_s AOD$ and $BOD$ are supplementary.}

\step[\indent\textbf{Proof.}]{$AOB$ is a straight line.}{Hyp.}

\step{$\therefore \angle AOB$ is a st.~$\angle$.}{§~66}

\step[\indent But]{$\angle AOD + \angle BOD =$ the st.~$\angle AOB$.}{Ax.~9}

\step{$\therefore$ the $\angle_s AOD$ and $BOD$ are supplementary.}{§~77}

\hfill\qed

\end{proof}

\begin{point}%
\defn{Adjacent angles that are supplements of each
other are called \emph{supplementary-adjacent angles}\label{suppladj}.}

Since the angular magnitude about a point is neither increased
nor diminished by the number of lines which radiate
from the point, it follows that,
\end{point}

\pp{\cor[1]{The sum of all the angles about a point in a
plane is equal to a perigon, or two straight angles.}}

\pp{\cor[2]{The sum of all the angles about a point in a
plane, on the same side of a straight line passing through the
point, is equal to a straight angle, or two right angles.}}
\scanpage{026.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{\textsc{Conversely:} If two adjacent angles are supplementary,
their exterior sides are in the same straight
line.}

\figc{026aaZ90}{Let the adjacent angles $OCA$ and $OCB$ be supplementary.}

\prove{$AC$ and $CB$ are in the same straight line.}

\textbf{Proof.} Suppose $CF$ to be in the same line with $AC$.

\step[\indent Then]{$\angle_s OCA$ and $OCF$ are supplementary,}{§~86}

\pnote{(if two adjacent angles have their exterior sides in a straight line, these
angles are supplementary).}

\step[\indent But]{$\angle_s OCA$ and $OCB$ are supplementary.}{Hyp.}

\step{$\therefore \angle_s OCF$ and $OCB$ have the same supplement.}{}

\eq{$\therefore \angle OCF$}{$= \angle OCB$.}{§~85}

\step{$\therefore CB$ and $CF$ coincide.}{§~60}

\step{$\therefore AC$ and $CB$ are in the same straight line.}{\qed}

Since Propositions II.\ and III.\ are true, their opposites are
true. Hence,~\hfill§~33

\end{proof}

\pp{\cor[1]{If the exterior sides of two adjacent angles are
not in a straight line, these angles are not supplementary.}}

\pp{\cor[2]{If two adjacent angles are not supplementary,
their exterior sides are not in the same straight line.}}
\scanpage{027.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If one straight line intersects another straight
line, the vertical angles are equal.}

\figc{027aaZ93}{Let the lines $OP$ and $AB$ intersect at $C$.}

\proveq{$\angle OCB$}{$= \angle ACP$.}

\step[\indent\textbf{Proof.}]{$\angle OCA$ and $\angle OCB$ are supplementary.}{§~86}

\step{$\angle OCA$ and $\angle ACP$ are supplementary,}{§~86}

\pnote{(if two adjacent angles have their exterior sides in a straight line, these
angles are supplementary).}

\step{$\therefore \angle_s OCB$ and $ACP$ have the same supplement.}{}

\eq{$\therefore \angle OCB$}{$= \angle ACP$.}{§~85}

\eq[\indent In like manner,]{$\angle ACO$}{$= \angle PCB$.}{\qed}

\end{proof}

\pp{\cor{If one of the four angles formed by the intersection
of two straight lines is a right angle, the other three
angles are right angles.}}

\ex{Find the complement and the supplement of an angle of~$49°$.}

\ex{Find the number of degrees in an angle if it is double its complement;
if it is one fourth of its complement.}

\ex{Find the number of degrees in an angle if it is double its supplement;
if it is one third of its supplement.}
\scanpage{028.png}%

\proposition{Theorem}

\begin{proof}%
\obs{Two straight lines drawn from a point in a perpendicular
to a given line, cutting off on the given line
equal segments from the foot of the perpendicular, are
equal and make equal angles with the perpendicular.}

\figc{028aaZ95}{Let $CF$ be a perpendicular to the line $AB$, and $CE$ and $CK$ two
straight lines cutting off on $AB$ equal segments $FE$ and $FK$ from $F$.}

\prove{$CE = CK$; and $\angle FCE = \angle FCK$.}

\textbf{Proof.} Fold over $CFA$, on $CF$ as an axis, until it falls on the
plane at the right of $CF$.

\step{$FA$ will fall along $FB$,}{}

\pnote{(since $\angle CFA = \angle CFB$, each being a rt.~$\angle$, by hyp.).}

\step{Point $E$ will fall on point $K$,}{}

\pnote{(since $FE = FK$, by hyp.).}

\eq{$\therefore CE$}{$= CK$,}{§~60}

\pnote{(their extremities being the same points);}

\eq{and $\angle FCE$}{$= \angle FCK$,}{§~60}

\pnote{(since their vertices coincide, and their sides coincide, each with each).}

\hfill\qed

\end{proof}

\ex{Find the number of degrees in the angle included by the hands
of a clock at $1$~o'clock. $3$~o'clock. $4$~o'clock. $6$~o'clock.}
\scanpage{029.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Only one perpendicular can be drawn to a given
line from a given external point.}

\figc{029aaZ96}{Let $AB$ be the given line, $P$ the given external point, $PC$ a perpendicular
to $AB$ from $P$, and $PD$ any other line from $P$ to $AB$.}

\proveq{$PD$ is not}{$\perp$ to $AB$.}

\textbf{Proof.} Produce $PC$ to $P'$, making $CP'$ equal to $PC$.

\step{Draw $DP'$.}{}

\step{By construction, $PCP'$ is a straight line.}{}

\step{$\therefore PDP'$ is not a straight line,}{§~46}

\pnote{(only one straight line can be drawn from one point to another).}

\step{Hence, $\angle PDP'$ is not a straight angle.}{}

\step{Since $PC$ is $\perp$ to $DC$, and $PC = CP'$,}{}

\step{$AC$ is $\perp$ to $PP'$ at its middle point.}{}

\step{$\therefore \angle PDC = \angle P'DC$,}{§~95}

\pnote{(two straight lines from a point in a $\perp$ to a line, cutting off on the line equal
segments from the foot of the $\perp$, make equal $\angle_s$ with the $\perp$)}

\step{Since $\angle PDP'$ is not a straight angle,}{}

\step{$\angle PDC$, the half of $\angle PDP'$, is not a right angle.}{}

\step{$\therefore PD$ is not $\perp$ to $AB$.}{\qed}

\end{proof}
\scanpage{030.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The perpendicular is the shortest line that can be
drawn to a straight line from an external point.}

\figc{030aaZ97}{Let $AB$ be the given straight line, $P$ the given point, $PC$ the perpendicular,
and $PD$ any other line drawn from $P$ to $AB$.}

\proveq{$PC$}{$< PD$.}

\textbf{Proof.} Produce $PC$ to $P'$, making $CP' = PC$.

\step{Draw $DP'$.}{}

\eq[\indent Then]{$PD$}{$= DP'$,}{§~95}

\pnote{(two straight lines drawn from a point in a $\perp$ to a line, cutting off on the
line equal segments from the foot of the $\perp$, are equal).}

\eq{$\therefore PD + DP'$}{$= 2PD$,}{}

\eq[and]{$PC + CP'$}{$= 2PC$.}{Const.}

\eq[\indent But]{$PC + CP'$}{$< PD + DP'$.}{§~49}

\eq{$\therefore 2 PC$}{$< 2 PD$.}{}

\eq{$\therefore PC$}{$< PD$.}{Ax.~7}

\hfill\qed

\end{proof}

\pp{\cor{The shortest line that can be drawn from a
point to a given line is perpendicular to the given line.}}

\pp{\defn{The \textbf{distance}\label{distance2} of a point from a line is the length
of the perpendicular from the point to the line.}}
\scanpage{031.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The sum of two lines drawn from a point to the
extremities of a straight line is greater than the sum of
two other lines similarly drawn, but included by them.}

\figc{031aa100}{Let $CA$ and $CB$ be two lines drawn from the point $C$ to the extremities
of the straight line $AB$. Let $OA$ and $OB$ be two lines similarly
drawn, but included by $CA$ and $CB$.}

\proveq{$CA + CB$}{$> OA + OB$.}

\textbf{Proof.} Produce $AO$ to meet the line $CB$ at $E$.

\eq[\indent Then]{$CA + CE$}{$> OA + OE$,}{}

\eq[and]{$BE + OE$}{$> OB$,}{§~49}

\pnote{(a straight line is the shortest line from one point to another).}

\step{Add these inequalities, and we have}{}

\eq{$CA + CE + BE + OE$}{$> OA + OE + OB$.}{Ax.~4}

\step{Substitute for $CE + BE$ its equal $CB$, then}{}

\eq{$CA + CB + OE$}{$> OA + OE + OB$.}{}

\step{Take away $OE$ from each side of the inequality.}{}

\eq{$CA + CB$}{$> OA + OB$.}{Ax.~5}

\hfill\qed
\end{proof}
\scanpage{032.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Of two straight lines drawn from the same point
in a perpendicular to a given line, cutting off on the
line unequal segments from the foot of the perpendicular,
the more remote is the greater.}

\figc{032aa101}{Let $OC$ be perpendicular to $AB$, $OG$ and $OE$ two straight lines to
$AB$, and $CE$ greater than $CG$.}

\proveq{$OE$}{$> OG$.}

\textbf{Proof.}  Take $CF$ equal to $CG$, and draw $OF$.

\eq[\indent Then]{$OF$}{$= OG$,}{§~95}

\pnote{(two straight lines drawn from a point in a $\perp$ to a line, cutting off on the
line equal segments from the foot of the $\perp$, are equal).}

\step{Produce $OC$ to $D$, making $CD = OC$.}{}

\step{Draw $ED$ and $FD$.}{}

\step[\indent Then]{$OE = ED$, and $OF = FD$.}{§~95}

\eq[\indent But]{$OE + ED$}{$> OF + FD$,}{§~100}

\eq{$\therefore 2OE > 2OF$, $OE$}{$> OF$, and $OE > OG$.}{\qed}

\end{proof}

\pp{\cor{Only two equal straight lines can be drawn
from a point to a straight line; and of two unequal lines,
the greater cuts off on the line the greater segment from the
foot of the perpendicular.}}
\scanpage{033.png}%


\pagebreak
\section{PARALLEL LINES.}


\pp{\defn{Two \indexbf{parallel lines} are lines that lie in the same
plane and cannot meet however far they are produced.}}

\proposition{Theorem.}

\begin{proof}%
\obs{Two straight lines in the same plane perpendicular
to the same straight line are parallel.}

\figc{033aa104}{Let $AB$ and $CD$ be perpendicular to $AC$.}

\prove{$AB$ and $CD$ are parallel.}

\textbf{Proof.} If $AB$ and $CD$ are not parallel, they will meet if
sufficiently prolonged; and we shall have two perpendicular
lines from their point of meeting to the same straight line;
but this is impossible,~\hfill§~96

\pnote{(only one perpendicular can be drawn to a given line from a given
external point).}

\step{$\therefore AB$ and $CD$ are parallel.}{\qed}

\end{proof}

\pp{\ax{Through a given point only one straight line
can be drawn parallel to a given straight line.}\label{axiomparallel}}

\begin{point}%
\cor{Two straight lines in the same plane parallel to
a third straight line are parallel to each other.}

For if they could meet, we should have two straight lines
from the point of meeting parallel to a straight line; but this
is impossible.~\hfill§~105
\end{point}
\scanpage{034.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If a straight line is perpendicular to one of two
  parallel lines, it is perpendicular to the other also.}

\figc{034aa107}{Let $AB$ and $EF$ be two parallel lines, and let $HK$ be
  perpendicular to $AB$, and cut $EF$ at $C$.}

\proveq{$HK$}{is $\perp$ to $EF$.}

\textbf{Proof.} Suppose $MN$ drawn through $C \perp$ to $HK$.

\eq[\indent Then]{$MN$} {is $\parallel$ to $AB$.}{§~104}

\eq[\indent But]{$EF$} {is $\parallel$ to $AB$.}{Hyp.}

\step{$\therefore EF$ coincides with $MN$.}{§~105}

\eq[\indent But]{$MN$} {is $\perp$ to $HK$.}{Const.}

\eq{$\therefore EF$}{is $\perp$ to $HK$,}{}

\eq[that is,]{$HK$}{is $\perp$ to $EF$.}{\qed}

\end{proof}

\pp{\defn{A straight line that cuts two or more straight
lines is called a \indexbf{transversal} of those lines.}}

\figc{034bb109}{}
\begin{point}%
If the transversal $EF$ cuts $AB$ and $CD$, the angles
$a$, $d$, $g$, $f$ are called \emph{interior}\label{interior} angles; $b$, $c$, $h$,
$e$ are called \emph{exterior}\label{exterior} angles.

The angles $d$ and $f$, and $a$ and $g$, are called
\emph{alternate-interior}\label{altint} angles; the angles $b$ and $h$, and $c$ and
$e$, are called \emph{alternate-exterior}\label{altext} angles.

The angles $b$ and $f$, $c$ and $g$, $e$ and $a$, $h$ and $d$, are
called \emph{exterior-interior}\label{extint} angles.
\end{point}
\scanpage{035.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two parallel lines are cut by a transversal, the
alternate-interior angles are equal.}

\figc{035aa110}{Let $EF$ and $GH$ be two parallel lines cut by the transversal $BC$.}

\proveq{$\angle EBC$}{$= \angle BCH$.}

\textbf{Proof.} Through $O$, the middle point of $BC$, suppose $AD$
drawn $\perp$ to $GH$.

\step[\indent Then]{$AD$ is likewise $\perp$ to $EF$,}{§~107}

\pnote{(a straight line $\perp$ to one of two $\parallel_s$ is $\perp$ to the other),}

\step[that is,]{$CD$ and $BA$ are both $\perp$ to $AD$.}{}

Apply the figure $COD$ to the figure $BOA$, so that $OD$ shall
fall along $OA$.

\step[\indent Then]{$OC$ will fall along $OB$,}{§~93}

\pnote{(since $\angle COD = \angle BOA$, being vertical $\angle_s$);}

\step[and]{$C$ will fall on $B$,}{}

\pnote{(since $OC = OB$, by construction).}

\step[\indent Then]{the $\perp CD$ will fall along the $\perp BA$,}{§~96}

\pnote{(only one $\perp$ can be drawn to a given line from a given external point).}

\step{$\therefore \angle OCD$ coincides with $\angle OBA$, and is equal to it,}{§~60}

\pnote{(two angles are equal, if their vertices coincide and their sides coincide, each
with each).}

\hfill\qed

\end{proof}
\scanpage{036.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{\textsc{Conversely:} When two straight lines in the same
plane are cut by a transversal, if the alternate-interior
angles are equal, the two straight lines are parallel.}

\figc{036aa111}{Let $EF$ cut the straight lines $AB$ and $CD$ in the points $H$ and $K$,
and let the angles $AHK$ and $HKD$ be equal.}

\proveq{$AB$ is}{$\parallel$ to $CD$.}

\textbf{Proof}. Suppose $MN$ drawn through $H \parallel$ to $CD$.

\eq[\indent Then]{$\angle MHK$}{$= \angle HKD$,}{§~110}

\pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).}

\eq[\indent But]{$\angle AHK$}{$= \angle HKD$.}{Hyp.}

\eq{$\therefore \angle MHK$}{$= \angle AHK$.}{Ax.~1}

\step{$\therefore MN$ and $AB$ coincide.}{§~60}

\eq[\indent But]{$MN$ is}{$\parallel$ to $CD$.}{Const.}

\step{$\therefore AB$, which coincides with $MN$, is $\parallel$ to $CD$.}{\qed}

\end{proof}

\ex{Find the complement and the supplement of an angle that contains
$37°\ 53'\ 49''$.}

\ex{If the complement of an angle is one third of its supplement,
how many degrees does the angle contain?}
\scanpage{037.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two parallel lines are cut by a transversal, the
exterior-interior angles are equal.}

\figc{037aa112}{Let $AB$ and $CD$ be two parallel lines cut by the transversal $EF$,
in the points $H$ and $K$.}

\proveq{$\angle EHB$}{$= \angle HKD$.}

\eq[\indent\textbf{Proof.}]{$\angle EHB$}{$= \angle AHK$,}{§~93}

\pnote{(being vertical $\angle_s$).}

\eq{$\angle AHK$}{$= \angle HKD$,}{§~110}

\pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).}

\eq{$\therefore \angle EHB$}{$= \angle HKD$.}{Ax.~1}

\eq[\indent In like manner]{$\angle EHA$}{$= \angle HKC$.}{\qed}

\end{proof}

\pp{\cor{The alternate-exterior angles $EHB$ and $CKF$,
and also $AHE$ and $DKF$, are equal.}}

\proposition{Theorem.}

\begin{point}%
\obs{\textsc{Conversely:} When two straight lines in a plane
are cut by a transversal, if the exterior-interior angles
are equal, these two straight lines are parallel.}

(Proof similar to that in §~111.)
\end{point}
\scanpage{038.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two parallel lines are cut by a transversal,
  the two interior angles on the same side of the transversal are
  supplementary.}

\figc{038aa115}{Let $AB$ and $CD$ be two parallel lines cut by the transversal
  $EF$ in the points $H$ and $K$.}

\prove{$\angle_s BHK$ and $HKD$ are supplementary.}

\step[\indent\textbf{Proof.}]{$\angle EHB + \angle BHK = \text{a st.\ }\angle$,}{§~86}

\pnote{(being sup.-adj.~$\angle_s$).}

\step[\indent But]{$\angle EHB = \angle HKD$,}{§~112}

\pnote{(being ext.-int.~$\angle_s$ of $\parallel$ lines).}

\step{\( \therefore \angle BHK + \angle HKD = \text{a st.\ }\angle \).}{}

\step{$\therefore \angle_s BHK$ and $HKD$ are supplementary.}{§~77}

\hfill\qed

\end{proof}

\proposition{Theorem.}

\begin{proof}%
\obs{\textsc{Conversely:} When two straight lines in a
  plane are cut by a transversal, if two interior angles on the same
  side of the transversal are supplementary, the two straight lines
  are parallel.}

(Proof similar to that in §~111.)

\end{proof}
\scanpage{039.png}%

\section{TRIANGLES.}

\begin{point}%
A \indexbf{triangle} is a portion of a plane bounded by three
straight lines; as, $ABC$ (Fig.~1).

\figcc{039aa117}{039bb118}
The bounding lines are called the
\textbf{sides}\label{trisides} of the triangle, and their sum is
called its \indexbf{perimeter}; the angles included
by the sides are called the \textbf{angles} of the
triangle\label{anglestri}, and the vertices of these angles,
the \textbf{vertices} of the triangle\label{trivertices}.
\end{point}

\begin{point}%
\textbf{Adjacent angles}\label{adjacent2} of a rectilinear
figure are two angles that have one side
of the figure common; as, angles $A$
and $B$ (Fig.~2).
\end{point}

\begin{point}%
An \textbf{exterior angle} of a triangle\label{exteriortri} is an angle included by
one side and another side produced; as, $ACD$ (Fig.~2). The
interior angle $ACB$ is adjacent to the exterior angle; the interior
angles, $A$ and $B$, are called \textbf{opposite interior angles}.
\end{point}

\figc{039ce119}{}

\begin{point}%
A triangle is called a \indexbf{scalene triangle} when no two of
its sides are equal; an \indexbf{isosceles triangle}, when two of its sides
are equal; an \indexbf{equilateral triangle}, when its three sides are equal.

\figc{039fi120}{}
\end{point}
\scanpage{040.png}%

\begin{point}%
A triangle is called a \indexbf{right triangle}, when one of its
angles is a right angle; an \indexbf{obtuse triangle}, when one of its
angles is an obtuse angle; an \textbf{acute triangle}, when all three
of its angles are acute angles; an \indexbf{equiangular triangle}, when
its three angles are equal.
\end{point}

\begin{point}%
In a right triangle, the side opposite the right angle is
called the \indexbf{hypotenuse}, and the other two sides the \indexbf{legs}.
\end{point}

\begin{point}%
The side on which a triangle is supposed to stand is
called the base\label{basetri} of the triangle. In the isosceles triangle, the
equal sides are called the legs, and the other side, the base\label{baseiso}; in
other triangles, any one of the sides may be taken as the base.
\end{point}

\begin{point}%
The angle opposite the base of a triangle is called the
\indexbf{vertical angle}, and its vertex, the \textbf{vertex} of the triangle\label{trivertex}.
\end{point}

\begin{point}%
The \textbf{altitude} of a triangle\label{alttri} is the perpendicular from the
vertex to the base, or to the base produced; as, $AD$ (Fig.~1).
\end{point}

\begin{point}%
The three perpendiculars from the vertices of a triangle
to the opposite sides (produced if necessary) are called the
\textbf{altitudes} of the triangle; the three bisectors of the angles are
called the \textbf{bisectors} of the triangle\label{tribisectors}; and the three lines from
the vertices to the middle points of the opposite sides are
called the \textbf{medians}\label{trimedians} of the triangle.
\end{point}

\begin{point}%
If two triangles have the angles of the one equal, respectively,
to the angles of the other, the equal angles are called
\indexbf{homologous angles}, and the sides opposite the equal angles are
called \indexbf{homologous sides}.
\end{point}

\begin{point}%
Two triangles are equal in all respects if they can be
made to coincide (§~60). The homologous sides of \indexemph{equal triangles}
are equal, and the homologous angles are equal.
\end{point}
\scanpage{041.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The sum of the three angles of a triangle is equal
to two right angles.}

\figc{041aa129}{Let $A$, $B$, and $BCA$ be the angles of the triangle $ABC$.}

\prove{$\angle A+\angle B+\angle BCA = 2$ rt.~$\angle_s$.}

\textbf{Proof.} Suppose $CE$ drawn $\parallel$ to $AS$, and prolong $AC$ to $F$.

\step[\indent Then]{$\angle ECF + \angle ECB + \angle BCA = 2$ rt.~$\angle_s$,}{§~89}

\pnote{(the sum of all the $\angle_s$ about a point on the same side of a straight line
passing through the point is equal to $2$~rt.~$\angle_s$).}

\eq[\indent But]{$\angle A$}{$= \angle ECF$,}{§~112}

\pnote{(being ext.-int.~$\angle_s$ of the $\parallel$ lines $AB$ and $CE$),}

\eq[and]{$\angle B$}{$= \angle BCE$,}{§~110}

\pnote{(being alt.-int.~$\angle_s$ of the $\parallel$ lines $AB$ and $CE$).}

Put for the $\angle_s ECF$ and $BCE$ their equals, the $\angle_s A$ and $B$.

\step[\indent Then]{$\angle A +\angle B + \angle BCA = 2$ rt.~$\angle_s$.}{\qed}

\end{proof}

\pp{\cor[1]{The sum of two angles of a triangle is less
than two right angles.}}

\pp{\cor[2]{If the sum of two angles of a triangle is
taken from two right angles, the remainder is equal to the
third angle.}}

\pp{\cor[3]{If two triangles have two angles of the one
equal to two angles of the other, the third angles are equal.}}
\scanpage{042.png}%

\pp{\cor[4]{If two right triangles have an acute angle of
the one equal to an acute angle of the other, the other acute
angles are equal.}}

\pp{\cor[5]{In a triangle there can be but one right angle,
or one obtuse angle.}}

\pp{\cor[6]{In a right triangle the two acute angles are
together equal to one right angle, or~$90°$.}}

\pp{\cor[7]{In an equiangular triangle, each angle is one
third of two right angles, or~$60°$.}}

\pp{\cor[8]{An exterior angle of a triangle is equal to
the sum of the two opposite interior angles, and therefore
greater than either of them.}}

\proposition{Theorem.}

\begin{proof}%
\obs{The sum of two sides of a triangle is greater than
the third side, and their difference is less than the third
side.}

\figc{042aa138}{In the triangle $ABC$, let $AC$ be the longest side.}

\prove{$AB + BC > AC$, and $AC - BC < AB$}.

\step[\indent\textbf{Proof.}]{$AB + BC > AC$,}{§~49}

\pnote{(a straight line is the shortest line from one point to another).}

\step{Take away $BC$ from both sides.}{}

\step[\indent Then]{$AB > AC - BC$,}{Ax.~5}

\step[or]{$AC - BC < AB$.}{\qed}

\end{proof}
\scanpage{043.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Two triangles are equal if two angles and the
included side of the one are equal, respectively, to two
angles and the included side of the other.}

\figc{043ab139}{In the triangles $ABC$, $DEF$, let the angle $A$ be equal to the angle
$D$, $B$ to $E$, and the side $AB$ to $DE$.}

\proveq{$\triangle ABC$}{$= \triangle DEF$.}

\textbf{Proof.} Apply the $\triangle ABC$ to the $\triangle DEF$ so that $AB$ shall
coincide with its equal, $DE$.

\step[\indent Then]{$AC$ will fall along $DF$, and $BC$ along $EF$,}{}

\pnote{(for $\angle A = \angle D$, and $\angle B = \angle E$, by hyp.).}

\step{$\therefore C$ will fall on $F$,}{§~48}

\pnote{(two straight lines can intersect in only one point).}

\step{$\therefore$ the two $\triangle_s$ coincide, and are equal.}{§~60}

\hfill\qed

\end{proof}

\pp{\cor[1]{Two triangles are equal if a side and any two
angles of the one are equal to the homologous side and two
angles of the other.}~\hfill§~132}

\pp{\cor[2]{Two right triangles are equal if the hypotenuse
and an acute angle of the one are equal, respectively, to
the hypotenuse and an acute angle of the other.}~\hfill§~133}

\pp{\cor[3]{Two right triangles are equal if a leg and
an acute angle of the one are equal, respectively, to a leg
and the homologous acute angle of the other.}~\hfill§~133}
\scanpage{044.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Two triangles are equal if two sides and the included
angle of the one are equal, respectively, to two sides
and the included angle of the other.}

\figc{044ab143}{In the triangles $ABC$ and $DEF$, let $AB$ be equal to $DE$, $AC$ to $DF$,
and the angle $A$ to the angle $D$.}

\proveq{$\triangle ABC$}{$= \triangle DEF$.}

\textbf{Proof.} Apply the $\triangle ABC$ to the $\triangle DEF$ so that $AB$ shall
coincide with its equal, $DE$.

\step{Then $AC$ will fall along $DF$,}{}

\pnote{(for $\angle A = \angle D$, by hyp.);}

\step{and $C$ will fall on $F$,}{}

\pnote{(for $AC = DF$, by hyp.).}

\step{$\therefore CB = FE$,}{}

\pnote{(their extremities being the same points).}

\step{$\therefore$ the two $\triangle_s$ coincide, and are equal.}{\qed}

\end{proof}

\pp{\cor{Two right triangles are equal if their legs are
equal, each to each.}}

\note{In §~139 we have given two angles and the included side, in
§~143 two sides and the included angle; hence, by interchanging the
words \emph{sides} and \emph{angles}, either theorem is changed to the other. This
is called the \emph{Principle of Duality}\label{princduality}, or the \emph{Principle of Reciprocity}\label{princreciprocity}. The
reciprocal of a theorem is not always true, just as the converse of a
theorem is not always true.}
\scanpage{045.png}%

\proposition{Theorem.}

\begin{proof}%
\textit{In an isosceles triangle the angles opposite the
equal sides are equal.}

\figc{045aa145}{Let $ABC$ be an isosceles triangle, having $AB$ and $AC$ equal.}

\proveq{$\angle B$}{$= \angle C$.}

\textbf{Proof.} Suppose $AD$ drawn so as to bisect the $\angle BAC$.

\step{In the $\triangle_s ADB$ and $ADC$,}{}

\eq{$AB$}{$=AC$,}{Hyp.}

\eq{$AD$}{$=AD$,}{Iden.}

\eq{and $\angle BAD$}{$= \angle CAD$.}{Const.}

\eq{$\therefore \triangle ADB$}{$= \triangle ADC$,}{§~143}

\pnote{(two $\triangle_s$ are equal if two sides and the included $\angle$ of the one are equal,
respectively, to two sides and the included $\angle$ of the other).}

\eq{$\therefore \angle B$}{$= \angle C$,}{§~128}

\pnote{(being homologous angles of equal triangles).}

\hfill\qed

\end{proof}

\pp{\cor{An equilateral triangle is equiangular, and each
angle is two thirds of a right angle.}}

\ex{If the equal sides of an isosceles triangle are produced, the
angles on the other side of the base are equal.}
\scanpage{046.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two angles of a triangle are equal, the sides
opposite the equal angles are equal, and the triangle is
isosceles.}

\figc{046aa147}{In the triangle $ABC$, let the angle $B$ be equal to the angle~$C$.}

\proveq{$AB$}{$= AC$.}

\step[\indent\textbf{Proof.}]{Suppose $AD$ drawn $\perp$ to $BC$.}{}

In the rt.~$\triangle_s ADB$ and $ADC$,

\eq{$AD$}{$= AD$,}{Iden.}

\eq{and $\angle B$}{$= \angle C$.}{Hyp.}

\eq{$\therefore$ rt.~$\triangle ADB$}{$=$ rt.~$\triangle ADC$,}{§~142}

\pnote{(having a leg and an acute $\angle$ of the one equal, respectively, to a leg and
the homologous acute $\angle$ of the other).}

\eq{$\therefore AB$}{$= AC$,}{§~128}

\pnote{(being homologous sides of equal $\triangle_s$).}

\hfill\qed

\end{proof}

\pp{\cor[1]{An equiangular triangle is also equilateral.}}

\pp{\cor[2]{The perpendicular from the vertex to the
base of an isosceles triangle bisects the base, and bisects the
vertical angle of the triangle.}}
\scanpage{047.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Two triangles are equal if the three sides of the
one are equal, respectively, to the three sides of the other.}

\figc{047ab150}{In the triangles $ABC$ and $A'B'C'$, let $AB$ be equal to $A'B'$, $AC$ to
$A'C'$, $BC$ to $B'C'$.}

\proveq{$\triangle ABC$}{$= \triangle A'B'C'$.}

\textbf{Proof.} Place $\triangle A'B'C'$ in the position $\triangle AB'C$ having its
greatest side $\triangle A'C'$ in coincidence with its equal $\triangle AC$, and its
vertex at $B'$, opposite $B$; and draw $BB'$.

\eq[\indent Since]{$AB$}{$=AB'$}{Hyp.}

\eq{$\angle ABB'$}{$= \angle AB'B$}{§~145}

\pnote{(in an isosceles $\triangle$ the $\angle_s$ opposite the
equal sides are equal).}

\eq[\indent Since]{$CB$}{$= CB'$,}{Hyp.}

\eq{$\angle CBB'$}{$= \angle CB'B$.}{§~145}

\eq{$\therefore \angle ABB' + \angle CBB'$}{$= \angle AB'B + \angle CB'B$.}{Ax.~2}

\eq[\indent Hence,]{$\angle ABC$}{$= \angle AB'C$.}{}

\eq{$\therefore \triangle ABC$}{$= \triangle AB'C$,}{§~143}

\pnote{(two $\triangle_s$ are equal if two sides and the included $\angle$
 of the one are equal, respectively, to two sides and the included $\angle$ of the other).}

\eq{$\therefore \triangle ABC$}{$= \triangle A'B'C'$.}{\qed}

\end{proof}
\scanpage{048.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Two right triangles are equal if a leg and the
hypotenuse of the one are equal, respectively, to a leg
and the hypotenuse of the other.}

\figc{048ac151}{In the right triangles $ABC$ and $A'B'C'$, let $AB$ be equal to $A'B'$, and
$AC$ to $A'C'$.}

\proveq{$\triangle ABC$}{$= \triangle A'B'C'$.}

\textbf{Proof.} Apply the $\triangle ABC$ to the $\triangle A'B'C'$, so that $AB$ shall
coincide with $A'B'$, $A$ falling on $A'$, $B$ on $B'$, and $C$ and $C'$ on
opposite sides of $A'B'$.

\step[\indent Then]{$BC$ will fall along $C'B'$ produced,}{}

\pnote{(for $\angle ABC = \angle A'B'C'$, each being a rt.~$\angle$.).}

\eq[\indent Since]{$AC$}{$= A'C'$,}{Hyp.}

\step{the $\triangle A'CC'$ is an isosceles triangle.}{§~120}

\eq{$\therefore \angle C$}{$= \angle C'$,}{§~145}

\pnote{($\angle_s$ opposite the equal sides of an isosceles $\triangle$ are equal).}

\step{$\therefore \triangle_s ABC$ and $A'B'C'$ are equal,}{§~141}

\pnote{(two right $\triangle_s$ are equal if they have the hypotenuse and an acute $\angle$ of, the
one equal to the hypotenuse and an acute $\angle$ of the other).}

\hfill\qed

\end{proof}

\ex{How many degrees are there in each of the acute angles of an
isosceles right triangle?}
\scanpage{049.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two sides of a triangle are unequal, the angles
opposite are unequal, and the greater angle is opposite
the greater side.}

\figc{049aa152}{In the triangle $ACB$, let $AB$ be greater than~$AC$.}

\prove{$\angle ACB$ is greater than $\angle B$.}

\step[\indent\textbf{Proof.}]{On $AB$ take $AE$ equal to $AC$.}{}

\step{Draw $EC$.}{}

\eq{$\angle AEC$}{$= \angle ACE$}{§~145}

\pnote{(being $\angle_s$ opposite equal sides).}

\step[\indent But]{$\angle AEC$ is greater than $\angle B$}{§~137}

\pnote{(an exterior $\angle$ of a $\triangle$ is greater than either opposite interior $\angle$),}

\step[and]{$\angle ACB$ is greater than $\angle ACE$.}{Ax.~8}

\step{Substitute for $\angle ACE$ its equal $\angle AEC$,}{}

\step[then]{$\angle ACB$ is greater than $\angle AEC$.}{}

Since $\angle AEC$ is greater than $\angle B$, and $\angle ACB$ is greater
than $\angle AEC$,

\step{$\angle ACB$ is greater than $\angle B$.}{\qed}

\end{proof}

\ex{If any angle of an isosceles triangle is equal to two thirds of a
right angle~($60°$), what is the value of each of the two remaining angles?}

\ex{One angle of a triangle is~$34°$. Find the other angles, if one
of them is twice the other.}
\scanpage{050.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{\textsc{Reciprocally:} If two angles of a triangle are
unequal, the sides opposite are unequal, and the greater
side is opposite the greater angle.}

\figc{050aa153}{In the triangle $ACB$, let the angle $C$ be greater than the angle~$B$.}

\prove{$AB > AC$.}

\step[\indent\textbf{Proof.}]{Now $AB=AC$, or $< AC$, or $>AC$.}{}
\label{41}

\step{But $AB$ is not equal to $AC$;}{}

\step{for then the $\angle C$ would be equal to the $\angle B$,}{§~145}

\pnote{(being $\angle_s$ opposite equal sides).}

\step{And $AB$ is not less than $AC$;}{}

\step{for then the $\angle C$ would be less than the $\angle B$.}{§~152}

Both these conclusions are contrary to the hypothesis that
the $\angle C$ is greater than the $\angle B$.

\step{Hence, $AB$ cannot be equal to $AC$ or less than $AC$.}{}

\step{$\therefore AB > AC$.}{\qed}

\end{proof}

\ex{If the vertical angle of an isosceles triangle is equal to~$30°$,
find the exterior angle included by a side and the base produced.}

\ex{If the vertical angle of an isosceles triangle is equal to~$36°$,
find the angle included by the bisectors of the base angles.}
\scanpage{051.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two triangles have two sides of the one equal,
respectively, to two sides of the other, but the included
angle of the first triangle greater than the included
angle of the second, then the third side of the first is
greater than the third side of the second.}

\figc{051ac154}{In the triangles $ABC$ and $ABE$, let $AB$ be equal to $AB$, $BC$ to $BE$;
but let the angle $ABC$ be greater than the angle $ABE$.}

\proveq{$AC$}{$> AE$.}

\textbf{Proof.} Place the $\triangle_s$ so that $AB$ of the one shall fall on
$AB$ of the other, and $BE$ within the $\angle ABC$.

Suppose $BF$ drawn to bisect the $\angle EBC$, and draw $EF$.

The $\triangle_s EBF$ and $CBF$ are equal.~\hfill§~143

\eq[\indent For]{$BF$}{$= BF$,}{Iden.}

\eq{$BE$}{$=BC$,}{Hyp.}

\eq[and]{$\angle EBF$}{$=\angle CBF$.}{Const.}

\eq{$\therefore EF$}{$=FC$.}{§~128}

\eq[\indent Now]{$AF+FE$}{$> AE$.}{§~138}

\eq{$\therefore AF+FC$}{$> AE$.}{}

\eq{$\therefore AC$}{$> AE$.}{\qed}

\end{proof}
\scanpage{052.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{\textsc{Conversely}: If two sides of a triangle are equal,
respectively, to two sides of another, but the third
side of the first triangle is greater than the third side
of the second, then the angle opposite the third side of
the first triangle is greater than the angle opposite the
third side of the second.}

\figc{052ab155}{In the triangles $ABC$ and $DEF$, let $AB$ be equal to $DE$, $AC$ to $DF$;
but let $BC$ be greater than EF.}

\prove{the $\angle A$ is greater than the $\angle D$.}

\textbf{Proof}. Now the $\angle A$ is equal to the $\angle D$, or less than the $\angle D$,
or greater than the $\angle D$.

\step{But the $\angle A$ is not equal to the $\angle D$;}{}

\step{for then the $\triangle ABC$ would be equal to the $\triangle DEF$,}{§~143}

\pnote{(having two sides and the included $\angle$ of the one equal, respectively, to two
sides and the included $\angle$ of the other),}

\step{and $BC$ would be equal to $EF$.}{}

And the $\angle A$ is not less than the $\angle D$, for then $BC$ would
be less than $EF$.~\hfill§~154

Both these conclusions are contrary to the hypothesis that
$BC$ is greater than $EF$.

Since the $\angle A$ is not equal to the $\angle D$ or less than the $\angle D$,

\step{the $\angle A$ is greater than the $\angle D$.}{\qed}

\end{proof}
\scanpage{053.png}%


\section{LOCI OF POINTS.}

\begin{point}%
If it is required to find a point which shall fulfil a
\emph{single} geometric condition, the point may have an \emph{unlimited
number of positions}. If, however, all the points are in the
same plane, the required point will be confined to a \emph{particular
line}, or \emph{group of lines}.

A point in a plane at a given distance from a fixed straight
line of indefinite length in that plane, is evidently in one of
two straight lines, so drawn as to be everywhere at the given
distance from the fixed line, one on one side of the fixed line,
and the other on the other side of it.

A point in a plane equidistant from two parallel lines in
that plane is evidently in a straight line drawn between the
two given parallel lines and everywhere equidistant from them.
\end{point}

\begin{point}%
All points in a plane that satisfy a single geometrical
condition lie, in general, in a line or group of lines; and this
line or group of lines is called the \textbf{locus} of the points that
satisfy the given condition.
\end{point}

\begin{point}%
To prove \emph{completely} that a certain line is the locus
of points that fulfil a given condition, it is necessary to
prove

1. \textit{Any point in the line satisfies the given condition;
and any point not in the line does not satisfy the given condition.}

Or, to prove

2. \textit{Any point that satisfies the given condition lies in the
line; and any point in the line satisfies the given condition}.
\end{point}

\note{The word \emph{locus} (pronounced lo\'{ }kus) is a Latin word that signifies
\emph{place}. The plural of locus is loci (pronounced lo\'{ }si).}

\pp{\defn{A line which bisects a given line and is perpendicular
to it is called the \textbf{perpendicular bisector} of the line.}}
\scanpage{054.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The perpendicular bisector\label{perpbisector} of a given line is the
locus of points equidistant from the extremities of the
line.}

\figc{054aa160}{Let $PR$ be the perpendicular bisector of the line $AB$, $O$ any point in
$PR$, and $C$ any point not in $PR$.}

\step{Draw $OA$ and $OB$, $CA$ and $CB$.}{}

\prove[To prove ]{$OA$ and $OB$ equal, $CA$ and $CB$ unequal.}

\eq[\indent\textbf{Proof.}]{\textbf{1. }$\triangle OPA$}{$= \triangle OPB$,}{§~144}

\step{for $PA = PB$ by hypothesis, and $OP$ is common,}{}

\pnote{(two right $\triangle_s$ are equal if their legs are equal, each to each).}

\eq{$\therefore OA$}{$= OB$.}{§~128}

\textbf{2.~}Since $C$ is not in the $\perp$, $CA$ or $CB$ will cut the $\perp$.

\step{Let $CA$ cut the $\perp$ at $D$, and draw $DB$.}{}

Then, by the first part of the proof $DA = DB$.

\step[\indent But]{$CB < CD + DB$.}{§~138}

\step{$\therefore CB < CD + DA$.}{}

\step[\indent That is, ]{$CB < CA$.}{}

$\therefore PR$ is the locus of points equidistant from $A$ and $B$.~\hfill§~158,1

\hfill\qed

\end{proof}

\pp{\cor{Two points each equidistant from the extremities
of a line determine the perpendicular bisector of the line.}}
\scanpage{055.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The bisector of a given angle is the locus of
  points equidistant from the sides of the angle.}

\vspace{-1ex}
\figc{055aa162}{Let $O$ be any point equidistant from the sides of the angle $PAQ$.}

\prove{$O$ is in the bisector of the $\angle PAQ$.}

\step [\indent\textbf{Proof.}] {Draw $AO$.} {}

\step {Suppose $OF$ drawn $\perp$ to $AP$ and $OG \perp$
  to $AQ$.} {}

\step {In the rt.~$\triangle_s AFO$ and $AGO$,} {}

\eq {$AO $}{$= AO$,} {Iden.}

\eq {$OF $}{$= OG$,} {Hyp.}

\eq {$\therefore \triangle AFO $} {$= \triangle AGO$.} {§~151}

\eq {$\therefore \angle FAO $} {$= \angle GAO$.} {§~128}

\step {$\therefore O$ is in the bisector of the $\angle PAQ$.}{}

\textbf{Let $\mathbf{D}$ be any point in the bisector of the angle $\mathbf{PAQ}$.}

\prove {$D$ is equidistant from $AP$ and $AQ$.}

\step[\indent\textbf{Proof.}]{Suppose $DB$ drawn $\perp$ to $AP$ and $DC \perp$ to~$AQ$.}{}

\step {In the rt.~$\triangle_s ABD$ and $ACD$,} {}

\eq {$AD $} {$= AD$,} {Iden.}

\eq {$\angle DAB $} {$= \angle DAC$,} {Hyp.}

\eq {$\therefore \triangle ABD $} {$= \triangle ACD$.} {§~141}

\eq {$\therefore DB $} {$= DC$.} {§~128}

\step{$\therefore D$ is equidistant from $AP$ and $AQ$.}{}

\step{$\therefore$ the bisector of the $\angle PAQ$ is the locus of points that are
equidistant from its sides.}{§~158, 2}
\end{proof}
\scanpage{056.png}%


\section{QUADRILATERALS.}

\begin{point}%
A \indexbf{quadrilateral} is a portion of a plane bounded by
four straight lines. The bounding lines are the \textbf{sides}, the
angles formed by these sides are the \textbf{angles}, and the vertices
of these angles are the \textbf{vertices}, of the quadrilateral.
\end{point}

\begin{point}%
A \indexbf{trapezium} is a quadrilateral which has no two sides
parallel.
\end{point}

\begin{point}%
A \indexbf{trapezoid} is a quadrilateral which has two sides, and
only two sides, parallel.
\end{point}

\begin{point}%
A \indexbf{parallelogram} is a quadrilateral which has its opposite
sides parallel.
\end{point}

\figc{056ac166}{}

\begin{point}%
A \indexbf{rectangle} is a parallelogram which has its angles
right angles.
\end{point}

\begin{point}%
A \indexbf{square} is a rectangle which has its sides equal.
\end{point}

\begin{point}%
A \indexbf{rhomboid} is a parallelogram which has its angles
oblique angles.
\end{point}

\begin{point}%
A \indexbf{rhombus} is a rhomboid which has its sides equal.
\end{point}

\figc{056dg170}{}

\begin{point}%
The side upon which a parallelogram stands, and the
opposite side, are called its lower and upper \emph{bases}\label{basepar}.
\end{point}
\scanpage{057.png}%

\begin{point}%
Two parallel sides of a trapezoid are called its \textbf{bases}\label{basetrap},
the other two sides its \textbf{legs}\label{legstrap}, and the line joining the middle
points of the legs is called the \textbf{median} of the trapezoid\label{mediantrap}.
\end{point}

\figc{057aa174}{}
\begin{point}%
A trapezoid is called an \indexbf{isosceles
trapezoid} if its legs are equal.
\end{point}

\begin{point}%
The \textbf{altitude} of a parallelogram\label{altpar}
or trapezoid\label{alttrap} is the perpendicular distance
between its bases, as $PQ$.
\end{point}

\begin{point}%
A \textbf{diagonal}\label{diagonal1} of a quadrilateral is a straight line joining
two opposite vertices, as $AC$.
\end{point}

\proposition{Theorem.}

\begin{proof}%
\obs{Two angles whose sides are parallel, each to each,
are either equal or supplementary.}

\figc{057bb176}{Let $BA$ be parallel to $HD$, and $BC$ be parallel to $MN$.}

\prove[To prove ]{$\angle_s a$, $a'$ and $c$ equal; $a$ and $c'$ supplementary.}

\step[\indent\textbf{Proof.}]{Let $HD$ and $BC$ prolonged intersect at $x$.}{}

\step[\indent Then]{$\angle a = \angle x$, and $\angle a' = \angle x$.}{§~112}

\step{$\therefore \angle a = \angle a'$.}{Ax.~1}

\step[\indent Also]{$\angle c = \angle a'$ (§~93). $\therefore \angle c = \angle a$.}{Ax.~1}

\step[\indent Now]{$\angle a'$ and $\angle c'$ are supplementary.}{§~89}

\step{Put $\angle a$ for its equal, $\angle a'$.}{}

\step[\indent Then]{$\angle a$ and $\angle c'$ are supplementary.}{\qed}

\end{proof}

\pp{\cor{The opposite angles of a parallelogram are
equal, and the adjacent angles are supplementary.}}
\scanpage{058.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The opposite sides of a parallelogram are equal.}

\figc{058aa178}{Let the figure $ABCE$ be a parallelogram.}

\prove[To prove ]{$BC = AE$, and $AB = EC$.}

\step[\indent\textbf{Proof.}]{Draw the diagonal $AC$.}{}

\step{$\triangle ABC = \triangle CEA$.}{§~139}

\step{For $AC$ is common,}{}

\step{$\angle BAC = \angle ACE$, and $\angle ACB = \angle CAE$,}{§~110}

\pnote{(being alt-int.  $\angle_s$ of $\parallel$ lines).}

\step{$\therefore BC = AE$, and $AB = CE$,}{§~128}

\pnote{(being homologous sides of equal $\triangle_s$).}

\hfill\qed

\end{proof}

\pp{\cor[1]{A diagonal divides a parallelogram into two
equal triangles.}}

\pp{\cor[2]{Parallel lines comprehended between parallel
lines are equal.}}

\figc{058bb181}{}
\begin{point}%
\cor[3]{Two parallel lines
are everywhere equally distant.}

For if $AB$ and $DC$ are parallel,
$\perp_s$ dropped from \emph{any} points in $AB$ to $DC$, are equal, §~180.
Hence, \emph{all} points in $AB$ are equidistant from~$DC$.
\end{point}
\scanpage{059.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If the opposite sides of a quadrilateral are
  equal, the figure is a parallelogram.}

\figc{059aa182}{Let the figure $ABCE$ be a quadrilateral, having $BC$ equal to
  $AE$ and $AB$ to $EC$.}

\prove{the figure $ABCE$ is a $\Par$.}

\step[\indent\textbf{Proof.}]{Draw the diagonal $AC$.}{}

In the $\triangle_s ABC$ and $CEA$,

\eq{$BC$}{$= AE$,}{Hyp.}

\eq{$AB$}{$= CE$,}{Hyp.}

\eq[and]{$AC$}{$= AC$,}{Iden.}

\eq{$\therefore \triangle ABC$}{$= \triangle CEA$,}{§~150}

\pnote{(having three sides of the one equal, respectively,
to the three sides of the other).}

\eq{$\therefore \angle ACB$}{$= \angle CAE$,}{§~128}

\eq[and]{$\angle BAC$}{$= \angle ACE$,}{}

\pnote{(being homologous $\angle_s$ of equal $\triangle_s$).}

\eq{$\therefore BC$}{is $\parallel$ to $AE$,}{}

\eq[and]{$AB$}{is $\parallel$ to $EC$,}{§~111}

\pnote{(two lines in the same plane cut by a transversal are parallel,
if the alt.-int.~$\angle_s$ are equal).}

\step{$\therefore$ the figure $ACBE$ is a $\Par$,}{§~166}

\pnote{(having its opposite sides parallel).}\hfill\llap{\qed}

\end{proof}
\scanpage{060.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two sides of a quadrilateral are equal and parallel, then the
other two sides are equal and parallel, and the figure is a parallelogram.}

\figc{060aa183}{Let the figure $ABCE$ be a quadrilateral, having the side $AE$ equal and
parallel to $BC$.}

\prove{$AB$ is equal and parallel to $EC$.}

\step[\indent\textbf{Proof.}]{Draw $AC$.}{}

The $\triangle_s ABC$ and $CEA$ are equal,~\hfill§~143

\pnote{(having two sides and the included $\angle$ of each equal,
respectively).}

\step[\indent For]{$AC$ is common,}{}

\eq{$BC$}{$=AE$}{Hyp.}

\eq[and]{$\angle BCA$}{$= \angle CAE$,}{§~110}

\pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).}

\eq{$\therefore AB$}{$=EC$,}{}

\eq[and]{$\angle BAC$}{$= \angle ACE$,}{§~128}

\pnote{(being homologous parts of equal $\triangle_s$).}

\step{$\therefore AB$ is $\parallel$ to $EC$,}{§~111}

\pnote{(two lines are $\parallel$, if the alt.-int. $\angle_s$ are equal).}

\step{$\therefore$ the figure $ABCE$ is a $\Par$,}{§~166}

\pnote{(the opposite sides being parallel).}

\hfill\qed

\end{proof}
\scanpage{061.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The diagonals of a parallelogram bisect each other.}

\figc{061aa184}{Let the figure $ABCE$ be a parallelogram, and let the diagonals $AC$
and $BE$ cut each other at $O$.}

\prove{$AO = OC$, and $BO = OE$.}

\textbf{Proof.} In the $\triangle_s AOE$ and $COB$,

\eq{$AE$}{$=BC$,}{§~178}

\pnote{(being opposite sides of a $\Par$).}

\eq{$\angle OAE$}{$=\angle OCB$,}{§~110}

\eq{and $\angle OEA$}{$= \angle OBC$,}{}

\pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).}

\eq{$\therefore\triangle AOE$}{$=\triangle COB$,}{§~139}

\pnote{(having two $\angle_s$ and the included side of the one equal, respectively, to two
$\angle_s$ and the included side of the other).}

\step{$\therefore AO=OC$, and $BO=OE$,}{§~128}
\pnote{(being homologous sides of equal $\triangle_s$).}

\hfill\qed

\end{proof}

\ex{The median from the vertex to the base of an isosceles triangle
is perpendicular to the base, and bisects the vertical angle.}

\ex{If two straight lines are cut by a transversal so that the alternate-exterior
angles are equal, the two straight lines are parallel.}

\ex{If two parallel lines are cut by a transversal, the two exterior
angles on the same side of the transversal are supplementary.}

\ex{If two straight lines are cut by a transversal so as to make the
exterior angles on the same side of the transversal supplementary, the two
lines are parallel.}
\scanpage{062.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Two parallelograms are equal, if two sides and
the included angle of the one are equal, respectively, to
two sides and the included angle of the other.}

\figc{062ab185}{In the parallelograms $ABCD$ and $A'B'C'D'$, let $AB$ be equal to $A'B'$,
$AD$ to $A'D'$, and angle $A$ to $A'$.}

\prove{the $\Par_s$ are equal}.

\textbf{Proof.} Place the $\Par$ $ABCD$ on the $\Par$ $A'B'C'D'$, so that $AD$
will fall on and coincide with its equal, $A'D'$.

\step{Then $AB$ will fall on $A'B'$, and $B$ on $B'$;}{}

\pnote{(for $\angle A = \angle A'$, and $AB = A'B'$, by hyp.)}

Now, $BC$ and $B'C'$ are both $\parallel$ to $A'D'$ and drawn through $B'$.

\step{$\therefore BC$ and $B'C'$ coincide,}{§~105}

\pnote{(through a given point only one line can be drawn $\parallel$
  to a given line).}

Also $DC$ and $D'C'$ are $\parallel$ to $A'B'$ and drawn through $D'$.

\step{$\therefore DC$ and $D'C'$ coincide.}{§~105}

\step{$\therefore C$ falls on $C'$,}{§~48}

\pnote{(two lines can intersect in only one point),}

\step{$\therefore$ the two $\Par_s$ coincide, and are equal.}{\qed}

\end{proof}

\pp{\cor{Two rectangles having equal bases and altitudes
are equal.}}
\scanpage{063.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If three or more parallels intercept equal parts
on one transversal, they intercept equal parts on every
transversal.}

\figc{063aa187}{Let the parallels $AH$, $BK$, $CM$, $DP$ intercept equal parts $HK$, $KM$,
$MP$ on the transversal $HP$.}

\prove{they intercept equal parts $AB$, $BC$, $CD$ on
the transversal~$AD$.}

\textbf{Proof.} Suppose $AH$, $BF$, and $CG$ drawn $\parallel$ to $HP$.

\step{$\angle_s$ $AEB$, $BFC$, etc.\ $=\angle_s$ $HKE$, $KMF$, etc., respectively.}{§~112}

\step{But $\angle_s$ $HKE$, $KMF$, etc.\ are equal.}{§~112}

\step{$\therefore \angle_s$ $AEB$, $BFC$, etc.\ are equal.}{Ax.~1}

\step{Also $\angle_s$ $BAE$, $CBF$, etc.\ are equal.}{§~112}

\step{Now $AE = HK$, $BF = KM$, $CG = MP$,}{§~180}

\pnote{(parallels comprehended between parallels are equal).}

\step{$\therefore AE = BF = CG$.}{Ax.~1}

\step{$\therefore \triangle ABE = \triangle BCF = \triangle CDG$,}{§~139}

\pnote{(having two $\angle_s$ and the included side of each respectively equal).}

\step{$\therefore AB = BC = CD$.}{§~128}

\hfill\qed

\end{proof}
\scanpage{064.png}%

\figc{064aa188}{}
\begin{point}%
\cor[1]{If a line is parallel to the base of a triangle
and bisects one side, it bisects the other
side also.}

Let $DE$ be $\parallel$ to $BC$ and bisect $AB$. Suppose
a line is drawn through $A \parallel$ to $BC$.
Then this line is $\parallel$ to $DE$, by §~106. The
three parallels by hypothesis intercept
equal parts on the transversal $AB$, and therefore, by §~187,
they intercept equal parts on the transversal $AC$; that is, the
line $DE$ bisects $AC$.
\end{point}

\begin{point}%
\cor[2]{The line which joins the middle points of two
sides of a triangle is parallel to the third side, and is equal
to half the third side.}

A line drawn through $D$, the middle point of $AB$, $\parallel$ to $BC$,
passes through $E$, the middle point of $AC$, by §~188. Therefore
the line joining $D$ and $E$ coincides with this parallel and
is $\parallel$ to $BC$. Also, since $EF$ drawn $\parallel$ to $AB$ bisects $AC$, it
bisects $BC$, by §~188; that is, $BF=FC = \frac{1}{2}BC$. But $BDEF$
is a $\Par$ by §~166, and therefore $DE = BF = \frac{1}{2}BC$.
\end{point}

\figc{064bb190}{}
\begin{point}%
\cor[3]{The median of a trapezoid is parallel to the
bases, and is equal to half the sum
of the bases.}

Draw the diagonal $DB$. In the
$\triangle ADB$ join $E$, the middle point of
$AD$, to $F$, the middle point of $DB$.
Then, by §~189, $EF$ is $\parallel$ to $AB$ and $= \frac{1}{2}AB$. In the $\triangle DBC$
join $F$ to $G$, the middle point of $BC$. Then $FG$ is $\parallel$ to $DC$
and $=\frac{1}{2}DC$. $AB$ and $FG$, being $\parallel$ to $DC$, are $\parallel$ to each other.
But only one line can be drawn through $F \parallel$ to $AB$ (§~105).
Therefore $FG$ is the prolongation of $EF$. Hence, $EFG$ is
parallel to $AB$ and $DC$, and equal to $\frac{1}{2} (AB + DC)$.
\end{point}
\scanpage{065.png}%


\section{POLYGONS IN GENERAL.}

\begin{point}%
A \indexbf{polygon} is a portion of a plane bounded by straight
lines.

The bounding lines are the sides\label{polysides}, and their sum, the \textbf{perimeter}\label{perimeter2}
of the polygon. The angles included by the adjacent
sides are the \textbf{angles}\label{polyangles} of the polygon, and the vertices of these
angles are the \textbf{vertices} of the polygon\label{polyvertices}. The number of sides
of a polygon is evidently equal to the number of its angles.
\end{point}

\begin{point}%
A \textbf{diagonal}\label{diagonal2} of a polygon is a line joining the vertices
of two angles not adjacent; as, $AC$ (Fig.~1).

\figc{065ac192}{}
\end{point}

\begin{point}%
An \indexbf{equilateral polygon} is a polygon which has all its
sides equal.
\end{point}

\begin{point}%
An \indexbf{equiangular polygon} is a polygon which has all its
angles equal.
\end{point}

\begin{point}%
A \indexbf{convex polygon} is a polygon of which no side, when
produced, will enter the polygon.
\end{point}

\begin{point}%
A \indexbf{concave polygon} is a polygon of which two or more
sides, if produced, will enter the polygon.
\end{point}

\begin{point}%
Each angle of a convex polygon (Fig.~2) is called a
\emph{salient}\label{salient} angle, and is less than a straight angle.
\end{point}

\begin{point}%
The angle $EDF$ of the concave polygon (Fig.~3) is
called a \emph{re-entrant} angle, and is greater than a straight angle.

When the term polygon is used, a \emph{convex} polygon is meant.
\end{point}
\scanpage{066.png}%

\begin{point}%
Two polygons are \emph{equal} when they can be divided by
diagonals into the same number of triangles, equal each to each,
and similarly placed; for if the polygons are applied to each
other, the corresponding triangles will coincide, and hence the
polygons will coincide and be equal.
\end{point}

\begin{point}%
Two polygons are \indexemph{mutually equiangular}, if the angles
of the one are equal to the angles of the other, each to each,
when taken in the same order. Figs.~1 and 2.
\end{point}

\begin{point}%
The equal angles in mutually equiangular polygons are
called \emph{homologous} angles\label{homangles}; and the sides which are included
by homologous angles are called \emph{homologous} sides\label{homsides}.
\end{point}

\begin{point}%
Two polygons are \indexemph{mutually equilateral}, if the sides of
the one are equal to the sides of the other, each to each, when
taken in the same order. Figs.~1 and 2.

\figc{066ad203}{}
\end{point}

\begin{point}%
Two polygons may be mutually equiangular without
being mutually equilateral; as, Figs.~4 and 5.

And, \emph{except in the case of triangles}, two polygons may be
mutually equilateral without being mutually equiangular; as,
Figs.~6 and 7.

If two polygons are mutually equilateral and mutually equiangular
\emph{they are equal}, for they can be made to coincide.
\end{point}

\begin{point}%
A polygon of three sides is called a \emph{triangle}\label{triangle2}; one of
four sides, a \emph{quadrilateral}\label{quadrilateral2}; one of five sides, a \indexemph{pentagon}; one
of six sides, a \indexemph{hexagon}; one of seven sides, a \indexemph{heptagon}; one
of eight sides, an \indexemph{octagon}; one of ten sides, a \indexemph{decagon}; one of
twelve sides, a \indexemph{dodecagon}.
\end{point}
\scanpage{067.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The sum of the interior angles of a polygon is
equal to two right angles, taken as many times less two
as the figure has sides.}

\figc{067aa205}{Let the figure $ABCDEF$ be a polygon, having $n$ sides.}

\prove{$\angle A + \angle B + \angle C$, etc.\ $= (n-2)2$ rt.~$\angle_s$.}

\textbf{Proof.} From $A$ draw the diagonals $AC$, $AD$, and $AE$.

The sum of the $\angle_s$ of the $\triangle_s$ is equal to the sum of the $\angle_s$ of
the polygon.

\step{Now, there are $(n-2)$~$\triangle_s$,}{}

\step{and the sum of the $\angle_s$ of each $\triangle = 2$ rt.~$\angle_s$.}{§~129}

$\therefore$ the sum of the $\angle_s$ of the $\triangle_s$, that is, the sum of the $\angle_s$ of
the polygon is equal to $(n-2) 2$ rt.~$\triangle_s$.~\hfill\qed

\end{proof}

\pp{\cor{The sum of the angles of a quadrilateral equals
4 right angles; and if the angles are all equal, each is a right
angle. In general, each angle of an equiangular polygon of
$n$ sides is equal to $\displaystyle \frac{2(n-2)}{n}$ right angles.}}

\ex{How many diagonals can be drawn in a polygon of $n$ sides?}
\scanpage{068.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The exterior angles of a polygon, made by
  producing each of its sides in succession, are together equal to
  four right angles.}

\figc{068aa207}{Let the figure $ABCDE$ be a polygon, having its sides produced
  in succession.}

\prove[To prove ]{the sum of the ext.~$\angle_s = 4$ rt.~$\angle_s$.}

\textbf{Proof.} Denote the int.~$\angle_s$ of the polygon by $A$, $B$,
$C$, $D$, $E$, and the corresponding ext.~$\angle_s$ by $a$, $b$, $c$,
$d$, $e$.

\eq{$\angle A + \angle a$}{$= 2$ rt.~$\angle_s$,}{§~89}

\eq[and]{$\angle B + \angle b$}{$= 2$ rt.~$\angle_s$,}{}

\pnote{(being sup.-adj.~$\angle_s$).}

In like manner each pair of adj.~$\angle_s = 2$ rt.~$\angle_s$.

$\therefore$ the sum of the interior and exterior $\angle_s$ of a
polygon of $n$ sides is equal to $2n$ rt.~$\angle_s$.

%[** TN: ad hoc visual formatting]
But the sum of the interior~$\angle_s = (n-2) 2$ rt.~$\angle_s$\hfill§~205 \\
$\phantom{\text{\indent But the sum of the interior}~\angle_s} = 2 n$ rt.~$\angle_s - 4$ rt.~$\angle_s$.

\step{$\therefore$ the sum of the exterior $\angle_s = 4$ rt.~$\angle_s$.}{\qed}

\end{proof}

\ex{How many sides has a polygon if the sum of its
interior $\angle_s$ is twice the sum of its exterior $\angle_s$?  ten
times the sum of its exterior $\angle_s$?}
\scanpage{069.png}%


\section{SYMMETRY.}

\label{symmetry}
\begin{point}%
Two points are said to be \textbf{symmetrical} with respect to
a third point, called the \textbf{centre of symmetry}\label{centresym}, if this third point
bisects the straight line which joins them.

\figc{069ac208}{}

Two points are said to be \emph{symmetrical} with respect to a
straight line, called the \textbf{axis of symmetry}\label{axissym}, if this straight line
bisects at right angles the straight line which joins them.

Thus, $P$ and $P'$ are symmetrical with respect to $O$ as a centre, and $XX'$
as an axis, if $O$ bisects the line $PP'$, and if $XX'$ bisects $PP'$ at right angles.
\end{point}

\begin{point}%
A figure is symmetrical with respect to a point as a
centre of symmetry, if the point bisects every straight line
drawn through it and terminated by the boundary of the figure.
\end{point}

\begin{point}%
A figure is symmetrical with respect to a line as an
axis of symmetry if one of the parts of the figure coincides,
point for point, with the other part when it is folded over on
that line as an axis.
\end{point}

\figc{069dd211}{}
\begin{point}%
Two figures are said to be symmetrical
with respect to an axis if every point
of one has a corresponding symmetrical
point in the other.

Thus, if every point in the figure $A'B'C'$ has a
symmetrical point in $ABC$, with respect to $XX'$
as an axis, the figure $A'B'C'$ is symmetrical to
$ABC$ with respect to $XX'$ as an axis.
\end{point}
\scanpage{070.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{A quadrilateral which has two adjacent sides
  equal, and the other two sides equal, is symmetrical with respect to
  the diagonal joining the vertices of the angles formed by the equal
  sides, and the diagonals are perpendicular to each other.}

\figc{070aa212}{Let $ABCD$ be a quadrilateral, having $AB$ equal to $AD$, and
  $CB$ equal to $CD$, and having the diagonals $AC$ and $BD$.}

\prove{the diagonal $AC$ is an axis of symmetry, and that
  it is $\perp$ to the diagonal $BD$.}

\textbf{Proof.} In the $\triangle_s ABC$ and $ADC$,

\step{$AB = AD$, and $BC = DC$,}{Hyp.}

\eq[and]{$AC$}{$= AC$.}{Iden.}

\eq{$\therefore \triangle ABC$}{$= \triangle ADC$.}{§~150}

\step{$\therefore \angle BAC = \angle DAC$, and
        $\angle BCA = \angle DCA$.}{}

Hence, if $ABC$ is turned on $AC$ as an axis until it falls on $ADC$,
$AB$ will fall upon $AD$, $CB$ on $CD$, and $OB$ on $OD$.

\step{$\therefore$ the $\triangle ABC$ will coincide with the $\triangle
  ADC$.}{}

\step{$\therefore AC$ is an axis of symmetry (§~210) and is $\perp$ to
  $BD$.}{§~208}

\hfill\qed

\end{proof}
\scanpage{071.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If a figure is symmetrical with respect to two
axes perpendicular to each other, it is symmetrical with
respect to their intersection as a centre.}

\figc{071aa213}{Let the figure $ABCDEFGH$ be symmetrical with respect to the two
perpendicular axes $XX'$, $YY'$, which intersect at $O$.}

\prove{$O$ is the centre of symmetry of the figure.}

\textbf{Proof.}  Let $N$ be any point in the perimeter.

\step{Suppose $NMI$ drawn $\perp$ to $YY'$, $IKL \perp$ to $XX'$.}{}

\step[\indent Then]{$NI$ is $\parallel$ to $XX'$ and $IL$ is $\parallel$ to $YY'$.}{§~104}

\step{Draw $LO$, $ON$, and $KM$.}{}

\eq[\indent Now]{$KI$}{$= KL$,}{§~208}

\pnote{(the figure being symmetrical with respect to $XX'$).}

\eq[\indent But]{$KI$}{$=OM$.}{§~180}

\step{$\therefore KL=OM$, and $KLOM$ is a $\Par$.}{§~183}

\step{$\therefore LO$ is equal and parallel to $KM$.}{§~183}

\step{In like manner $ON$ is equal and parallel to $KM$.}{}

\step{$\therefore LON$ is a straight line.}{§~105}

$\therefore O$ bisects $LN$, \emph{any} straight line and therefore \emph{every} straight
line drawn through $O$ and terminated by the perimeter.

\step{$\therefore O$ is the centre of symmetry of the figure.}{\qed}

\end{proof}
\scanpage{072.png}%

\subsection{REVIEW QUESTIONS ON BOOK I.}

\begin{myenum}

\item What is the subject-matter of Geometry?

\item What is a geometric magnitude?

\item What is an axiom? a theorem? a converse theorem? an opposite
theorem? a contradictory theorem?

\item Define a straight line; a curved line; a broken line; a plane surface;
a curved surface.

\item How many points are necessary to determine a straight line?

\item How many straight lines are necessary to determine a point?

\item On what does the magnitude of an angle depend?

\item Define a straight angle; a right angle; an oblique angle.

\item Define adjacent angles; complementary angles; supplementary
angles; conjugate angles.

\item Define parallel lines and give the axiom of parallels.

\item If two lines in the same plane are parallel and cut by a transversal,
what pairs of angles are equal?  what pairs are supplementary?

\item Define a right triangle; an isosceles triangle; a scalene triangle.

\item To how many right angles is the sum of the angles of a triangle
equal? the sum of the acute angles of a right triangle?

\item To what angles is the exterior angle of a triangle equal?

\item What is the test of equality of two geometric magnitudes?

\item How does a reciprocal theorem differ from a converse theorem?

\item State the three cases in which two triangles are equal.

\item State the cases in which two right triangles are equal.

\item What is meant by a locus of points?

\item Where are the points located in a plane that are each equidistant
from two given points? from two intersecting lines?

\item Define a parallelogram; a trapezoid; an isosceles trapezoid.

\item When is a figure symmetrical with respect to a centre?

\item When is a figure symmetrical with respect to an axis?

\item Must a triangle be equiangular if equilateral? must a triangle be
equilateral if equiangular?

\item When are two polygons said to be mutually equiangular?

\item When are two polygons said to be mutually equilateral?

\item Can two polygons of more than three sides be mutually equiangular
without being mutually equilateral? mutually equilateral without being
mutually equiangular?

\item What line do two points each equidistant from the extremities of
a given straight line determine?

\end{myenum}
\scanpage{073.png}%

\subsection{METHODS OF PROVING THEOREMS.}

\begin{point}%
There are \emph{three} general methods of proving theorems,
the \textbf{synthetic}, the \textbf{analytic}, and the \textbf{indirect} methods.

The \emph{synthetic} method is the method employed in most of
the theorems already given, and consists in putting together
known truths in order to obtain a new truth.

The \emph{analytic} method is the reverse of the synthetic method.
It asserts that the conclusion is true if another proposition is
true, and so on step by step, until a known truth is reached.
Thus, proposition $A$ is true if proposition $B$ is true, and $B$ is
true if $C$ is true; but $C$ \emph{is} true, hence $A$ and $B$ are true.

If a known truth \emph{suggests} the required proof, it is best to
use the synthetic form at once. If no proof occurs to the
mind, it is necessary to use the analytic method to \emph{discover}
the proof, and then the synthetic proof may be given.

The \emph{indirect} method, or the method of \emph{reductio ad absurdum},
is illustrated on page \pageref{41}. It consists in proving a theorem to
be true by proving its contradictory to be false.
\end{point}

\begin{point}%
Generally auxiliary lines are required, as a line \emph{connecting
two points}; a line \emph{parallel to or perpendicular to a given
line}; a line \emph{produced by its own length}; a line \emph{making with
another line an angle equal to a given angle.}

\textbf{Two lines are proved equal} by proving them \emph{homologous sides
of equal triangles}; or \emph{legs of an isosceles triangle}; or \emph{opposite
sides of a parallelogram.}

\textbf{Two angles are proved equal} by proving them \emph{alternate-interior
angles or exterior-interior angles of parallel lines}; or \emph{homologous
angles of equal triangles}; or \emph{base angles of an isosceles triangle};
or \emph{opposite angles of a parallelogram.}

Two suggestions are of special importance to the beginner:
\begin{myenum}
\item \emph{Draw as accurate figures as possible.}
\item \emph{Draw as general figures as possible.}
\end{myenum}
\end{point}
\scanpage{074.png}%

\section{EXERCISES.}

\exheader{Prove by the analytic method:}

\figc{074aaZ19}{}
\begin{proofex}%
\obs{A median of a triangle is less than half the sum of the two adjacent
sides.}

\prove[To prove ]{the median $AD < \frac{1}{2}(AB + AC)$.}

\eq[\indent Now]{}{$AD < \frac{1}{2}(AB + AC)$,}{}

\eq[if]{}{$2AD < AB + AC$.}{}

This suggests producing $AD$ by its own length to $E$,
and joining $BE$.

\eq[\indent Then]{}{$AE=2AD$,}{}

\step[and]{$2AD<AB+AC$ if $AE<AB+AC$.}{}

\step[\indent But]{$AE<AB+BE$.}{§~138}

\step{$\therefore AE<AB+AC$ if $AC = BE$.}{}

\step[\indent And]{$AC=BE$ if $\triangle ACD = \triangle EBD$.}{§~128}

\eq[\indent But]{$\triangle ACD$}{$=\triangle EBD$.}{§~143}

\eq[\indent For]{$CD$}{$=DB$,}{Hyp.}

\eq{$AD$}{$=DE$,}{Const.}

\eq[and]{$\angle ADC$}{$=\angle BDE$.}{§~93}

\step{$\therefore AE < AB + AC$.}{}

\step{$\therefore AD < \frac{1}{2}(AB+AC)$.}{}


\end{proofex}


\figc{074bbZ20}{}
\begin{proofex}%
\obs{A straight line which bisects two sides of a triangle is parallel
to the third side.}

If $AD = DB$ and $AE = EC$, to prove $DE\parallel$ to $BC$.

Draw $CG\parallel$ to $BA$, and produce $DE$ to meet it at $G$.

\step{$DE$ is $\parallel$ to $BC$ if $BCGD$ is a $\Par$.}{§~166}

\step{$BCGD$ is a $\Par$ if $CG=BD$.}{§~183}

\step{$CG=BD$ if each is equal to $AD$.}{Ax.~1}

\eq[\indent Now]{$BD$}{$=AD$.}{Hyp.}

\eq[\indent And]{$CG$}{$=AD$ if $\triangle CGE = \triangle ADE$.}{§~128}

\eq[\indent But]{$\triangle CGE$}{$=\triangle ADE$.}{§~139}

\eq[\indent For]{$EC$}{$=AE$.}{Hyp.}

\eq{$\angle GEC$}{$=\angle AED$.}{§~93}

\eq{$\angle ECG$}{$=\angle DAE$.}{§~110}

\step{$\therefore DE$ is $\parallel$ to $BC$.}{}

\end{proofex}
\scanpage{075.png}%

%\pagebreak
Prove by the synthetic method:

\begin{proofex}%
\obs{The middle point of the hypotenuse of a
right triangle is equidistant from the three vertices.}

From $D$, the middle point, draw $DE \perp$ to $CB$.

$DE$ is $\parallel$ to $AC$ (why?), and $DE$ bisects $CB$ (why?).

$\therefore D$ is equidistant from $B$, $A$, and $C$.    (Why?)

\end{proofex}

\figcc{075aaZ21}{075bbZ22}
\begin{proofex}%
\obs{If one acute angle of a right triangle is double the other, the
hypotenuse is double the shorter leg.}

The median $CD = BD = AD$ (Ex.~21).

Then $\angle b = \angle a$; and $\angle c = \angle 2a$.    (Why?)

Now $a + 2 a = 90°$.    (Why?)

$\therefore \angle a = 30°$; $\angle 2a = 60°$; $\angle c = 60°$.

$\therefore \triangle ACD$ is equilateral (why?), and $AD$, half of
$AB = AC$.  $\therefore AB = 2AC$.

\end{proofex}

\begin{proofex}%
\obs{If two triangles have two sides of the one equal, respectively, to
two sides of the other, and the angles opposite two equal sides equal, the
angles opposite the other two equal sides are equal or supplementary, and
if equal the triangles are equal.}

Let $AC = A'C'$, $BC = B'C'$, and $\angle B = \angle B'$.

Place $\triangle A'B'C'$ on $\triangle ABC$ so that $B'C'$ shall coincide with $BC$, and
$\angle A'$ and $\angle A$ shall be on the same side of $BC$.

\figc{075ceZ23}{}

Since $\angle B'= \angle B$, $B'A'$ will fall along $BA$, and $A'$ will fall at $A$ or at
some other point in $BA$, as $D$. If $A'$ falls at $A$, the $\triangle_s A'B'C'$ and $ABC$
coincide and are equal.

If $A'$ falls at $D$, the $\triangle_s A'B'C'$ and $DBC$ coincide and are equal.

Since $CD = C'A'= CA$, $\angle A = \angle CDA$.    (Why?)

But $\angle_s CDA$ and $CDB$ are supplements.    (Why?)

$\therefore \angle_s A$ and $CDB$ are supplements.    (Why?)

Draw figures and show that the triangles are equal:

1. If the given angles $B$ and $B'$ are both right or both obtuse angles.

2. If the required angles $A$ and $A'$ are both acute, both right, or both
obtuse.

3. If $AC$ and $A'C'$ are not less than $BC$ and $B'C'$, respectively.
\end{proofex}
\scanpage{076.png}%

\filbreak
\figcc{076aaZ24}{076bbZ25}
\begin{proofex}%
\obs{The bisectors of the angles of a triangle meet
  in a point which is equidistant from the sides of the triangle.}

Let the bisectors $AD$ and $BE$ intersect at $O$.  Then $O$ being in
$AD$ is equidistant from $AC$ and $AB$. (Why?)  And $O$ being in $BE$
is equidistant from $BC$ and $AB$.  Hence, $O$ is equidistant from
$AC$ and $BC$, and therefore in the bisector $CF$. (Why?)

\end{proofex}

\begin{proofex}%
\obs{The perpendicular bisectors of the sides of a
  triangle meet in a point which is equidistant from the vertices of
  the triangle.}

Let the $\perp$ bisectors $EE'$ and $DD'$ intersect at $O$.  Then $O$
being in $EE'$ is equidistant from $A$ and $C$. (Why?)  And $O$ being
in $DD'$ is equidistant from $A$ and $B$.  Hence, $O$ is equidistant
from $B$ and $C$, and therefore is in the $\perp$ bisector
$FF'$. (Why?)

\end{proofex}

%\pagebreak
\begin{proofex}%
\obs{The perpendiculars from the vertices of a
  triangle to the opposite sides meet in a point.}

Let the $\perp_s$ be $AH$, $BP$, and $CK$.  Through $A$, $B$, $C$
suppose $B'C'$, $A'C'$, $A'B'$, drawn $\parallel$ to $BC$, $AC$, $AB$,
respectively.  Then $AH$ is $\perp$ to $B'C'$.  (Why?)  Now $ABCB'$
and $ACBC'$ are $\Par_s$ (why?) and $AB' = BC$, and $AC' = BC$. (Why?)
That is, $A$ is the middle point of $B'C'$.  In the same way, $B$ and
$C$ are the middle points of $A'C'$ and $A'B'$, respectively.
Therefore, $AH$, $BP$, and $CK$ are the $\perp$ bisectors of the sides
of the $\triangle A'B'C'$.  Hence, they meet in a point. (Why?)

\end{proofex}

\figcc{076ccZ26}{076ddZ27}
\begin{proofex}%
\obs{The medians of a triangle meet in a point which
  is two thirds of the distance from each vertex to the middle of the
  opposite side.}

Let the two medians $AD$ and $CE$ meet in $O$.  Take $F$ the middle
point of $OA$, and $G$ of $OC$.  Join $GF$, $FE$, $ED$, and $DG$.  In
$\triangle AOC$, $GF$ is $\parallel$ to $AC$ and equal to
$\frac{1}{2}AC$. (Why?)  $DE$ is $\parallel$ to $AC$ and equal to
$\frac{1}{2}AC$. (Why?)  Hence, $DGFE$ is a $\Par$.  (Why?)  Hence,
$AF = FO = OD$, and $CG = GO = OE$.  (Why?)  Hence, \emph{any median}
cuts off \emph{on any other median} two thirds of the distance from
the vertex to the middle of the opposite side.  Therefore, the median
from $B$ will cut off $AO$, two thirds of $AD$; that is, will pass
through $O$.

\end{proofex}

\note{If \emph{three} or more lines pass through the same
point, they are called \emph{concurrent} lines\label{concurrent}.}
\scanpage{077.png}%

\filbreak
\ex{If an angle is bisected, and if a line is drawn through the vertex
perpendicular to the bisector, this line forms equal angles with the sides
of the given angle.}

\figc{077adZ28}{}

\ex{The bisectors of two supplementary adjacent angles are perpendicular
to each other.}

\ex{If the bisectors of two adjacent angles are perpendicular to
each other, the adjacent angles are supplementary.}

\ex{The bisector of one of two vertical angles bisects the other.}

\ex{The bisectors of two vertical angles form one line.}

\ex{The bisectors of the two pairs of vertical angles formed by two
intersecting lines are perpendicular to each other.}

\filbreak
\begin{proofex}%
The bisector of the vertical angle of an isosceles
triangle bisects the base, and is perpendicular to the base.

\step{$\triangle ADC = \triangle BDC$ (§~143)}{}

\end{proofex}

\figcc{077eeZ34}{077ffZ35}

\ex{The perpendicular bisector of the base of an
isosceles triangle passes through the vertex and bisects
the angle at the vertex (§~160).}

\ex{If the perpendicular bisector of the base of a
triangle passes through the vertex, the triangle is isosceles.}

\ex{Any point in the bisector of the vertical angle of an isosceles
triangle is equidistant from the extremities of the base (Ex.~34, §~160).}

\ex{If the bisector of an angle of a triangle is perpendicular to the
opposite side, the triangle is isosceles.}

\ex{If two isosceles triangles are on the same base, a straight line
passing through their vertices is perpendicular to the base, and bisects
the base (§~161).}
\scanpage{078.png}%

\ex{Two isosceles triangles are equal when a side and an angle of
the one are equal, respectively, to the homologous side and angle of the
other.}

\figc{078aaZ41}{}
\ex{The bisector of an exterior angle of an isosceles
\phantomsection\label{page:69}% [** TN: Ref. to Exercise 41, p. 69]
triangle, formed by producing one of the legs through the
vertex, is parallel to the base. Why does $\angle DAC = \angle B +
\angle C$? Why is $\angle DAE = \angle ABC$? Why is $AE \parallel$ to $BC$?}

\ex{If the bisector of an exterior angle of a triangle is parallel to
one side, the triangle is isosceles.}

\figc{078bbZ43}{}
\begin{proofex}%
If one of the legs of an isosceles triangle is produced
through the vertex by its own length, the line joining
the end of the leg produced to the nearer end of the base is
perpendicular to the base.

\step{$\angle CBA = \angle A$, and $\angle CBD = \angle D$. (Why?)}{}

\step{$\therefore \angle ABD = \angle A + \angle D$.}{}

\end{proofex}

\ex{A line drawn from the vertex of the right angle of a right triangle
to the middle point of the hypotenuse divides the triangle into two
isosceles triangles.}

\ex{If the equal sides of an isosceles triangle are produced through
the vertex so that the external segments are equal, the extremities of
these segments will be equally distant from the extremities of the base,
respectively.}

\figc{078ccZ46}{}
\ex{If through any point in the bisector of an
angle a line is drawn parallel to either of the sides of
the angle, the triangle thus formed is isosceles.}

\ex{Through any point $C$ in the line $AB$ an intersecting line is
drawn, and from any two points in this line equidistant from $C$ perpendiculars
are dropped on $AB$ or $AB$ produced.  Prove that these perpendiculars
are equal.}

\ex{If the median drawn from the vertex of a triangle to the base
is equal to half the base, the vertical angle is a right angle.}

\figc{078ddZ49}{}
\ex{The lines joining the middle points of the sides of
a triangle divide the triangle into four equal triangles.}
\scanpage{079.png}%

\begin{proofex}%
The altitudes upon the legs of an isosceles triangle are equal.

\step{Rt.~$\triangle BEC =$ rt.~$\triangle CDB$ (§~141).}{}

\end{proofex}

\begin{proofex}%
If the altitudes upon two sides of a triangle are equal, the triangle
is isosceles.

\step{Rt.~$\triangle BEC =$ rt.~$\triangle CDB$ (§~151).}{}
\end{proofex}

\figc{079adZ51}{}

\begin{proofex}%
The medians drawn to the legs of an isosceles triangle are equal.

\step{$\triangle BEC = \triangle CDB$ (§~143).}{}

\end{proofex}

\begin{proofex}%
If the medians to two sides of a triangle are equal, the triangle
is isosceles.

\step{$BO = CO$, and $OE = OD$ (Ex.~27).}{}

\step{$\angle BOE = \angle COD$. \quad $\therefore \triangle BOE = \triangle COD$ (§~143).}{}

\end{proofex}

\begin{proofex}%
The bisectors of the base angles of an isosceles triangle are
equal.

\step{$\triangle BEC = \triangle CDB$ (§~139).}{}

\end{proofex}

\begin{proofex}%
\textsc{Opposite Theorem.} If a triangle is not isosceles, the bisectors
of the base angles are not equal.

Let $\angle ABC$ be greater than $\angle ACB$; then $KC > KB$. (Why?)

Now $CD > BE$, if $KD$ is greater than or equal to $KE$.

But suppose $KD < KE$. Lay off $KH = KD$ and $KG = KB$, join $HG$,
and draw $GF \parallel$ to $BE$.

$\triangle KDB = \triangle KHG$. (Why?) $\therefore \angle KHG = \angle KDB$. (Why?)

$\therefore \angle KEC$ is greater than $\angle KHG$. (Why?) $\therefore GF > HE$. (Why?)

$\angle GFC$ is greater than $\angle FCG$ ($\frac{1}{2}ACB$). $\therefore CG > GF$, and $> HE$.

$\therefore KC - KG > KE - KH$, or $KC + KD > KB + KE$, or $CD > BE$.

\end{proofex}

\ex{State the converse theorem of Ex.~54. Is the converse theorem
true?}

\figc{079eeZ57}{}
\begin{proofex}%
The perpendiculars dropped from the middle
point of the base upon the legs of an isosceles triangle are
equal.

\step{$\triangle BED = \triangle CFD$ (§~141).}{}

\end{proofex}

\begin{proofex}%
State and prove the converse.

\step{$\triangle BED = \triangle CFD$ (§~151).}{}

\end{proofex}
\scanpage{080.png}%

\filbreak
\begin{proofex}%
The difference of the distances from any
point in the base produced of an isosceles triangle to
the equal sides of the triangle is constant.

Rt. $\triangle DGC=$ rt.~$\triangle DFC$. (Why?) $\therefore DF = DG$.

$\therefore DE - DF = DE - DG = EG$, the $\perp$ distance
between the two $\parallel_s$, $BA$ and $CH$.

\end{proofex}

\figcc{080aaZ59}{080bbZ60}

\begin{proofex}%
The sum of the perpendiculars dropped from any point in the
base of an isosceles triangle to the legs is constant, and equal to the altitude
upon one of the legs.

Let $PE$ and $PD$ be the $\perp_s$ and $BF$ the altitude.

Draw $PG \perp$ to $BF$.

$EPGF$ is a parallelogram. (Why?) $\therefore GF = PE$.
It remains to prove $GB = PD$.

The rt.~$\triangle PGB =$ the rt.~$\triangle BDP$. (Why?)
\end{proofex}

%\pagebreak
\begin{proofex}%
The sum of the perpendiculars dropped from any point within
an equilateral triangle to the three sides is constant,
and equal to the altitude.

$AD$ is the altitude, $PE$, $PG$, and $PF$ the three perpendiculars.
Through $P$ draw $HK \parallel$ to $BC$, meeting
$AD$ at $M$.

\eq[\indent Then]{$MD$}{$= PE$. (Why?)}{}

\eq{$PG + PF$}{$= AM$ (Ex.~60).}{}


\end{proofex}

\figcc{080ccZ61}{080ddZ62}

\ex{$ABC$ and $ABD$ are two triangles on the same base
$AB$, and on the same side of it, the vertex of each triangle
being without the other. If $AC$ equals $AD$, show that $BC$
cannot equal $BD$ (§~154).}

\ex{The sum of the lines which join a point
within a triangle to the three vertices is less than the
perimeter, but greater than half the perimeter.}

\figcc{080eeZ63}{080ffZ64}
\ex{If from any point in the base of an isosceles triangle
parallels to the legs are drawn, a parallelogram is
formed whose perimeter is constant, being equal to the sum
of the legs of the triangle.}
\scanpage{081.png}%

%\pagebreak
\ex{The bisector of the vertical angle $A$ of a triangle
$ABC$, and the bisectors of the exterior angles at
the base formed by producing the sides $AB$ and $AC$,
meet in a point which is equidistant from the base and
the sides produced (§~162).}

\figcc{081aaZ65}{081bbZ66}

\begin{proofex}%
If the bisectors of the base angles of a triangle
are drawn, and through their point of intersection a line
is drawn parallel to the base, the length of this parallel
between the sides is equal to the sum of the segments of
the sides between the parallel and the base.

\step{$\angle EOB = \angle OBC = \angle OBE$. \quad $\therefore BE = EO$.}{}
\end{proofex}

\begin{proofex}%
The bisector of the vertical angle of a triangle makes with the
perpendicular from the vertex to the base an angle equal
to half the difference of the base angles.

Let $\angle B$ be greater than $\angle A$.

\eq{$\angle DCE$}{$= 90° - \angle A - \angle ACD$.}{}

\eq{$\angle ACD$}{$= 90° - \frac{1}{2}\angle A - \frac{1}{2}\angle B$.}{}

\step{$\therefore \angle DCE = 90° - \angle A -
    (90° - \frac{1}{2}\angle A - \frac{1}{2}\angle B) =
  \frac{1}{2}\angle B - \frac{1}{2}\angle A$.}{}

\end{proofex}

\figcc{081ccZ67}{081ddZ68}
\ex{If the diagonals of a quadrilateral bisect each
other, the figure is a parallelogram.

Prove $\triangle AOB = \triangle COD$.}

\figc{081eeZ69}{}
\ex{The diagonals of a rectangle are equal.

Prove $\triangle ABC = \triangle BAD$.}

\ex{If the diagonals  of a parallelogram are
equal, the figure is a rectangle.}

\ex{The diagonals of a rhombus are perpendicular to each other,
and bisect the angles of the rhombus.}

\ex{The diagonals of a square are perpendicular to each other, and
bisect the angles of the square.}

\figc{081ffZ73}{}
\begin{proofex}%
Lines from two opposite vertices of a parallelogram to the
middle points of the opposite sides trisect the diagonal.

\step{$EBFD$ is a $\Par$ (why?), and $DF$ is $\parallel$ to $EB$.}{}

\step{$AM = MN$, and $MN = CN$ (§~188).}{}
\end{proofex}
\scanpage{082.png}%

\begin{proofex}%
The lines joining the middle points of the sides of any quadrilateral,
taken in order, enclose a parallelogram.

Prove $HG$ and $EF \parallel$ to $AC$; and $FG$ and $EH \parallel$ to $BD$ (§~189).

Then $HG$ and $EF$ are each equal to $\frac{1}{2}AC$.
\end{proofex}

\figc{082adZ74}{}

\ex{The lines joining the middle points of the sides of a rhombus,
taken in order, enclose a rectangle. (Proof similar to that of Ex.~74.)}

\ex{The lines joining the middle points of the sides of a rectangle
(not a square), taken in order, enclose a rhombus.}

\ex{The lines joining the middle points of the sides of a square,
taken in order, enclose a square.}

\begin{proofex}%
The lines joining the middle points of the sides of an isosceles
trapezoid, taken in order, enclose a rhombus or a square.

$SHR$ and $QFP$ drawn $\perp$ to $AB$ are parallel. $\therefore PQSR$ is a $\Par$, and by
Const.~is a rectangle or a square.

$\therefore EFGH$ is a rhombus or a square (Exs.~76, 77).

\figc{082ehZ78}{}


\end{proofex}

\ex{The bisectors of the angles of a rhomboid enclose a rectangle.}

\ex{The bisectors of the angles of a rectangle enclose a square.}

\ex{If two parallel lines are cut by a transversal, the bisectors of
the interior angles form a rectangle.}

\filbreak
\figc{082iiZ82}{}
\begin{proofex}%
The median of a trapezoid passes through the
middle points of the two diagonals.

The median $EF$ is $\parallel$ to $AB$ and bisects $AD$ (§~190).

$\therefore$ it bisects $DB$.

Likewise $EF$ bisects $BC$ and $BD$.

\end{proofex}
\scanpage{083.png}%

\begin{proofex}%
The lines joining the middle points of the diagonals of a trapezoid
is equal to half the difference of the bases.

\step{$\triangle BFG = \triangle DFC$. (Why?)
  $\therefore EF = \frac{1}{2}AG$ (§~180).}{}

\step{$CF=FG$, $DC=BG$.}{}

\step{$\therefore AG=AB-DC$. $\therefore EF=\frac{1}{2}(AB-DC)$}{}

\figc{083adZ83}{}

\end{proofex}

\begin{proofex}%
In an isosceles trapezoid each base makes equal angles with the legs.

Draw $CE \parallel$ to $DB$.  $CE=DB$. (Why?)  $\angle A = \angle CEA$,
$\angle B = \angle CEA$, $\angle_s C$ and $D$ have equal supplements.

\end{proofex}

\ex{If the angles at the base of a trapezoid are equal, the other angles are equal, and the trapezoid is isosceles.}


\begin{proofex}%
In an isosceles trapezoid the opposite angles are supplementary:

\step{$\angle C = \angle D$ (Ex.~84)}{}
\end{proofex}


\begin{proofex}%
The diagonals on an isosceles trapezoidal are equal.

Prove $\triangle ACD = \triangle BDC$.
\end{proofex}

%\pagebreak
\begin{proofex}%
If the diagonals of a trapezoid are equal, the trapezoid is isosceles.

Draw $CE$ and $DF \perp$ to $AB$.

\eq{$\triangle ADF$}{$= \triangle BCE$.}{(Why?)}

\eq{$\therefore \angle ADF$}{$= \angle CBA$.}{}

\eq{$\triangle ABC$}{$= \triangle BAD$.}{}

\end{proofex}

\figcc{083eeZ88}{083ffZ89}

\begin{proofex}%
If from the diagonal $DB$, of a square $ABCD$, $BE$
is cut off equal to $BC$, and $EF$ is drawn perpendicular
to $BD$ meeting $DC$ at $F$, then $DE$ is equal to $EF$ and
also to $FC$.

$\angle EDF = 45°$, and $\angle DFE = 45°$; and $DE=DF$.
Rt.~$\triangle BEF =$ rt.~$\triangle BCF$ (§~151); and $EF=FC$.

\end{proofex}

\ex{Two angles whose sides are so perpendicular, each to each, are either equal or supplementary.}
\scanpage{084.png}%


\chapter{BOOK II\@. THE CIRCLE.}

\section{DEFINITIONS.}

\begin{point}%
A \textbf{circle}\label{circle} is a portion of a plane bounded by a curved
line, all points of which are equally distant from a point within
called the \textbf{centre}\label{centrecirc}. The bounding line is called the \textbf{circumference}\label{circumference}
of the circle.
\end{point}

\begin{point}%
A \textbf{radius} is a straight line from the centre to the circumference;
and a \textbf{diameter}\label{diameter} is a straight line through the
centre, with its ends in the circumference.

By the definition of a circle, \emph{all its radii are equal}. All its
diameters are equal, since a diameter is equal to two radii.
\end{point}

\begin{point}%
\textbf{Postulate.} A circumference can be described from any
point as a centre, with any given radius.
\end{point}

\begin{point}%
A \indexbf{secant} is a straight line of unlimited length which
intersects the circumference in two points; as, $AD$ (Fig.~1).
\end{point}

\figc{084aa220}{}

\begin{point}%
A \indexbf{tangent} is a straight line of unlimited length which
has one point, and only one, in common
with the circumference; as, $BC$ (Fig.~1).
In this case the circle is said to be tangent
to the straight line. The common
point is called the \indexbf{point of contact}, or
\indexbf{point of tangency}.
\end{point}

\begin{point}%
Two \emph{circles} are tangent to each
other, if both are tangent to a straight line at the same point;
and are said to be tangent \emph{internally} or \emph{externally}, according
as one circle lies wholly \emph{within} or \emph{without} the other.
\end{point}
\scanpage{085.png}%

\pp{An \textbf{arc}\label{arc} is any part of the circumference; as, $BC$ (Fig.~3).
Half a circumference is called a \indexbf{semicircumference}. Two arcs
are called \textbf{conjugate arcs}, if their sum is a circumference.}

\pp{A \textbf{chord}\label{chord} is a straight line that has its extremities in
the circumference; as, the straight line $BC$ (Fig.~3).}

\pp{A chord subtends two conjugate arcs. If the arcs are
unequal, the less is called the \indexbf{minor} arc, and the greater the
\indexbf{major} arc. A minor arc is generally called simply an arc.}

\figc{085ac224}{}

\pp{A \indexbf{segment} of a circle is a portion of the circle bounded
by an arc and its chord (Fig.~2).}

\pp{A \indexbf{semicircle} is a segment equal to half the circle (Fig.~2).}

\pp{A \indexbf{sector} of a circle is a portion of the circle bounded
by two radii and the arc which they intercept. The angle
included by the radii is called the \emph{angle of the sector} (Fig.~2).}

\pp{A \indexbf{quadrant} is a sector equal to a quarter of the circle
(Fig.~2).}

\pp{An angle is called a \textbf{central angle}\label{central}, if its vertex is at the
centre and its sides are radii of the circle; as, $\angle AOD$ (Fig.~2).}

\begin{point}%
An angle is called an \textbf{inscribed angle}\label{inscribedcirc}, if its vertex is in
the circumference and its sides are chords; as, $\angle ABC$ (Fig.~3).

An angle is \emph{inscribed in a segment}\label{inscribedseg}, if its vertex is in the
arc of the segment and its sides pass through the extremities
of the arc.
\end{point}
\scanpage{086.png}%

\pp{A polygon is \emph{inscribed in a circle}\label{polyinscribed}, if its sides are chords;
and a circle is \emph{circumscribed about a polygon}\label{circcircumscribed}, if all the vertices
of the polygon are in the circumference (Fig.~3).}

\pp{A circle is \emph{inscribed in a polygon}\label{circinscribed}, if the sides of the
polygon are tangent to the circle; and a polygon is \emph{circumscribed
about}\label{polycircumscribed} a circle if its sides are tangents (Fig.~4).}

\begin{point}%
\emph{Two circles are equal, if they have equal radii.}

For they will coincide, if their centres are made to coincide.

\textsc{Conversely:} \emph{Two equal circles have equal radii.}
\end{point}

\pp{\emph{Two circles are concentric}\label{concentric}, if they have the same centre.}

\filbreak
\section{ARCS, CHORDS, AND TANGENTS.}

\proposition{Theorem.}

\begin{proof}%
\obs{A straight line cannot meet the circumference of
a circle in more than two points}.

\figc{086aa235}{Let $HK$ be any line meeting the circumference $HKM$ in $H$ and $K$.}

\prove{$HK$ cannot meet the circumference in any
other point}.

\textbf{Proof.} If possible, let $HK$ meet the circumference in $P$.

\step{Then the radii $OH$, $OP$, and $OK$ are equal.}{§~217}

\step{$\therefore P$ does not lie in the straight line $HK$.}{§~102}

\step{$\therefore HK$ meets the circumference in only two points.}{\llap{\qed}}

\end{proof}
\scanpage{087.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{In the same circle or in equal circles, equal
central angles intercept equal arcs; and of two unequal
central angles the greater intercepts the greater arc.}

\figc{087ab236}{In the equal circles whose centres are $O$ and $O'$, let the angles
$AOB$ and $A'O'B'$ be equal, and angle $AOC$ be greater than angle  $A'O'C'$.}

\prove{1. $\arc AB = \arc A'B'$;}

\prove[\phantom{To prove that~}]{2. $\arc AC > \arc A'B'$.}

\textbf{Proof. 1.} Place the $\odot A'B'P'$ on the $\odot ABP$ so that the
$\angle A'O'B'$ shall coincide with its equal, the $\angle AOB$.

\step{Then $A'$ falls on $A$, and $B'$ on $B$.}{§~233}

\step{$\therefore$ $\arc A'B'$ coincides with $\arc AB$.}{§~216}

\textbf{2.~} Since the $\angle AOC$ is greater than the $\angle A'O'B'$,
it is greater than the $\angle AOB$, the equal of the $\angle A'O'B'$.

\step{Therefore, $OC$ falls without the $\angle AOB$.}{}

\step{$\therefore$ $\arc AC > \arc AB$.}{Ax.~8}

\step{$\therefore$ $\arc AC > \arc A'B'$, the equal of $\arc AB$.}{\qed}

\end{proof}

\begin{proof}%
\obs{\textsc{Conversely:} In the same circle or in equal circles,
equal arcs subtend equal central angles; and of two unequal
arcs the greater subtends the greater central angle.}
\scanpage{088.png}%

\proveq{\textup{1.} $\angle AOB$}{$=\angle A'O'B'$;}

\proveq[\indent]{\settowidth{\TmpLen}{\textit{To prove that}}\rule{\TmpLen}{0pt}\textup{2.} $\angle AOC$}{is greater than $\angle A'O'B'$.}

\textbf{Proof. 1.} Place the $\odot A'B'P'$ on the $\odot ABP$ so that $O'A$
shall fall on its equal $OA$, and the arc $A'B'$ on its equal $AB$.

\step{Then $O'B'$ will coincide with $OB$.}{§~47}

\step{$\therefore \angle A'O'B'=\angle AOB$.}{§~60}

\textbf{2.~}Since $\arc AC>A'B'$, it is greater than $\arc AB$, the equal
of $A'B'$, and $OB$ will fall within the $\angle AOC$.

\eq{}{$\therefore \angle AOC$ is greater than $\angle AOB$.}{Ax.~8}

\eq{}{$\therefore \angle AOC$ is greater than $\angle A'O'B'$.}{\qed}
\end{proof}


\pp{\cor[1]{In the same circle or in equal circles, two sectors
that have equal angles are equal; two sectors that have
unequal angles are unequal, and the greater sector has the
greater angle.}}

\pp{\cor[2]{In the same circle or in equal circles, equal
sectors have equal angles; and of two unequal sectors the
greater has the greater angle.}}

\begin{point}%
\textbf{Law of Converse Theorems.}\label{converse2} It was stated in §~32 that the converse
of a theorem is not necessarily true. If, however, a theorem is in
fact a group of three theorems, and if \emph{one of the hypotheses} of the group
\emph{must} be true, and \emph{no two of the conclusions can be true at the same time},
then the converse of the theorem is \emph{necessarily} true.

Proposition II. is a group of three theorems. It asserts that the arc
$AB$ is equal to the arc $A'B'$, if the angle $AOB$ is equal to the angle
$A'O'B'$; that the arc $AB$ is greater than the arc $A'B'$, if the angle $AOB$
is greater than the angle $A'O'B'$; that the arc $AB$ is less than the arc
$A'B'$, if the angle $AOB$ is less than the angle $A'O'B'$.

One of these hypotheses must be true; for the angle $AOB$ must be
equal to, greater than, or less than, the angle $A'O'B'$.

No two of the conclusions can be true at the same time, for the arc $AB$
cannot be both equal to and greater than the arc $A'B'$; nor can it be both
equal to and less than the arc $A'B'$; nor both greater than and less than
the arc $A'B'$. In such a case, the converse theorem is \emph{necessarily} true,
and no proof like that given in the text is required to establish it.
\end{point}
\scanpage{089.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{In the same circle or in equal circles, equal arcs
  are subtended by equal chords; and of two unequal arcs the greater
  is subtended by the greater chord.}

\figc{089ab241}{In the equal circles whose centres are $O$ and $O'$, let the arcs
  $AB$ and $A'B'$ be equal, and the arc $AF$ greater than arc $A'B'$.}

\proveq{\textup{1. }chord $AB$}{$=$ chord $A'B'$;}

\proveq[]{\textup{2. }chord $AF$}{$>$ chord $A'B'$.}

\step[\indent\textbf{Proof.}]{Draw the radii $OA$, $OB$, $OF$, $O'A'$, $O'B'$.}{}

\step[\indent 1.]{The $\triangle_s AOB$ and $A'O'B'$ are equal.}{§~143}

\step{For $OA = O'A'$, and $OB = O'B'$,}{§~233}

\pnote{(radii of equal circles),}

\step{and $\angle AOB = \angle A'O'B'$,}{§~237}

\pnote{(in equal $\odot_s$ equal arcs subtend equal central $\angle_s$).}

\step{$\therefore$ chord $AB = $ chord $A'B'$.}{§~128}

\step{}{}

\step[\indent 2.]{In the $\triangle_s AOF$ and $A'O'B'$,}{}

\step{$OA = O'A'$, and $OF = O'B'$.}{§~233}

\step{But the $\angle AOF$ is greater than the $\angle A'O'B'$,}{§~237}

\pnote{(in equal $\odot_s$, the greater of two unequal arcs subtends the
  greater $\angle$).}

\step{$\therefore$ chord $AF >$ chord $A'B'$.}{§~154}

\hfill\qed

\end{proof}


\pp{\cor{In the same circle or in equal
  circles, the greater of two unequal major arcs is subtended by the
  less chord.}}
\scanpage{090.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{\textsc{Conversely:} In the same circle or in equal
circles, equal chords subtend equal arcs; and of two
unequal chords the greater subtends the greater arc.}

\figc{090ab243}{In the equal circles whose centres are $O$ and $O'$, let the chords $AB$
and $A'B'$ be equal, and the chord $AF$ greater than $A'B'$.}

\prove{\quad\upshape{1.} $\arc AB = \arc A'B'$;}

\prove[\phantom{To prove that~}]{\quad\upshape{2.} $\arc AF > \arc A'B'$.}

\step[\indent\textbf{Proof.}]{Draw the radii $OA$, $OB$, $OF$, $O'A'$, $O'B'$.}{}

\step[\indent 1.]{The $\triangle_s OAB$ and $O'A'B'$ are equal.}{§~150}

\step{For $OA = O'A'$, and $OB = O'B'$,}{§~233}

\step{and chord $AB =$ chord $A'B'$.}{Hyp.}

\step{$\therefore \angle AOB = \angle A'O'B'$.}{§~128}

\step{$\therefore \arc AB = \arc A'B'$,}{§~236}

\pnote{(in equal $\odot_s$ equal central $\angle_s$ intercept equal arcs).}


\step[\indent 2.]{In the $\triangle_s OAF and O'A'B'$,}{}

\step{$OA = O'A'$ and $OF = O'B'$.}{§~233}

\step{But chord $AF >$ chord $A'B'$.}{Hyp.}

\step{$\therefore$ the $\angle AOF$ is greater than the $\angle A'O'B'$.}{§~155}

\step{$\therefore \arc AF > \arc A'B'$,}{§~236}

\pnote{(in equal $\odot_s$ the greater central $\angle$ intercepts the greater arc).}

\hfill\qed

\end{proof}

\pp{\cor{In the same circle or in equal circles, the greater
of two unequal chords subtends the less major arc.}}
\scanpage{091.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{A diameter perpendicular to a chord bisects the
  chord and the arcs subtended by it.}

\figc{091aa245}{Let $ES$ be a diameter perpendicular to the chord $AB$ at $M$.}

\prove{$AM = BM$, $AS = BS$, and $AE = BE$.}

\textbf{Proof.} Draw $OA$ and $OB$ from $O$, the centre of the circle.

\step{The rt.~$\triangle_s OAM$ and $OBM$ are equal.}{§~151}

\eq[\indent For]{$OM$}{$= OM$,}{Iden.}

\eq[and]{$OA$}{$= OB$.}{§~217}

\step{$\therefore AM = BM$, and $\angle AOS = \angle BOS$.}{§~128}

\eq[\indent Likewise]{$\angle AOE$}{$= \angle BOE$.}{§~85}

\step{$\therefore AS = BS$, and $AE = BE$.}{§~236}

\hfill\qed

\end{proof}


\pp{\cor[1]{A diameter bisects the
  circumference and the circle.}}

\pp{\cor[2]{A diameter which bisects a chord
  is perpendicular to it.}}

\pp{\cor[3]{The perpendicular bisector of a
  chord passes through the centre of the circle, and bisects the arcs
  of the chord.}}
\scanpage{092.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{In the same circle or in equal circles, equal
  chords are equally distant from the centre. \textsc{Conversely:}
  Chords equally distant from the centre are equal.}

\figc{092aa249}{Let $AB$ and $CF$ be equal chords of the circle $ABFC$.}

\prove{$AB$ and $CF$ are equidistant from the centre $O$.}

\textbf{Proof.} Draw $OP \perp$ to $AB$, $OH \perp$ to $CF$, and join
$OA$ and $OC$.

\step{$OP$ bisects $AB$, and $OH$ bisects $CF$.}{§~245}

\step{The rt.~$\triangle_s OPA$ and $OHC$ are equal.}{§~151}

\eq{$AP$}{$= CH$,}{Ax.~7}

\eq[and]{$OA$}{$= OC$,}{§~217}

\eq[\indent Hence,]{$OP$}{$= OH$.}{§~128}

\step{$\therefore AB$ and $CF$ are equidistant from $O$.}{}

\step{}{}

\eq[\indent \textsc{Conversely:}]{\textbf{Let }$OP$}{$=OH$.}{}

\proveq[\indent To prove]{$AB$}{$= CF$.}

\textbf{Proof.} The rt.~$\triangle_s OPA$ and $OHC$ are equal.~\hfill§~151

\eq[\indent For]{$OA$}{$= OC$,}{§~217}

\eq[and]{$OP$}{$= OH$,}{Hyp.}

\eq[\indent Hence,]{$AP$}{$= CH$.}{§~128}

\eq{$\therefore AB$}{$= CF$.}{Ax.~6}

\hfill\qed

\end{proof}
\scanpage{093.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{In the same circle or in equal circles, if two chords
are unequal, they are unequally distant from the centre;
and the greater chord is at the less distance.}

\figc{093aa250}{In the circle whose centre is $O$, let the chords $AB$ and $CD$ be
unequal, and $AB$ the greater; and let $OE$ be perpendicular to $AB$ and
$OF$ perpendicular to $CD$.}

\proveq{$OE$}{$< OF$.}

\textbf{Proof.} Suppose $AG$ drawn equal to $CD$, and $OH \perp$ to $AG$.

\step{Draw $EH$.}{}

\step{$OE$ bisects $AB$, and $OH$ bisects $AG$.}{§~245}

\eq[By hypothesis,]{$AB$}{$> CD$.}{}

\step{$\therefore AB > AG$, the equal of $CD$.}{}

\eq{$\therefore AE$}{$> AH$.}{Ax.~7}

\step{$\therefore \angle AHE$ is greater than $\angle AEH$.}{§~152}

$\therefore \angle OHE$, the complement of $\angle AHE$, is less than $\angle OEH$,
the complement of $\angle AEH$.\hfill~Ax.~5

\eq{$\therefore OE$}{$< OH$.}{§~153}

\eq[\indent But]{$OH$}{$=OF$.}{§~249}

\eq{$\therefore OE$}{$< OF$.}{\qed}

\end{proof}

\ex{The perpendicular bisectors of the sides of an inscribed polygon
are concurrent (pass through the same point).}
\scanpage{094.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{\textsc{Conversely:} In the same circle or in
  equal circles, if two chords are unequally distant from the centre,
  they are unequal; and the chord at the less distance is the greater.}

\figc{094aa251}{In the circle whose centre is $O$, let $AB$ and $CD$ be
  unequally distant from $O$; and let $OE$, the perpendicular to $AB$,
  be less than $OF$, the perpendicular to $CD$.}

\proveq{$AB$}{$> CD$.}

\textbf{Proof.} Suppose $AG$ drawn equal to $CD$,
  and $OH \perp$ to $AG$.

\eq[\indent Then]{$OH$}{$= OF$}{§~249}

\eq[\indent Hence,]{$OE$}{$< OH$.}{}

\step{Draw $EH$.}{}

\step{$\angle OHE$ is less than $\angle OEH$.}{§~152}

$\therefore \angle AHE$, the complement of $\angle OHE$, is greater
than $\angle AEH$, the complement of $\angle OEH$.\hfill~Ax.~5

\eq{$\therefore AE$}{$> AH$.}{§~153}

\step[\indent But]{$AE = \frac{1}{2}AB$, and $AH = \frac{1}{2}AG$.}{§~245}

\eq{$\therefore AB$}{$> AG$.}{Ax.~6}

\eq[\indent But]{$CD$}{$= AG$.}{Const.}

\eq{$\therefore AB$}{$> CD$.}{\qed}


\end{proof}

\pp{\cor{A diameter of a circle is greater
  than any other chord.}}
\scanpage{095.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{A straight line perpendicular to a radius at its
  extremity is a tangent to the circle.}

\figc{095aa253}{Let $MB$ be perpendicular to the radius $OA$ at $A$.}

\prove{$MB$ is a tangent to the circle.}

\textbf{Proof.} From $O$ draw any other line to $MB$, as $OH$.

\eq[\indent Then]{$OH$}{$> OA$.}{§~97}

\step{$\therefore$ the point $H$ is without the circle.}{§~216}

Hence, \emph{every point}, except $A$, of the line $MB$ is without the
circle, and therefore $MB$ is a tangent to the circle at $A$.~\hfill§~220

\hfill\qed

\end{proof}

\begin{point}%
\cor[1]{A tangent to a circle is
  perpendicular to the radius drawn to the point of contact.}

For $OA$ is the shortest line from $O$ to $MB$, and is therefore
$\perp$ to $MB$ (§~98); that is, $MB$ is $\perp$ to $OA$.
\end{point}

\begin{point}%
\cor[2]{A perpendicular to a tangent at
  the point of contact passes through the centre of the circle.}

For a radius is $\perp$ to a tangent at the point of contact, and
therefore a $\perp$ erected at the point of contact coincides with
this radius and passes through the centre.
\end{point}

\pp{\cor[3]{A perpendicular from the centre
  of a circle to a tangent passes through the point of contact.}}
\scanpage{096.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Parallels intercept equal arcs on a circumference.}

\figc{096ac257}{\textnormal{\textsc{Case 1.~}} Let $AB$ \textnormal{(Fig.~1)} be a tangent at
  $F$ parallel to $CD$, a secant.}

\proveq{$\arc CF$}{$= \arc DF$.}

\step[\indent\textbf{Proof.}]{Suppose $FF'$ drawn $\perp$ to $AB$.}{}

\step{Then $FF'$ is a diameter of the circle.}{§~255}

\step{And $FF'$ is also $\perp$ to $CD$.}{§~107}

\step{$\therefore CF = DF$, and $CF' = DF'$.}{§~245}

\step{}{}

\textsc{Case 2.~}\textbf{Let $AB$ and $CD$} (Fig.~2) \textbf{be
  parallel secants.}

\proveq{$\arc AC$}{$= \arc BD$.}

\step[\indent\textbf{Proof.}]{Suppose $EF \parallel$ to $CD$ and tangent to the
circle at $M$.}{}

\eq[\indent Then]{$\arc AM$}{$= \arc BM$,}{Case~1}

\eq[and]{$\arc CM$}{$= \arc DM$.}{}

%proofrule
\eq{$\therefore \arc AC$}{$= \arc BD$.}{Ax.~3}

\step{}{}

\textsc{Case 3.~}\textbf{Let $AB$ and $CD$} (Fig.~3) \textbf{be
  parallel tangents at $E$ and $F$.}

\proveq{$\arc EGF$}{$= \arc EHF$.}

\step[\indent\textbf{Proof.}]{Suppose $GH$ drawn $\parallel$ to $AB$.}{}

\eq[\indent Then]{$\arc EG$}{$= \arc EH$,}{Case~1}

\eq[and]{$\arc GF$}{$= \arc HF$.}{}

%proofrule
\eq{$\therefore \arc EGF$}{$= \arc EHF$.}{Ax.~2}

\hfill\qed

\end{proof}
\scanpage{097.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Through three points not in a straight line one
circumference, and only one, can be drawn.}

\figc{097aa258}{Let $A$, $B$, $C$ be three points not in a straight line.}

\prove{one circumference, and only one, can be drawn
through $A$, $B$, and~$C$.}

\step[\indent\textbf{Proof.}]{Draw $AB$ and $BC$.}{}

At the middle points of $AB$ and $BC$ suppose $\perp_s$ erected.

These $\perp_s$  will intersect at some point $O$, since $AB$ and $BC$
are not in the same straight line.

The point $O$ is in the perpendicular bisector of $AB$, and is
therefore equidistant from $A$ and $B$; the point $O$ is also in
the perpendicular bisector of $BC$, and is therefore equidistant
from $B$ and $C$.~\hfill§~160

Therefore, $O$ is equidistant from $A$, $B$, and $C$; and a circumference
described from $O$ as a centre, with a radius $OA$,
will pass through the three given points.

The centre of a circumference passing through the three
points must be in both perpendiculars, and hence at their
intersection. As two straight lines can intersect in only one
point, $O$ is the centre of the only circumference that can pass
through the three given points.~\hfill\qed

\end{proof}

\pp{\cor{Two circumferences can intersect in only two
points. \textup{For, if two circumferences have three points common,
they coincide and form one circumference.}}}
\scanpage{098.png}%

\pp{\defn{A \textbf{tangent from an external point
  to a circle}\label{tangent2} is the part of the tangent between the external point
  and the point of contact.}}


\proposition{Theorem.}

\begin{proof}%
\obs{The tangents to a circle drawn from an external
  point are equal, and make equal angles with the line joining the
  point to the centre.}

\figc{098aa261}{Let $AB$ and $AC$ be tangents from $A$ to the circle whose
  centre is $O$, and let $AO$ be the line joining $A$ to the centre
  $O$.}

\prove{$AB = AC$, and $\angle BAO = \angle CAO$.}

\step[\indent\textbf{Proof.}]{Draw $OB$ and $OC$.}{}

\step{$AB$ is $\perp$ to $OB$, and $AC \perp$ to $OC$,}{§~254}

\pnote{(a tangent to a circle is $\perp$
    to the radius drawn to the point of contact).}

\step{The rt.~$\triangle_s OAB$ and $OAC$ are equal.}{§~151}

For $OA$ is common, and the radii $OB$ and $OC$ are equal.~\hfill§~217

\step{$\therefore AB=AC$, and $\angle BAO = \angle CAO$.}{§~128}

\hfill\qed

\end{proof}

\pp{\defn{The line joining the centres of two
circles is called the \indexbf{line of centres}.}}

\pp{\defn{A tangent to two circles is called a
\indexbf{common external tangent} if it does not cut the line of
centres, and a \indexbf{common internal tangent} if it cuts the line of centres.}}
\scanpage{099.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two circles intersect each other, the line of
  centres is perpendicular to their common chord at its middle point.}

\figc{099aa264}{Let $C$ and $C'$ be the centres of the two circles, $AB$ the
  common chord, and $CC'$ the line of centres.}

\prove{$CC'$ is $\perp$ to $AB$ at its middle point.}

\step[\indent\textbf{Proof.}]{Draw $CA$, $CB$, $C'A$, and $C'B$.}{}

\step{$CA = CB$, and $C'A = C'B$.}{§~217}

\step{$\therefore C$ and $C'$ are two points, each
  equidistant from $A$ and $B$.}{}

\step{$\therefore CC'$ is the perpendicular bisector of $AB$.}{§~161}

\hfill\qed

\end{proof}

\begin{proofex}%
Describe the relative position of two circles if the
line of centres:
\begin{myenum}
\item is greater than the sum of the radii;
\item is equal to the sum of the radii;
\item is less than the sum but greater than the difference of the
  radii;
\item is equal to the difference of the radii;
\item is less than the difference of the radii.
\end{myenum}

Illustrate each case by a figure.

\end{proofex}

\ex{The straight line drawn from the middle point of a
chord to the middle point of its subtended arc is perpendicular to the
chord.}

\ex{The line which passes through the middle points of
two parallel chords passes through the centre of the circle.}
\scanpage{100.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two circles are tangent to each other, the line
of centres passes through the point of contact.}

\figc{100aa265}{Let the two circles, whose centres are $C$ and $C'$, be tangent to the
straight line $AB$ at $Q$, and $CC'$ the line of centres.}

\prove{$O$ is in the straight line $CC'$.}

\textbf{Proof.}  A $\perp$ to $AB$, drawn through the point $O$, passes through
the centres $C$ and $C'$,~\hfill§~255

\pnote{(a $\perp$ to a tangent at the point of contact passes through the centre
of the circle).}

$\therefore$ the line $CC'$, having two points in common with this $\perp$
must coincide with it.~\hfill§~47

\step{$\therefore O$ is in the straight line $CC'$.}{\qed}

\end{proof}

\begin{proofex}%
Describe the relative position of two circles if they may have:
\begin{myenum}
\item two common external and two common internal tangents;
\item two common external tangents and one common internal tangent;
\item two common external tangents and no common internal tangent;
\item one common external and no common internal tangent;
\item no common tangent.
\end{myenum}

Illustrate each case by a figure.

\end{proofex}

\ex{The line drawn from the centre of a circle to the point of intersection
of the two tangents is the perpendicular bisector of the chord joining
the points of contact.}
\scanpage{101.png}%


\section{MEASUREMENT.}

\begin{point}%
To \textbf{measure} a quantity of any kind is to find \emph{the number
of times} it contains a known quantity of the \emph{same kind}, called
the \textbf{unit of measure}.

The \emph{number} which shows the number of times a quantity
contains the unit of measure is called the \indexbf{numerical measure} of
that quantity.
\end{point}

\begin{point}%
No quantity is great or small except by comparison
with another quantity of the \emph{same kind}. This comparison is
made by finding the numerical measures of the two quantities
in terms of a common unit, and then dividing one of the
measures by the other.

The quotient is called their \indexbf{ratio}. In other words the ratio
of two quantities of the same kind is the \emph{ratio} of their \emph{numerical
measures} expressed in terms of a common unit.

The ratio of $a$ to $b$ is written $a : b$, or $\dfrac{a}{b}$.
\end{point}

\begin{point}%
Two quantities that can be expressed in \emph{integers} in
terms of a common unit are said to be \textbf{commensurable}\label{commensurable}, and the
exact value of their ratio can be found. The common unit is
called their \emph{common measure}, and each quantity is called a
\emph{multiple} of this common measure.

Thus, a common measure of $2\frac{1}{2}$~feet and $3\frac{2}{3}$~feet is $\frac{1}{6}$~of a foot, which is
contained $15$~times in $2\frac{1}{2}$~feet, and $22$~times in $3\frac{2}{3}$~feet. Hence, $2\frac{1}{2}$~feet
and $3\frac{2}{3}$~feet are multiples of $\frac{1}{6}$~of a foot, since $2\frac{1}{2}$~feet may be obtained by
taking $\frac{1}{6}$~of a foot $15$~times, and $3\frac{2}{3}$~feet by taking $\frac{1}{6}$~of a foot $22$~times. The
ratio of $2\frac{1}{2}$~feet to $3\frac{2}{3}$~feet is expressed by the fraction~$\frac{15}{22}$.
\end{point}

\begin{point}%
Two quantities of the same kind that cannot \emph{both} be
expressed in \emph{integers} in terms of a common unit, are said to be
\textbf{incommensurable}, and the \emph{exact value} of their ratio cannot be
found. But by taking the unit sufficiently small, an \emph{approximate
value} can be found that shall differ from the true value
of the ratio by less than any assigned value, however small.
\scanpage{102.png}%

Thus, suppose the ratio, $\dfrac{a}{b} = \sqrt{2}$.

Now $\sqrt{2} = 1.41421356\cdots$, a value greater than $1.414213$,
but less than $1.414214$.

If, then, a \emph{millionth part} of $b$ is taken as the unit of measure,
the value of $\dfrac{a}{b}$ lies between $1.414213$ and $1.414214$, and therefore
differs from either of these values by less than $0.000001$.

By carrying the decimal further, an approximate value may
be found that will differ from the true value of the ratio by
less than \emph{a billionth, a trillionth, or any other assigned value}.

In general, if $\dfrac{a}{b} > \dfrac{m}{n}$ but $< \dfrac{m+1}{n}$, then the error in taking
either of these values for $\dfrac{a}{b}$ is less than $\dfrac{1}{n}$, the difference
between these two fractions.     But by increasing $n$ indefinitely,
$\dfrac{1}{n}$ can be decreased indefinitely, and a value of the ratio
can be found within any required degree of accuracy.
\end{point}

\pp{The ratio of two incommensurable quantities is called
an \indexbf{incommensurable ratio}; and is a \emph{fixed value} which its successive
approximate values constantly approach.}


\section[THEORY OF LIMITS.]{THE THEORY OF LIMITS.}

\begin{point}%
When a quantity is regarded as having a \emph{fixed} value
throughout the same discussion, it is called a \indexbf{constant}; but
when it is regarded, under the conditions imposed upon it, as
having \emph{different successive} values, it is called a \indexbf{variable}.

If a variable, by having different successive values, can be
made to differ from a given constant by less than any assigned
value, however small, but cannot be made absolutely equal to
the constant, that constant is called the \indexbf{limit} of the variable,
and the variable is said to \textbf{approach the constant as its limit}.
\end{point}
\scanpage{103.png}%

\figc{103aa272}{}
\begin{point}%
Suppose a point to move from $A$ toward $B$, under
the conditions that the
first second it shall
move one half the distance from $A$ to $B$, that is, to $M$; the
next second, one half the remaining distance, that is, to $M'$;
and so on indefinitely.

Then it is evident that the moving point \emph{may approach as
near to $B$ as we choose, but will never arrive at $B$}. For, however
near it may be to $B$ at any instant, the next second it
will pass over half the distance still remaining; it must,
therefore, approach nearer to $B$, since \emph{half} the distance still
remaining is \emph{some} distance, but will not reach $B$, since \emph{half}
the distance still remaining is not the \emph{whole} distance.

Hence, the distance from $A$ to the moving point is an
increasing variable, which indefinitely approaches the constant
$AB$ as its \emph{limit}; and the distance from the moving point
to $B$ is a decreasing variable, which indefinitely approaches
the \emph{constant zero} as its \emph{limit}.
\end{point}

\figc{103bb273}{}
\begin{point}%
Again, suppose a square $ABCD$ inscribed in a circle,
and $E$, $F$, $H$, $K$ the middle points of the arcs subtended by the
sides of the square. If we draw the
lines $AE$, $EB$, $BF$, etc., we shall have an
inscribed polygon of double the number
of sides of the square.

The length of the perimeter of this
polygon, represented by the dotted lines,
is greater than that of the square, since
two sides replace each side of the square
and form with it a triangle, and two
sides of a triangle are together greater than the third side;
but less than the length of the circumference, for it is made
up of straight lines, each one of which is less than the part of
the circumference between its extremities.
\end{point}
\scanpage{104.png}%

By continually doubling the number of sides of each resulting
inscribed figure, the length of the perimeter will increase
with the increase of the number of sides, but will not become
equal to the length of the circumference.

The difference between the perimeter of the inscribed polygon
and the circumference of the circle can be made less than
any assigned value, but cannot be made equal to zero.

The length of the circumference is, therefore, the \emph{limit} of the
length of the perimeter as the \emph{number of sides} of the inscribed
figure is \emph{indefinitely increased.}~\hfill§~271

\begin{point}%
Consider the decimal $0.333 \cdots$ which may be written

\centerline{\( \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots \)}

The value of each fraction after the first is one tenth of the
preceding fraction, and by continuing the series we shall reach
a fraction less than \emph{any} assigned value, that is, the values of
the successive fractions \emph{approach zero as a limit.}

The \emph{sum} of these fractions is less than $\frac{1}{3}$; but the more terms
we take, the nearer does the sum \emph{approach $\frac{1}{3}$ as a limit.}
\end{point}

\begin{point}%
\textbf{Test for a limit.} In order to prove that a variable
approaches a constant as a limit, it is necessary to prove that
the difference between the variable and the constant:

\begin{myenum}
\item \emph{Can be made less than any assigned value, however small.}
\item \emph{Cannot be made absolutely equal to zero.}
\end{myenum}
\end{point}

\begin{point}%
\thm{If the limit of a variable $x$ is zero, then the
limit of $kx$, the product of the variable by any finite constant
$k$, is zero.}

1. Let $q$ be any assigned quantity, however small.

Then $\dfrac{q}{k}$ is not~$0$.  Hence $x$, which may differ as little as we please from~$0$,
may be taken less than $\dfrac{q}{k}$, and then $kx$ will be less than $q$.

2. Since $x$ cannot be~$0$, $kx$ cannot be~$0$.

\step{Therefore, the limit of $kx=0$}{§~275}
\end{point}
\scanpage{105.png}%

\begin{point}%
\cor{If the limit of a variable~$x$ is zero, then the
limit of the quotient of the variable by any finite constant~$k$,
is also zero.}

For $\dfrac{x}{k} = \dfrac{1}{k} × x$, which by §~276 can be made less than any assigned
value, however small, but cannot be made equal to zero.
\end{point}

\begin{point}%
\thm{The limit of the sum of a finite number of
variables \mbox{$x$, $y$, $z$, $\cdots$} is equal to the sum of their respective
limits \mbox{$a$, $b$, $c$, $\cdots$.}}

\sloppy
Let \mbox{$d$, $d'$, $d''$, $\cdots$} denote the differences between
\mbox{$x$, $y$, $z$, $\cdots$} and \mbox{$a$, $b$, $c$, $\cdots$,}
respectively. Then \mbox{$d+d'+d''+\cdots$} can be made less than any
assigned quantity $q$.

\fussy
For, if \mbox{$d$, $d'$, $d''$, $\cdots$} are $n$ in number and $d$ is the largest,

\step{$d+d'+d''+\cdots < nd$.}{(1)}

Since $d$ may be diminished at pleasure, we may make $d$ so small that

\step{$ d < \dfrac{q}{n}$; and therefore $nd < q$.}{}

But by (1), $d+d'+d''+\cdots < nd$, and therefore $< q$.

Therefore, the difference between ($x + y + z + \cdots$) and $(a + b + c + \cdots)$
can be made less than any assigned quantity, but not zero.

Therefore, the limit of $(x + y + z + \cdots) = a + b + c + \cdots$.~\hfill§~275
\end{point}

\begin{point}%
\thm{If the limit of a variable~$x$ is not zero, and
if $k$~is any finite constant, the limit of the product~$kx$ is
equal to the limit of~$x$ multiplied by~$k$.}

1. If $a$ denotes the limit of $x$, then $x$ cannot be equal to $a$.~\hfill§~271

\step{Therefore, $kx$ cannot be equal to $ka$.}{}

2. The limit of $(a - x) = 0$. Hence, the limit of $ka - kx=0$.~\hfill§~276

\step{Therefore, the limit of $kx = ka$.}{§~275}
\end{point}

\begin{point}%
\cor{The limit of the quotient of a variable $x$ by
any finite constant $k$ is the limit of $x$ divided by $k$.}

For $\dfrac{x}{k} = \dfrac{1}{k} × x$, and $\dfrac{\text{the
 limit of }x}{k} = \dfrac{1}{k}$ $×$ the limit of $x$.
\end{point}
\scanpage{106.png}%

\begin{point}%
\thm{The limit of the product of two or more
variables is the product of their respective limits, provided
no one of these limits is zero.}

If $x$ and $y$ are variables, $a$ and $b$ their respective limits, we may put
$x = a - d$, $y = b - d'$; then $d$ and $d'$ are variables which can be made
less than any assigned quantity, but not zero.~\hfill§~275

Now, %[**TN: ad hoc visual formatting]
\vspace*{-1.5\baselineskip}
\begin{align*}
xy &= (a - d) (b - d') \\
   &= ab - ad' - bd + dd' \\
\therefore ab - xy &= ad' + bd - dd'.
\end{align*}

Since every term on the right contains $d$ or $d'$, the whole right member
can be made less than any assigned quantity, but not zero.~\hfill§~278

Hence, $ab - xy$ can be made less than any assigned quantity, but not
zero.

\step{Therefore, the limit of $xy = ab$.}{§~275}

\step{Similarly, for three or more variables.}{}
\end{point}

\begin{point}%
\cor[1]{The limit of the nth power of a variable is
the nth power of its limit.}

For the limit of the product of the variables $x$, $y$, $z$, $\cdots$ to $n$ factors is
the product of their respective limits, the constants $a$, $b$, $c$, $\cdots$ to $n$ factors~(§~281).
If the $n$ factors $xyz\cdots$ are each equal to $x$, and the $n$ factors
$abc\cdots$ are each equal to $a$, we have $xyz\cdots = x^n$, and $abc\cdots = a^n$.

\step{Therefore, the limit of $x^n = a^n$.}{}
\end{point}

\begin{point}%
\cor[2]{The limit of the nth root of a variable is the
nth root of its limit.}

For if the limit of $x = a$, we may put this in the following form,

\step{the limit of \( \sqrt[n]{x^n} = \sqrt[n]{a^n} \);}{}

\noindent that is, the limit of \( \sqrt[n]{xxx\cdots\text{ to $n$~factors }} \) is
  \( \sqrt[n]{aaa\cdots\text{ to $n$~factors}} \).

Now, $xxx\cdots$ is a variable since each factor is a variable,
and $aaa\cdots$ is a constant since each factor is a constant.

If we denote $xxx\cdots$ to $n$~factors by the variable~$y$, and $aaa\cdots$ to $n$
factors by the constant~$b$, we have

% the vphantoms make the two sqrt boxes the same height, to line up better
\step{the limit of \( \displaystyle \sqrt[n]{\vphantom{b}y} = \sqrt[n]{\vphantom{y}b} \).}{}
\end{point}
\scanpage{107.png}%

\begin{point}%
\thm{If two variables are constantly equal, and
each approaches a limit, the limits are equal.}

Let $x$ and $y$ be two variables, $a$ and $b$ their respective limits, $d$ and $d'$
the respective differences between the variables and their limits. Then,
if the variables are \emph{increasing} toward their limits

\step{$a = x + d$, and $b = y + d'$.}{}

Since the equation $x = y$ is always true, we have by subtraction

\step{$a - b = d - d'$.}{}

Since $a$ and $b$ are constants, $a - b$ is a constant; therefore, $d - d'$,
which is equal to $a - b$, is a constant.

But the only constant which is less than any assigned value is~$0$.
Therefore, $d - d' = 0$. Therefore, $a - b = 0$, and $a = b$.

If the variables $x$ and $y$ are \emph{decreasing} toward their limits $a$ and $b$,
respectively, then

\step{$a = x - d$ and $b = y - d'$.}{}

Therefore, by subtraction

\step{$a - b = d' - d$.}{}

Therefore, by the same proof as for increasing variables

\step{$a = b$.}{}
\end{point}


\begin{point}%
\thm{If two variables have a constant ratio, and
each approaches a limit that is not zero, the limits have the
same ratio.}

Let $x$ and $y$ be two variables, $a$ and $b$ their respective limits.

\eq[\indent Let]{$\dfrac{x}{y}$}{$= r$, a constant; then $x = ry$.}{}

Since $x$ and $ry$ are two variables that are always equal,

\eq{the limit of $x$}{$=$ the limit of $ry$.}{§~284}

\eq[\indent Now,]{the limit of $ry$}{$= r$ $×$ limit of $y$.}{§~279}

But the limit of $x$ is $a$, and the limit of $y$ is $b$.

\eq[\indent Therefore,]{$a$}{$= rb$; that is, $\dfrac{a}{b} = r$.}{}
\end{point}
\scanpage{108.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To find the ratio of two straight lines.}

\figc{108aa286}{Let $AB$ and $CD$ be two straight lines.}

\prove[To find ]{the ratio of $AB$ and $CD$.}

\step{Apply $CD$ to $AB$ as many times as possible.}{}

\step{Suppose twice, with a remainder $EB$.}{}

\step{Then apply $EB$ to $CD$ as many times as possible.}{}

\step{Suppose three times, with a remainder $FD$.}{}

\step{Then apply $FD$ to $EB$ as many times as possible.}{}

\step{Suppose once, with a remainder $HB$.}{}

\step{Then apply $HB$ to $FD$ as many times as possible.}{}

\step{Suppose once, with a remainder $KD$.}{}

\step{Then apply $KD$ to $HB$ as many times as possible.}{}

\step{Suppose $KD$ is contained just twice in $HB$.}{}

\setlength{\eqalign}{.33\dentwidth}
\eq[\indent Then]{$HB$}{$= 2 KD$;}{}

\eq{$FD$}{$= HB + KD = 3 KD$;}{}

\eq{$EB$}{$= FD + HB = 5 KD$;}{}

\eq{$CD$}{$= 3 EB + FD = 18 KD$;}{}

\eq{$AB$}{$= 2 CD + EB = 41 KD$;}{}

\eq{$\therefore \dfrac{AB}{CD}$}{$=
          \dfrac{41 KD}{18 KD} = \dfrac{41}{18}$.}{\qef}

\setlength{\eqalign}{.5\dentwidth}

\end{proof}

\note{By the same process the ratio of two arcs of the same circle or of equal circles can be found.
\par
If the lines or arcs are incommensurable, an approximate value of the ratio can be found by the same method.}
\scanpage{109.png}%

\clearpage
\section{MEASURE OF ANGLES.}

\proposition{Theorem.}

\begin{proof}%
\obs{In the same circle or in equal circles, two
  central angles have the same ratio as their intercepted arcs.}

\figc{109ac287}{In the equal circles whose centres are $C$ and $C'$, let $ACB$
  and $A'C'B'$ be the angles, $AB$ and $A'B'$ the intercepted arcs.}

\proveq{$\dfrac{\angle A'C'B'}{\angle ACB}$}
     {$=\dfrac{\arc A'B'}{\arc AB}$.}

\textsc{Case~1.} \emph{When the arcs are commensurable} (Figs.~1 and
2).

\textbf{Proof.} Let the arc~$m$ be a common measure of $A'B'$ and
$AB$.

\step{Suppose $m$ to be contained $4$~times in $A'B'$,}{}

\step{and $7$~times in $AB$.}{}

\eq[\indent Then]{$\dfrac{\arc A'B'}{\arc AB}$}{$=\dfrac{4}{7}$.}{}

At the several points of division on $AB$ and $A'B'$ draw radii.

These radii will divide $\angle ACB$ into $7$~parts, and $\angle A'C'B'$
into $4$~parts, equal each to each,~\hfill§~237

\pnote{(in the same $\odot$, or equal
    $\odot_s$, equal arcs subtend equal central $\angle_s$).}

\eq{$\therefore \dfrac{\angle A'C'B'}{\angle ACB}$}
  {$=\dfrac{4}{7}$.}{}

\eq{$\therefore \dfrac{\angle A'C'B'}{\angle ACB}$}
  {$=\dfrac{\arc A'B'}{\arc AB}$.}{Ax.~1}
\scanpage{110.png}%

\step{}{}

\filbreak
\textsc{Case 2.}  \emph{When the arcs are incommensurable} (Figs.~2 and 3).

\textbf{Proof.} Divide $AB$ into any number of equal parts, and apply
one of these parts to $A'B'$ as many times as $A'B'$ will contain it.

Since $AB$ and $A'B'$ are incommensurable, a certain number
of these parts will extend from $A'$ to some point, as $D$, leaving
a remainder $DB'$ less than one of these parts. Draw $C'D$.

\step{By construction $AB$ and $A'D$ are commensurable.}{}

\eq{$\therefore \dfrac{\angle A'C'D}{\angle ACB}$}
  {$=\dfrac{\arc A'D}{\arc AB}$.}{Case~1}

By increasing the \emph{number} of equal parts into which $AB$ is
divided we can diminish at pleasure the \emph{length} of each part,
and therefore make $DB'$ less than any assigned value, however
small, since $DB'$ is always less than one of the equal parts into
which $AB$ is divided.

We cannot make $DB'$ equal to zero, since, by hypothesis, $AB$
and $A'B'$ are incommensurable.~\hfill§~269

Hence, $DB'$ approaches zero as a limit, if the number of
parts of $AB$ is indefinitely increased.~\hfill§~275

And the corresponding angle $DC'B'$ approaches zero as a
limit.

Therefore, the arc $A'D$ approaches the arc $A'B'$ as a limit,~\hfill§~271\linebreak
and the $\angle A'C'D$ approaches the $\angle A'C'B'$ as a limit.

\step[\indent Therefore,]{$\dfrac{\arc A'D}{\arc AB}$ approaches
  $\dfrac{\arc A'B'}{\arc AB}$ as a limit,}{§~280}

\step[and]{$\dfrac{\angle A'C'D}{\angle ACB}$ approaches
  $\dfrac{\angle A'C'B'}{\angle ACB}$ as a limit.}{§~280}

\step[\indent But]{$\dfrac{\angle A'C'D}{\angle ACB}$ is constantly equal to
  $\dfrac{\arc A'D}{\arc AB}$,}{}

\step{as $A'D$ varies in value and approaches $A'B'$ as a limit.}{}

\step{$\therefore \dfrac{\angle A'C'B'}{\angle ACB} =
  \dfrac{\arc A'B'}{\arc AB}$.}{§~284}

\hfill\qed

\end{proof}
\scanpage{111.png}%

\begin{point}%
A circumference is divided into $360$~equal parts, called
  \emph{degrees}; and therefore a unit angle at the centre intercepts
  a unit arc on the circumference.  Hence, the \emph{numerical measure
  of a central angle} expressed in terms of the unit angle is equal to
  the \emph{numerical measure of its intercepted arc} expressed in
  terms of the unit arc.  This must be understood to be the meaning
  when it is said that

\emph{A central angle is measured by its intercepted arc.}
\end{point}

\proposition{Theorem.}

\begin{proof}%
\obs{An inscribed angle is measured by half the arc
  intercepted between its sides.}

\figc{111ac289}{1. Let the centre $C$ \textnormal{(Fig.~1)} be in one of the sides of
  the angle.}

\prove{the $\angle B$ is measured by $\frac{1}{2}$ the
  arc $PA$.}

\step[\indent\textbf{Proof.}]{Draw $CA$.}{}

\eq{$CA$}{$= CB$.}{§~217}

\eq{$\therefore \angle B$}{$= \angle A$.}{§~145}

\eq[\indent But]{$\angle PCA$}{$= \angle B + \angle A$.}{§~137}

\eq{$\therefore \angle PCA$}{$= 2 \angle B$.}{}

\step[\indent But]{$\angle PCA$ is measured by $\arc PA$,}{§~288}

\pnote{(a central $\angle$ is measured
    by its intercepted arc).}

\step{$\therefore \angle B$ is measured by $\frac{1}{2} \arc PA$.}{}
\scanpage{112.png}%

\step{}{}

\textbf{2. Let the centre $C$ \textnormal{(Fig.~2)} fall within the angle $PBA$.}

\prove{the $\angle PBA$ is measured by $\frac{1}{2}$ the arc $PA$.}

\step[\indent\textbf{Proof.}]{Draw the diameter $BCE$.}{}

\setlength{\eqalign}{.33\dentwidth}
\eq[\indent Then]{$\angle EBA$}{is measured by $\frac{1}{2} \arc AE$,}{}

\eq[and]{$\angle EBP$}{is measured by $\frac{1}{2} \arc EP$.}{Case~1}

\eq{$\therefore \angle EBA + \angle EBP$}{is measured by $\frac{1}{2}(\arc AE +\arc EP)$,}{}

\eq[or]{$\angle PBA$}{is measured by $\frac{1}{2} \arc PA$.}{}


\step{}{}

\textbf{3. Let the centre $C$} (Fig.~3) \textbf{fall without the angle $PBA$.}

\prove{the $\angle PBA$ is measured by $\frac{1}{2}$ the arc $PA$.}

\step[\indent\textbf{Proof.}]{Draw the diameter $BCF$.}{}

\eq[\indent Then]{$\angle FBA$}{is measured by $\frac{1}{2} \arc FA$,}{}

\eq[and]{$\angle FBP$}{is measured by $\frac{1}{2} \arc FP$.}{Case~1}

\eq{$\therefore \angle FBA - \angle FBP$}{is measured by $\frac{1}{2}(\arc FA - \arc FP)$,}{}

\eq[or]{$\angle PBA$}{is measured by $\frac{1}{2} \arc PA$.}{\qed}

\setlength{\eqalign}{.5\dentwidth}


\end{proof}

\figc{112ac290}{}

\pp{\cor[1]{An angle inscribed in a semicircle is a right
angle. \textup{For it is measured by half a semicircumference (Fig.~4).}}}

\pp{\cor[2]{An angle inscribed in a segment greater than
a semicircle is an acute angle. \textup{For it is measured by an arc
less than half a semicircumference; as, $\angle CAD$ (Fig.~5).}}}

\pp{\cor[3]{An angle inscribed in a segment less than a
semicircle is an obtuse angle. \textup{For it is measured by an arc
greater than half a semicircumference; as, $\angle CBD$ (Fig.~5).}}}

\pp{\cor[4]{Angles inscribed in the same segment or in
equal segments are equal \textup{(Fig.~6).}}}
\scanpage{113.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{An angle formed by two chords intersecting
within the circumference is measured by half the sum
of the intercepted arcs.}

\figc{113aa294}{Let the angle $COD$ be formed by the chords $AC$ and $BD$.}

\prove{the $\angle COD$ is measured by $\frac{1}{2}(CD + AB)$.}

\step[\indent\textbf{Proof.}]{Suppose $AE$ drawn $\parallel$ to $BD$.}{}

\eq{}{Then $\arc AB = \arc DE$,}{§~257}

\pnote{(parallels intercept equal arcs on a circumference).}

\eq{}{Also $\angle COD = \angle CAE$,}{§~112}

\pnote{(ext.-int.~$\angle_s$ of $\parallel_s$).}

\step{But $\angle CAE$ is measured by $\frac{1}{2}(CD + DE)$,}{§~289}

\pnote{(an inscribed $\angle$ is measured by half its intercepted arc).}

Put $\angle COD$ for its equal, the $\angle CAE$, and $\arc AB$ for its
equal, the arc $DE$; then $\angle COD$ is measured by $\frac{1}{2}(CD + AB)$.

\hfill\qed

\end{proof}

\ex{The opposite angles of an inscribed quadrilateral are supplementary.}

\ex{If through a point within a circle two perpendicular chords
are drawn, the sum of either pair of the opposite arcs which they intercept
is equal to a semicircumference.}

\ex{The line joining the centre of the square described upon the
hypotenuse of a right triangle to the vertex of the right angle bisects the
right angle.}
\scanpage{114.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{An angle included by a tangent and a chord drawn
  from the point of contact is measured by half the intercepted arc.}

\figc{114aa295}{Let $MAH$ be the angle included by the tangent $MO$ to the
  circle at $A$ and the chord $AH$.}

\prove{the $\angle MAH$ is measured by $\frac{1}{2}$ the
  arc $AEH$.}

\step[\indent\textbf{Proof.}]{Suppose $HF$ drawn $\parallel$ to $MO$.}{}

\step{Then $\arc AF = \arc AEH$,}{§~257}

\pnote{(parallels intercept equal arcs
    on a circumference).}

\step{Also $\angle MAH = \angle AHF$,}{§~110}

\pnote{(alt.-int.~$\angle_s$ of $\parallel_s$).}

\step{$\angle AHF$ is measured by $\frac{1}{2} AF$,}{§~289}

\pnote{(an inscribed $\angle$ is measured by half its intercepted arc).}

Put $\angle MAH$ for its equal, the $\angle AHF$, and $\arc AEH$ for
its equal, the arc $AF$; then $\angle MAH$ is measured by
$\frac{1}{2} \arc AEH$.

Likewise, the $\angle OAH$, the supplement of the $\angle MAH$, is
measured by half the arc $AFH$, the conjugate of the arc $AEH$.

\hfill\qed

\end{proof}

\ex{Two circles are tangent externally at $A$, and a
common external tangent touches them at $B$ and $C$, respectively.
Show that angle $BAC$ is a right angle.}
\scanpage{115.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{An angle formed by two secants, two tangents, or a
  tangent and a secant, drawn to a circle from an external point, is
  measured by half the difference of the intercepted arcs.}

\figc{115ac296}{}

The proof of this theorem is left as an exercise for the student.

\end{proof}

\figcc{115dd297}{115ee297}
\begin{point}%
\textbf{Positive and Negative Quantities.} In
measurements it is convenient to mark the distinction between two
quantities that are measured in \emph{opposite directions}, by calling
one of them \indexbf{positive} and the other \indexbf{negative}.

Thus, if $OA$ is considered positive, then $OC$ may be considered
negative, and if $OR$ is considered positive, then $OD$ may be
considered negative.

When this distinction is applied to angles, an angle is considered to
be \emph{positive}, if the rotating line that describes it moves in
the opposite direction to the hands of a clock (counter clockwise),
and to be \emph{negative}, if the rotating line moves in the same
direction as the hands of a clock (clockwise).

Arcs corresponding to positive angles are considered \emph{positive},
and arcs corresponding to negative angles are considered
\emph{negative}.

Thus, the angle $ACB$ described by a line rotating about $C$ from $CA$
to $CB$ is positive, and the arc $AB$ is positive; the angle $ACB'$
described by the line rotating about $C$ from $CA$ to $CB'$ is
negative, and the arc $AB'$ is negative.
\end{point}
\scanpage{116.png}%

\pp{\textbf{The Principle of Continuity.}\label{princcont} By marking the distinction between
quantities measured in opposite directions, a theorem may often be so
stated as to include two or more particular theorems.}

The following theorem furnishes a good illustration:

\begin{point}%
\textit{The angle included between two lines of unlimited length that cut
or touch a circumference is measured by half the sum of the intercepted arcs.}

Here the word \emph{sum} means the algebraic sum and includes both the arithmetical
sum and the arithmetical difference of two quantities.

\figc{116ag299}{}

1. If the lines intersect at the centre, the two intercepted arcs are equal,
and half the sum will be one of the arcs (§~288).

2. If the lines intersect between the centre and the circumference, the
angle is measured by half the sum of the arcs (§~294).

3. If the lines intersect on the circumference, one of the arcs becomes
zero and we have an inscribed angle (§~289), or an angle formed by a
tangent and a chord (§~295). In each case the angle is measured by half
the sum of the intercepted arcs.

4. If the lines intersect without the circumference, then the arc $ab$
is negative and the algebraic sum is the arithmetical difference of the
included arcs.

When the reasoning employed to prove a theorem is continued in the
manner just illustrated to include two or more theorems, we are said to
reason by the \emph{Principle of Continuity.}
\end{point}

\subsection{REVIEW QUESTIONS ON BOOK II.}

\begin{myenum}
\item  What do we call the locus of points in a plane that are equidistant
from a fixed point in the plane?

\item What does the chord of a segment become when the segment is a
semicircle?

\item What kind of an angle do the radii of a sector include when the
sector is a semicircle?

\item  What is the difference between a chord and a secant?

\item What part of a tangent is meant by a tangent to a circle from an
external point?
\scanpage{117.png}%

\item Two chords are equal in equal circles under either of two conditions.
What are the two conditions?

\item Points that lie in a straight line are called \emph{collinear}; points that
lie in a circumference are called \emph{concyclic}. How many collinear points
can be concyclic?

\item What is meant by the statement that a central angle is measured
by the arc intercepted between its sides?

\item What is an inscribed angle?   What is its measure?

\item What kind of an angle is the angle inscribed in a semicircle? in a
segment less than a semicircle? in a segment greater than a semicircle?

\item What is the measure of an angle included by two intersecting
chords? by two secants intersecting without the circle?

\item What is the measure of an angle included by a tangent and a
chord drawn to the point of contact?

\item When are two quantities of the same kind incommensurable?

\item When are two quantities of the same kind commensurable?

\item Define a variable and the limit of a variable.

\item Does the series $\frac{1}{2}$ in., $\frac{3}{4}$ in., $\frac{7}{8}$ in., $\frac{15}{16}$ in., etc., constitute a variable?
Is the variable increasing or decreasing?

\item What is the limit of this variable?

\item What is the test of a limit?
\end{myenum}

%\section{EXERCISES.}

\subsection{THEOREMS.}

\ex{An angle formed by a tangent and a chord is equal to the
angle inscribed in the opposite segment.}

\ex{Two chords drawn perpendicular to a third chord at its
extremities are equal.}

\ex{The sum of two opposite sides of a circumscribed quadrilateral
is equal to the sum of the other two sides.}

\figcc{117aa104}{117bb105}
\begin{proofex}%
If the sum of two opposite angles of a quadrilateral
is equal to two right angles, a circle may be circumscribed
about the quadrilateral.

Let $\angle A + \angle C = 2 \text{ rt.\ } \angle_s$. Pass a circumference through
$D$, $A$, and $B$, and prove that this circumference passes through $C$.

\end{proofex}

\begin{proofex}%
The shortest line that can be drawn from a point within a
circle to the circumference is the shorter segment of the
diameter through that point.

Let $A$ be the given point. Prove $AB$ shorter than any
other line $AD$ from $A$ to the circumference.

\end{proofex}
\scanpage{118.png}%

\ex{The longest line that can be drawn from a point within a
circle to the circumference is the longer segment of the diameter through
that point.}

\figc{118aa107}{}
\ex{The shortest line that can be drawn
from a point without a circle to the circumference
will pass through the centre of the circle if produced.}

\ex{The longest line that can be drawn from a point without a
circle to the concave arc of the circumference passes through the centre of
the circle.}

\figc{118be108}{}

\ex{The shortest chord that can be drawn through a point within
a circle is perpendicular to the diameter at that point.}

\begin{proofex}%
If two intersecting chords make equal angles with the diameter
drawn through the point of intersection, the two chords are equal.

% special case, two centered columns
\hspace{\stretch{1}}Rt.~$\triangle COM =$ rt.~$\triangle CON$.
\hspace{\stretch{1}}$\therefore OM = ON$.
\hspace{\stretch{1}}


\end{proofex}

\ex{The angles subtended at the centre of a circle by any two
opposite sides of a circumscribed quadrilateral are supplementary.}

\begin{proofex}%
The radius of a circle inscribed in an equilateral triangle is
equal to one third the altitude of the triangle.

$\triangle OEF$ is equiangular and equilateral; $\angle FEA = \angle FAE$.

% special case, two centered columns
\hspace{\stretch{1}}$\therefore AF = EF$.
\hspace{\stretch{1}}$\therefore AF = FO = OD$.
\hspace{\stretch{1}}

\end{proofex}

\ex{The radius of a circle circumscribed about an equilateral
triangle is equal to two thirds the altitude of the triangle (Ex.~27).}

\ex{A parallelogram inscribed in a circle is a rectangle.}

\ex{A trapezoid inscribed in a circle is an isosceles trapezoid.}

\ex{All chords of a circle which touch an interior concentric
circle are equal, and are bisected at the point of contact.}

\ex{If the inscribed and circumscribed circles of a triangle are
concentric, the triangle is equilateral (Ex.~116).}
\scanpage{119.png}%

\ex{If two circles are tangent to each other the tangents to them
from any point of the common internal tangent are equal.}

\ex{If two circles touch each other and a line is drawn through
the point of contact terminated by the circumferences, the tangents at its
ends are parallel.}

\figcc{119aa120}{119bb121}
\begin{proofex}%
If two circles touch each other and two
lines are drawn through the point of contact terminated
by the circumferences, the chords joining the
ends of these lines are parallel.

$\angle A = \angle MPC$ and $\angle B = \angle NPD$. \quad $\therefore \angle A = \angle B$.

\end{proofex}

\ex{If two circles intersect and a line is drawn
through each point of intersection terminated by the circumferences,
the chords joining the ends of these lines
are parallel.}

\figcc{119cc122}{119dd123}
\begin{proofex}%
Through one of the points of intersection of two circles a
diameter of each circle is drawn.    Prove that the line
joining the ends of the diameters passes through the other
point of intersection.

\step{$\angle ABC = \angle ABD = 90°$}{§~290}

\end{proofex}

\ex{If two common external tangents or two common internal
tangents are drawn to two circles, the segments
intercepted between the points of contact are equal.}

\ex{The diameter of the circle inscribed in
a right triangle is equal to the difference between
the sum of the legs and the hypotenuse.}

\figcc{119ee125}{119ff126}
\begin{proofex}%
If one leg of a right triangle is the diameter
of a circle, the tangent at the point where the circumference
cuts the hypotenuse bisects the other leg.

\begin{center}
$\angle BOE = \angle DOE$.\qquad $\angle BOE = \angle OAD$.

$\therefore OE$ and $AC$ are $\parallel$.\qquad $\therefore BE = EC$ (§~188).
\end{center}


\end{proofex}

\ex{If, from any point in the circumference of
a circle, a chord and a tangent are drawn, the perpendiculars
dropped on them from the middle point of the
subtended arc are equal. $\angle BAM = \angle CAM$.}

\ex{The median of a trapezoid circumscribed about a circle is equal
to one fourth the perimeter of the trapezoid (Ex.~103).}
\scanpage{120.png}%

\ex{Two fixed circles touch each other externally and a circle of
variable radius touches both externally. Show that the difference of the
distances from the centre of the variable circle to the centres of the fixed
circles is constant.}

\ex{If two fixed circles intersect, and circles are drawn to touch
both, show that either the sum or the difference of the distances of their
centres from the centres of the fixed circles is constant, according as they
touch (i)~one internally and one externally, (ii)~both internally or both
externally.}

\figcc{120aa130}{120bb131}
\begin{proofex}%
If two straight lines are drawn through
any point in a diagonal of a square parallel to the sides
of the square, the points where these lines meet the
sides lie on the circumference of a circle whose centre
is the point of intersection of the diagonals.

$\triangle POE = \triangle POF$ (§~143). \quad
  $\therefore OE = OF$. \quad $\triangle POE' = \triangle POF'$.

\end{proofex}

\begin{proofex}%
If $ABC$ is an inscribed equilateral triangle and
$P$ is any point in the arc $BC$, then $PA = PB + PC$.

Take $PM = PB$. $\triangle ABM = \triangle CBP$ (§~143) and $AM
= PC$.

\end{proofex}

\ex{The tangents drawn through the vertices of an inscribed
rectangle, which is not a square, enclose a rhombus.}

\figc{120cd132}{}

\begin{proofex}%
The bisectors of the angles included by the opposite sides
(produced) of an inscribed quadrilateral intersect at right angles.

\eq{Arc $AF - \arc BM$}{$= \arc DF - \arc CM$}{}

\eq{and $\arc AH - \arc DN$}{$= \arc BH - \arc CN$.}{}

\eq{$\therefore \arc FH + \arc MN$}{$= \arc HM + \arc FN$.}{}

\eq{$\therefore \angle FIH$}{$= \angle HIM$.}{}

\textbf{Discussion.}  This problem is impossible, if any two sides of the quadrilateral
are parallel.

\end{proofex}
\scanpage{121.png}%


\section{PROBLEMS OF CONSTRUCTION.}

\note{Hitherto we have supposed the figures constructed. We now
proceed to explain the methods of constructing simple problems, and afterwards
to apply these methods to the solution of more difficult problem.}

\proposition{Problem.}

\begin{proof}%
\obs{To let fall a perpendicular upon a given line from
a given external point.}

\figc{121aa300}{Let $AB$ be the given straight line, and $C$ the given external point.}

\prove[]{To let fall a $\perp$ to the line $AB$ from the point $C$}.

From $C$ as a centre, with a radius sufficiently great, describe
an arc cutting $AB$ in two points, $H$ and $K$.

From $H$ and $K$ as centres, with equal radii greater than $\frac{1}{2}HK$,

\step{describe two arcs intersecting at $O$.}{}

\step{Draw $CO$,}{}

\step{and produce it to meet $AB$ at $M$.}{}

\step{$CM$ is the $\perp$ required.}{}

\textbf{Proof.} Since $C$ and $O$ are two points each equidistant from
$H$ and $K$, they determine a $\perp$ to $HK$ at its middle point.~\hfill§~161

\hfill\qef

\end{proof}

\note{\emph{Given} lines of the figures are represented by full lines, \emph{resulting}
lines by long-dashed, and \emph{auxiliary} lines by short-dashed lines.}
\scanpage{122.png}%

\proposition{Problem.}

\begin{proof}%
\obs{At a given point in a straight line, to erect a
perpendicular to that line.}

\figc{122ab301}{1. Let $O$ be the given point in $AC$. \textnormal{Fig.~1.}}

\step{Take $OH$ and $OB$ equal.}{}

From $H$ and $B$ as centres, with equal radii greater than $OB$,
describe two arcs intersecting at $R$. Join $OR$.

Then the line $OR$ is the $\perp$ required.

\textbf{Proof.} $O$ and $R$, two points each equidistant from $H$ and $B$,
determine the perpendicular bisector of $HB$.~\hfill§~161

\lett{2. Let $B$ be the given point. \textnormal{Fig.~2.}}

Take any point $C$ without $AB$; and from $C$ as a centre,
with the distance $CB$ as a radius, describe an arc intersecting
$AB$ at $E$.

Draw $EC$, and prolong it to meet the arc again at $D$.

Join $BD$, and $BD$ is the $\perp$ required.

\step[\indent\textbf{Proof.}]{The $\angle B$ is a right angle.}{§~290}

\step{$\therefore BD$ is $\perp$ to $AB$.}{\qef}

\textbf{Discussion.} The point $C$ must be so taken that it will not
be in the required perpendicular.

\end{proof}
\scanpage{123.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To bisect a given straight line.}

\figc{123aa302}{To bisect the given straight line $AB$.}

From $A$ and $B$ as centres, with equal radii greater than
$\frac{1}{2} AB$, describe arcs intersecting at $C$ and $E$.

\step{Join $CE$.}{}

\step{Then $CE$ bisects $AB$.}{§~161}

\hfill\qef

\end{proof}

\proposition{Problem.}

\begin{proof}%
\obs{To bisect a given arc.}

\figc{123bb303}{To bisect the given arc $AB$.}

\step{Draw the chord $AB$.}{}

From $A$ and $B$ as centres, with equal radii greater than
$\frac{1}{2} AB$, describe arcs intersecting at $D$ and $E$.
\scanpage{124.png}%

\step{Draw $DE$.}{}

\step{Then $DE$ is the $\perp$ bisector of the chord $AB$.}{§~161}

\step{$\therefore DE$ bisects the arc $ACB$.}{§~248}

\hfill\qef

\end{proof}

\proposition{Problem.}

\begin{proof}%
\obs{To bisect a given angle.}

\figc{124aa304}{Let $AEB$ be the given angle.}

From $E$ as a centre, with any radius, as $EA$, describe an
arc cutting the sides of the $\angle E$ at $A$ and $B$.

From $A$ and $B$ as centres, with equal radii greater than half
the distance from $A$ to $B$, describe two arcs intersecting at $D$.

\step{Draw $DE$.}{}

\step{Then $DE$ bisects the arc $AB$ at $C$.}{§~303}

\step{$\therefore DE$ bisects the angle $E$.}{§~237}

\hfill\qef

\end{proof}

\ex{To construct an angle of $45°$; of $135°$.}

\ex{To construct an equilateral triangle, having given one side.}

\ex{To construct an angle of $60°$; of $150°$.}

\ex{To trisect a right angle.}
\scanpage{125.png}%

\proposition{Problem.}

\begin{proof}%
\obs{At a given point in a given straight line, to
construct an angle equal to a given angle.}

\figc{125ab305}{At $C$ in the line $CM$, construct an angle equal to the given angle $A$.}

From $A$ as a centre, with any radius, $AE$, describe an arc
cutting the sides of the $\angle A$ at $E$ and $F$.

\step{From $C$ as a centre, with a radius equal to $AE$,}{}

\step{describe an arc $HG$ cutting $CM$ at $H$.}{}

\step{From $H$ as a centre, with a radius equal to the chord $EF$,}{}

\step{describe an arc intersecting the arc $HG$ at $O$.}{}

\step{Draw $CO$, and $\angle$ $HCO$ is the required angle.}{Why?}

\hfill\qef

\end{proof}


\proposition{Problem.}

\begin{proof}%
\obs{To draw a straight line parallel to a given
straight line through a given external point.}

\figc{125cc306}{Let $AB$ be the given line, and $C$ the given point.}
\scanpage{126.png}%

\step{Draw $ECD$, making any convenient $\angle EDB$.}{}

\step{At the point $C$ construct $\angle ECF$ equal to $\angle EDB$.}{§~305}

\step{Then the line $HCF$ is $\parallel$ to $AB$.}{Why?}

\hfill\qef


\end{proof}


\proposition{Problem.}

\begin{proof}%
\obs{To divide a given straight line into a given
number of equal parts.}

\figc{126aa307}{Let $AB$ be the given straight line.}

From $A$ draw the line $AO$, making any convenient angle
with $AB$.

Take any convenient length, and apply it to $AO$ as many
times as the line $AB$ is to be divided into parts.

From $C$, the last point thus found on $AO$, draw $CB$.

Through the points of division on $AO$ draw parallels to
the line $CB$.~\hfill§~306

\step{These lines will divide $AB$ into equal parts.}{§~187}

\hfill\qef

\end{proof}

\ex{To construct an equilateral triangle, having given the perimeter.}

\ex{To divide a line into four equal parts by two different
methods.}

\begin{proofex}%
Through a given point to draw a line which shall make equal
angles with the two sides of a given angle.

Through the given point draw a $\perp$ to the bisector of the given $\angle$.

\end{proofex}

\ex{To draw a line through a given point, so that it shall form
with the sides of a given angle an isosceles triangle (Ex.~140).}
\scanpage{127.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To find the third angle of a triangle when two
of the angles are given.}

\figc{127ab308}{Let $A$ and $B$ be the two given angles.}

\step{At any point $H$ in any line $EF$,}{}

\step{construct $\angle a$ equal to $\angle A$, and $\angle b$ equal to $\angle B$.}{§~305}

\step[\indent Then]{$\angle c$ is the $\angle$ required.}{Why?}

\hfill\qef

\end{proof}


\proposition{Problem.}

\begin{proof}%
\obs{To construct a triangle when two sides and the
included angle are given.}

\figc{127cd309}{Let $b$ and $c$ be the two sides of the triangle and $E$ the included angle.}

\step{Take $AB$ equal to the side $c$.}{}

At $A$, construct $\angle BAD$ equal to the given $\angle E$.~\hfill§~305
\scanpage{128.png}%

\step{On $AD$ take $AC$ equal to $b$, and draw $CB$.}{}

\step{Then $\triangle ACB$ is the $\triangle$ required.}{\qef}


\end{proof}


\proposition{Problem.}

\begin{proof}%
\obs{To construct a triangle when a side and two
angles of the triangle are given.}

\figc{128ab310}{Let $c$ be the given side, $A$ and $B$ the given angles.}

\step{Take $EC$ equal to the side $c$.}{}

\step{At $E$ construct the $\angle CEH$ equal to $\angle A$.}{§~305}

\step{At $C$ construct the $\angle ECK$ equal to $\angle B$.}{}

\step{Produce $EH$ and $CK$ until they intersect at $O$.}{}

\step{Then $\triangle COE$ is the $\triangle$ required.}{\qef}

\textsc{Remark.} If one of the given angles is opposite to the given side, find
the third angle by §~308, and proceed as above.

\textbf{Discussion.} The problem is impossible when the two given
angles are together equal to or greater than two right angles.

\end{proof}

\ex{To construct an equilateral triangle, having given the altitude.}

\exheader{To construct an isosceles triangle, having given:}

\ex{The base and the altitude.}

\ex{The altitude and one of the legs.}

\ex{The angle at the vertex and the altitude.}
\scanpage{129.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To construct a triangle when two sides and the
angle opposite one of them are given.}

\figc{129ab311}{Let $a$ and $b$ be the given sides, and $A$ the angle opposite $a$.}

\textsc{Case 1.} \emph{If $a$~is less than~$b$.}

\step{Construct $\angle DAE$ equal to the given $\angle A$}{§~305}

\step{On $AD$ take $AB$ equal to $b$.}{}

\step{From $B$ as a centre, with a radius equal to $a$,}{}

\step{describe an arc intersecting the line $AE$ at $C$ and $C'$.}{}

\step{Draw $BC$ and $BC'$.}{}

Then both the $\triangle_s ABC$ and $ABC'$ fulfil the conditions, and
hence we have two constructions.

This is called the \emph{ambiguous} case.

\figcc{129cc311}{129dd311}
\textbf{Discussion.} If the side $a$ is equal to
the $\perp BH$, the arc described from $B$ will touch $AE$, and there will be but
one construction, the right $\triangle ABH$.

If the given side $a$ is less than the
$\perp$ from $B$, the arc described from $B$
 will not intersect or touch $AE$, and
hence the problem is impossible.
\scanpage{130.png}%

If the $\angle A$ is right or obtuse, the problem is impossible; for
the side opposite a right or obtuse angle is the greatest side.~\hfill§~153
\filbreak
\textsc{Case 2.} \emph{If $a$ is equal to $b$.}

\figc{130aa311}{}
If the $\angle A$ is acute, and $a = b$, the arc described from $B$ as
a centre, and with a radius equal to $a$, will
cut the line $AE$ at the points $A$ and $C$.
There is therefore but one solution: the
isosceles $\triangle ABC$.

\textbf{Discussion.} If the $\angle A$ is right or obtuse, the problem is
impossible; for equal sides of a $\triangle$ have equal $\angle_s$ opposite
them, and a $\triangle$ cannot have two right $\angle_s$ or two obtuse $\angle_s$.

\figccc{130bb311}{130cc311}{130dd311}
\textsc{Case 3.} \emph{If $a$ is greater than $b$.}

If the given $\angle A$ is acute, the arc described from $B$ will cut
the line $ED$ on opposite sides of $A$, at $C$ and $C'$. The $\triangle ABC$
answers the required conditions, but the
$\triangle$ $ABC'$ does not, for it does not contain
the acute $\angle A$. There is then only one
solution; namely, the $\triangle ABC$.

If the $\angle A$ is right, the arc described
from $B$ cuts the line $ED$ on opposite
sides of $A$, and we have two \emph{equal} right
$\triangle_s$ which fulfil the required conditions.

If the $\angle A$ is obtuse, the arc described
from $B$ cuts the line $ED$ on opposite
sides of $A$, at the points $C$ and $C'$. The
$\triangle ABC$ answers the required conditions,
but the  $\triangle ABC'$ does not, for it does not contain the obtuse
$\angle A$. There is then only one solution; namely, the $\triangle ABC$.~\hfill\qef

\end{proof}
\scanpage{131.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To construct a triangle when the three sides of
the triangle are given.}

\figc{131aa312}{Let the three sides be $c$, $a$, and $b$.}

Take $AB$ equal to $c$. From $A$ as a centre, with a radius
equal to $b$, describe an arc. From $B$ as a centre, with a radius
equal to $a$, describe an arc, intersecting the other arc at $C$.

\step{Draw $CA$ and $CB$.}{}

\step{$\triangle CAB$ is the $\triangle$ required.}{\qef}

\textbf{Discussion.} The problem is impossible when one side is equal
to or greater than the sum of the other two sides.

\end{proof}

\proposition{Problem.}

\begin{proof}%
\obs{To construct a parallelogram when two sides
and the included angle are given}.

\figc{131bb313}{Let $m$ and $o$ be the two sides, and $C$ the included angle.}

\step{Take $AB$ equal to $o$.}{}

\step{At $A$ construct $\angle BAD$ equal to $\angle C$.}{§~305}
\scanpage{132.png}%

Take $AH$ equal to $m$. From $H$ as a centre, with a radius
equal to $o$, describe an arc, and from $B$ as a centre, with a
radius equal to $m$, describe an arc, intersecting the other arc
at $E$; and draw $EH$ and $EB$.

\step{The quadrilateral $ABEH$ is the $\Par$ required.}{§~182}

\hfill\qef

\end{proof}

\proposition{Problem.}

\begin{proof}%
\obs{To circumscribe a circle about a given triangle.}

\figc{132aa314}{Let $ABC$ be the given triangle.}

\step{Bisect $AB$ and $BC$.}{§~302}

\step{At $E$ and $D$, the points of bisection, erect $\perp_s$.}{§~301}

Since $BC$ is not the prolongation of $AB$, these $\perp_s$ will intersect
at some point $O$.

From $O$, with a radius equal to $OB$, describe a circle.

\step{The $\odot ABC$ is the $\odot$ required.}{}

\step[\indent\textbf{Proof.}]{The point $O$ is equidistant from $A$ and $B$,}{}

\step{and also is equidistant from $B$ and $C$.}{§~160}

\step{$\therefore$ the point $O$ is equidistant from $A$, $B$, and $C$,}{}

\noindent and a $\odot$ described from $O$ as a centre, with a radius equal to
$OB$, will pass through the vertices $A$, $B$, and $C$.~\hfill\qef

\end{proof}

The same construction serves to describe a circumference
which shall pass through three points not in the same straight
line; also to find the centre of a given circle or of a given arc.

\note{The point $O$ is called the \emph{circum-centre}\label{circum-centre} of the triangle.}
\scanpage{133.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To inscribe a circle in a given triangle.}

\figc{133aa315}{Let $ABC$ be the given triangle.}

\step{Bisect the $\angle_s A$ and $C$.}{§~304}

\step{From $E$, the intersection of the bisectors,}{}

\step{draw $EH \perp$ to the side $AC$.}{§~300}

\step{From $E$ as centre, with radius $EH$, describe the $\odot KHM$.}{}

\step{The $\odot KHM$ is the $\odot$ required.}{}

\textbf{Proof.} Since $E$ is in the bisector of the $\angle A$, it is equidistant
from the sides $AB$ and $AC$; and since $E$ is in the bisector of
the $\angle C$, it is equidistant from the sides $AC$ and $BC$.~\hfill§~162

$\therefore$ a $\odot$ described from $E$ as centre, with a radius equal to $EH$,
will touch the sides of the $\triangle$ and be inscribed in it.~\hfill\qef

\end{proof}

\note{The point $E$ is called the \indexemph{in-centre} of the triangle.}

\figc{133bb316}{}
\begin{point}%
The intersections of the
bisectors of the exterior angles of
a triangle are the centres of three
circles, each of which will touch
one side of the triangle, and the
two other sides produced. These
three circles are called \emph{escribed}\label{escribed}
circles; and their centres are called
the \indexemph{ex-centres} of the triangle.
\end{point}
\scanpage{134.png}%

\proposition{Problem.}

\begin{proof}%
\obs{Through a given point, to draw a tangent to a
given circle.}

\textsc{Case 1.} \emph{When the given point is on the circumference.}

\figc{134aa317}{Let $C$ be the given point on the circumference whose centre is~$O$.}

\step{From the centre $O$ draw the radius $OC$.}{}

\step{Through $C$ draw $AM \perp$ to $OC$.}{§~301}

\step{Then $AM$ is the tangent required.}{§~253}

\textsc{Case 2.} \emph{When the given point is without the circle.}

\lett{Let $O$ be the centre of the given circle, $E$ the given point.}

\step{Draw $OE$.}{}

On $OE$ as a diameter, describe a circumference intersecting
the given circumference at the points $M$ and $H$.

\step{Draw $OM$ and $EM$.}{}

\step{Then $EM$ is the tangent required.}{}

\step[\indent\textbf{Proof.}]{$\angle OME$ is a right angle.}{§~290}

\step{$\therefore EM$ is tangent to the circle at $M$.}{§~253}

In like manner, we may prove $EH$ tangent to the given $\odot$.

\hfill\qef

\end{proof}

\ex{To draw a tangent to a given circle, so that it shall be parallel
to a given straight line.}
\scanpage{135.png}%

\proposition{Problem.}

\begin{proof}%
\obs{Upon a given straight line, to describe a segment
of a circle in which a given angle may be
inscribed.}

\figc{135aa318}{Let $AB$ be the given line, and $M$ the given angle.}

\step{Construct the $\angle ABE$ equal to the $\angle M$.}{§~305}

\step{Bisect the line $AB$ by the $\perp OF$.}{§~302}

\step{From the point $B$ draw $BO \perp$ to $EB$.}{§~301}

From $O$, the point of intersection of $FO$ and $BO$, as a centre
with a radius equal to $OB$, describe a circumference.

\step{The segment $AKB$ is the segment required.}{}

\step[\indent\textbf{Proof.}]{The point $O$ is equidistant from $A$ and $B$.}{§~160}

\step{$\therefore$ the circumference will pass through $A$.}{}

\step{But $BE$ is $\perp$ to $OB$.}{Const.}

\step{$\therefore BE$ is tangent to the $\odot$,}{§~253}

\pnote{(a straight line $\perp$ to a radius at its extremity is tangent to the $\odot$).}

\step{$\therefore \angle ABE$ is measured by $\frac{1}{2} \arc AB$,}{§~295}

\pnote{(being an $\angle$ formed by a tangent and a chord).}

But any $\angle$ as $\angle K$ inscribed in the segment $AKB$ is measured
by $\frac{1}{2} \arc AB$.\hfill§~289

$\therefore$ the $\angle M$ may be inscribed in the segment $AKB$.

\hfill\qef

\end{proof}
\scanpage{136.png}%


\subsection{SOLUTION OF PROBLEMS.}

\pp{If a problem is so simple that the solution is obvious
from a known theorem, we have only to make the construction
according to the theorem, and then give a synthetic proof,
if a proof is necessary, that the construction is correct, as in
the examples of the fundamental problems already given.}

\begin{point}%
But problems are usually of a more difficult type.
The application of known theorems to their solution is not
immediate, and often far from obvious. To discover the mode
of application is the first and most difficult part of the solution.
The best way to attack such problems is by a method
resembling the analytic proof of a theorem, called the \textbf{analysis}
of the problem.

1. \textbf{Suppose the construction made}, and let the figure represent
all parts concerned, both given and required.

2. Study the relations among the parts with the aid of
known theorems, and try to find some relation that will suggest
the construction.

3. If this attempt fails, introduce new relations by drawing
auxiliary lines, and study the new relations. If this attempt
fails, make a new trial, and so on till a clue to the right construction
is found.
\end{point}

\pp{A problem is \emph{determinate} if it has a \emph{definite} number
of solutions, \emph{indeterminate} if it has an \emph{indefinite} number of
solutions, and \emph{impossible} if it has \emph{no} solution. A problem is
sometimes determinate for certain relative positions or magnitudes
of the given parts, and indeterminate for other positions
or magnitudes of the given parts.}

\pp{The \textbf{discussion} of a problem consists in examining the
problem with reference to all possible conditions, and in determining
the conditions necessary for its solution.}
\scanpage{137.png}%

\begin{proofex}%
\obs{\textsc{Problem.} To construct a circle that
  shall pass through a given point and cut chords of a given length
  from two parallels.}

\figc{137aa147}{}
\textbf{Analysis.} Suppose the problem solved.  Let $A$ be the given
point, $BC$ and $DE$ the given parallels, $MN$ the given length, and
$O$ the centre of the required circle.

Since the circle cuts equal chords from two parallels its centre must
be equidistant from them.  Therefore, one locus for $O$ is $FG
\parallel$ to $BC$ and equidistant from $BC$ and $DE$.

Draw the $\perp$ bisector of $MN$, cutting $FG$ in $P$.  $PM$ is the
radius of the circle required.  With $A$ as centre and radius $PM$
describe an arc cutting $FG$ at $O$.  Then $O$ is the centre of the
required circle.

\textbf{Discussion.} The problem is impossible if the distance from
$A$ to $FG$ is greater than $PM$.

\end{proofex}

\figc{137bb148}{}
\begin{proofex}%
\obs{\textsc{Problem.} To construct a triangle,
  having given the perimeter, one angle, and the altitude from the
  vertex of the given angle.}

\textbf{Analysis.} Suppose the problem solved, and let $ABC$ be the
$\triangle$ required, $ACB$ the given $\angle$, and $CD$ the given
altitude.

Produce $AB$ both ways, and take $AE=AC$, and $BF=BC$, then $EF=$ the
given perimeter.  Join $CE$ and $CF$, forming the isosceles
$\triangle_s CAE$ and $CBF$.

In the $\triangle ECF$, $\angle E+\angle F+\angle ECF=180°$ (why?),
but $\angle ECF=\angle ECA+\angle FCB+\angle ACB$.

Since $\angle E=\angle ECA$ and $\angle F=\angle FCB$, we have $\angle
ECF=\angle E+\angle F+\angle ACB$.\quad $\therefore 2\angle E+2\angle
F+\angle ACB = 180°$.

\( \therefore \angle E+\angle F+\frac{1}{2}\angle ACB = 90° \), and\
\( \angle E+\angle F = 90° - \frac{1}{2}\angle ACB \).

By substitution, \(\angle ECF = 90° + \frac{1}{2}\angle ACB \).

$\therefore \angle ECF$ is known.

\end{proofex}

\textbf{Construction.} To find the point $C$, construct on $EF$ a
segment that will contain the $\angle ECF$ (§~318), and draw a
parallel to $EF$ at the distance $CD$, the given altitude.

To find the points $A$ and $B$, draw the $\perp$ bisectors of the
lines $CE$ and $CF$, and the points $A$ and $B$ will be vertices of
the required $\triangle$.  Why?
\scanpage{138.png}%


\section[EXERCISES.]{PROBLEMS OF CONSTRUCTION.}

\ex{Find the locus of a point at a given distance from a given
circumference.}

\exheader{Find the locus of the centre of a circle:}

\ex{Which has a given radius $r$ and passes through a given point $P$.}

\ex{Which has a given radius $r$ and touches a given line $AB$.}

\ex{Which passes through two given points $P$ and $Q$.}

\ex{Which touches a given straight line $AB$ at a given point $P$.}

\ex{Which touches each of two given parallels.}

\ex{Which touches each of two given intersecting lines.}

\begin{proofex}%
To find in a given line a point $X$ which is equidistant from
two given points.

The required point is the intersection of the given line with the perpendicular
bisector of the line joining the two given points (§~160).

\end{proofex}

\ex{To find a point $X$ equidistant from three given points.}

\figc{138aa158}{}
\ex{To find a point $X$ equidistant from two given
points and at a given distance from a third given point.}

\ex{To construct a circle which has a given radius
and passes through two given points.}

\ex{To find a point $X$ at given distances from two given points.}

\ex{To construct a circle which has its centre in a given line and
passes through two given points.}

\ex{To find a point $X$ equidistant from two given points and also
equidistant from two given intersecting lines (§§~160 and 162).}

\ex{To find a point $X$ equidistant from two given points and also
equidistant from two given parallel lines.}

\ex{To find a point $X$ equidistant from two given intersecting
lines and also equidistant from two given parallels.}

\figc{138bb165}{}
\ex{To find a point $X$ equidistant from two given
intersecting lines and at a given distance from a given point.}

\ex{To find a point $X$ which lies in one side of a
given triangle and is equidistant from the other two sides.}
\scanpage{139.png}%
% 169 = c
% 174 = d
% 175 = e
% 177 = f
% 176 = g

\filbreak
\figcc{139aa167}{139bb168}
\ex{A straight railway passes two miles from a
town.  A place is four miles from the town and one
mile from the railway. To find by construction the
places that answer this description.}

\ex{In a triangle $ABC$, to draw $DE$ parallel to
the base $BC$, cutting the sides of the triangle in $D$ and
$E$, so that $DE$ shall equal $DB + EC$ (§~162).}

\figc{139cc169}{}
\begin{proofex}%
To draw through two sides of a triangle a line
parallel to the third side so that the part intercepted between
the sides shall have a given length.
\null
\step{Take $BD = d$.}{}

\end{proofex}

\ex{Prove that the locus of the vertex of a right triangle, having
a given hypotenuse as base, is the circumference described upon the given
hypotenuse as diameter (§~290). }

\ex{Prove that the locus of the vertex of a triangle, having a given
base and a given angle at the vertex, is the arc which forms with the base
a segment capable of containing the given angle (§~318).}

\ex{Find the locus of the middle point of a chord of a given length
that can be drawn in a given circle.}

\ex{Find the locus of the middle point of a chord drawn from a
given point in a given circumference.}

\figccc{139dd174}{139ee175}{139gg176}
\ex{Find the locus of the middle point of a straight line drawn
from a given exterior point to a given circumference.}

\ex{A straight line moves so that it remains parallel to a given
line, and touches at one end a given circumference. Find the locus of
the other end.}

\ex{A straight rod moves so that its ends constantly
touch two fixed rods which are perpendicular to
each other.   Find the locus of its middle point. }
\scanpage{140.png}%

\ex{In a given circle let $AOB$ be a diameter, $OC$ any radius, $CD$
the perpendicular from $C$ to $AB$.   Upon $OC$ take $OM$ equal to $CD$. Find
the locus of the point $M$ as $OC$ turns about $O$.}

\figcc{139ff177}{140aa178}
\ex{To construct an equilateral triangle, having
given the radius of the circumscribed circle.}

\exheader{To construct on isosceles triangle, having given:}

\ex{The angle at the vertex and the base (§~160 and §~318).}

\ex{The base and the radius of the circumscribed circle.}

\ex{The base and the radius of the inscribed circle.}

\figc{140bb182}{}
\begin{proofex}%
The perimeter and the altitude.

Let $ABC$ be the $\triangle$ required, $EF$ the given perimeter.
The altitude $CD$ passes through the middle
of $EF$, and the  $\triangle_s AEC$, $BFC$ are isosceles.

\end{proofex}

\exheader{To construct a right triangle, having given:}

\ex{The hypotenuse and one leg.}

\ex{One leg and the altitude upon the hypotenuse.}

\ex{The median and the altitude drawn from the vertex of the
right angle.}

\ex{The hypotenuse and the altitude upon the hypotenuse.}

\ex{The radius of the inscribed circle and one leg.}

\ex{The radius of the inscribed circle and an acute angle.}

\ex{An acute angle and the sum of the legs.}

\ex{An acute angle and the difference of the legs.}

\figc{140cc191}{}
\ex{To construct an equilateral triangle, having
given the radius of the inscribed circle.}

\exheader{To construct a triangle, having given:}

\ex{The base, the altitude, and an angle at the base.}

\ex{The base, the altitude, and the $\angle$ at the vertex.}

\ex{The base, the corresponding median, and the $\angle$ at the vertex.}

\ex{The perimeter and the angles.}

\ex{One side, an adjacent $\angle$, and the sum of the other sides.}
\scanpage{141.png}%

\exheader{To construct a triangle, having given:}

\ex{One side, an adjacent $\angle$, and the difference of the other sides.}

\ex{The sum of two sides and the angles.}

\ex{One side, an adjacent $\angle$, and the radius of the circumscribed
circle.}

\ex{The angles and the radius of the circumscribed circle.}

\ex{The angles and the radius of the inscribed circle. }

\ex{An angle, and the bisector and the altitude drawn from the
vertex of the given angle.}

\ex{Two sides and the median corresponding to the other side.}

\ex{The three medians.}

\exheader{To construct a square, having given:}

\ex{The diagonal.}

\begin{proofex}%
The sum of the diagonal and one side.

Let $ABCD$ be the square required, $CA$ the diagonal.
Produce $CA$, making $AE = AB$. $\triangle_s ABC$ and $ABE$ are
isosceles and $\angle BAC = \angle BCA = 45°$.

\end{proofex}

\figcc{141aa206}{141bb207}
\ex{Given two perpendiculars, $AB$ and $CD$,
intersecting in $O$, and a straight line intersecting
these perpendiculars in $E$ and $F$; to construct a
square, one of whose angles shall coincide with one
of the right angles at $O$, and the vertex of the opposite
angle of the square shall lie in $EF$. (Two solutions.)}

\exheader{To construct a rectangle, having given:}

\ex{One side and the angle between the diagonals.}

\ex{The perimeter and the diagonal.}

\ex{The perimeter and the angle between the diagonals.}

\ex{The difference of two adjacent sides and the angle between
the diagonals.}

\exheader{To construct a rhombus, having given:}

\ex{The two diagonals.}

\ex{One side and the radius of the inscribed circle.}
\scanpage{142.png}%

\ex{One angle and the radius of the inscribed circle.}

\ex{One angle and one of the diagonals.}

\exheader{To construct a rhomboid, having given:}

\ex{One side and the two diagonals.}

\ex{The diagonals and the angle between them.}

\ex{One side, one angle, and one diagonal.}

\ex{The base, the altitude, and one angle.}

\exheader{To construct an isosceles trapezoid, having given:}

\ex{The bases and one angle.}

\ex{The bases and the altitude.}

\ex{The bases and the diagonal.}

\figc{142aa223}{}
\begin{proofex}%
The bases and the radius of the circumscribed circle.

Let $ABCD$ be the isosceles trapezoid required, $O$ the
centre of the circumscribed $\odot$. A diameter $\perp$ to $AB$ is $\perp$ to
$CD$, and bisects both $AB$ and $CD$. Draw $CG$ $\parallel$ to $FE$.
Then $EG = FC = \frac{1}{2}DC$.

\end{proofex}

\exheader{To construct a trapezoid, having given:}

\ex{The four sides.}

\ex{The two bases and the two diagonals.}

\ex{The bases, one diagonal, and the $\angle$ between the diagonals.}

\exheader{To construct a circle which has the radius $r$ and which also:}

\ex{Touches each of two intersecting lines $AB$ and $CD$.}

\ex{Touches a given line $AB$ and a given circle~$K$.}

\ex{Passes through a given point $P$ and touches a given line~$AB$.}

\ex{Passes through a given point~$P$ and touches a given circle~$K$.}

\exheader{To construct a circle which shall:}

\ex{Touch two given parallels and pass through a given point~$P$.}

\ex{Touch three given lines two of which are parallel.}

\ex{Touch a given line $AB$ at $P$ and pass through a given point~$Q$.}

\ex{Touch a given circle at $P$ and pass through a given point~$Q$.}

\ex{Touch two given lines and touch one of them at a given point~$P$.}
\scanpage{143.png}%

\ex{Touch a given line and touch a given circle at a point $P$.}

\ex{Touch a given line $AB$ at $P$ and also touch a given circle.}

\ex{To inscribe a circle in a given sector.}

\ex{To construct within a given circle three equal circles, so that
each shall touch the other two and also the given circle.}

\ex{To describe circles about the vertices of a given triangle as
centres, so that each shall touch the two others.}

\figc{143aa241}{}
\begin{proofex}%
To bisect the angle formed by two lines, without
producing the lines to their point of intersection.

Draw any line $EF \parallel$ to $BA$. Take $EG = EH$, and produce
$GH$ to meet $BA$ at $I$. Draw the $\perp$ bisector of $GI$.

\figccc{143bb242}{143cc243}{143dd244}


\end{proofex}

\ex{To draw through a given point $P$ between the sides of an
angle $BAC$ a line terminated by the sides of the angle and bisected at $P$.}

\begin{proofex}%
Given two points $P$, $Q$, and a line $AB$; to draw lines from $P$
and $Q$ which shall meet on $AB$ and make equal angles with $AB$.

Make use of the point which forms with $P$ a pair of points symmetrical
with respect to $AB$.

\end{proofex}

\ex{To find the shortest path from $P$ to $Q$ which shall touch a line~$AB$.}

\figc{143ee245}{}
\begin{proofex}%
To draw a common tangent to two given circles.

Let $r$ and $r'$ denote the radii of the circles, $O$ and $O'$ their centres.
With centre $O$ and radius
$r-r'$ describe a $\odot$.
From $O'$ draw the tangents
$O'M$, $O'N$. Produce
$OM$ and $ON$ to
meet the circumference
at $A$ and $C$. Draw the
radii $O'B$ and $O'D \parallel$,
respectively, to $OA$ and $OC$. Draw $AB$ and $CD$.

To draw the internal tangents use an auxiliary $\odot$ of radius $r + r'$.

\end{proofex}
\scanpage{144.png}%


\chapter{BOOK III\@.  PROPORTION\@.  SIMILAR POLYGONS.}

\section[THEORY OF PROPORTION.]{THE THEORY OF PROPORTION.}

\pp{A \indexbf{proportion} is an expression of equality between two
equal ratios; and is written in one of the following forms:
\[ a:b = c:d; \qquad a:b::c:d; \qquad \frac{a}{b}=\frac{c}{d}. \]
This proportion is read, ``$a$ is to $b$ as $c$ is to $d$''; or ``the ratio
of $a$ to $b$ is equal to the ratio of $c$ to $d$.''}

\begin{point}%
The \indexbf{terms} of a proportion are the four quantities compared;
the \emph{first} and \indexemph{third} terms are called the \indexbf{antecedents}, the
\emph{second} and \indexemph{fourth} terms, the \indexbf{consequents}; the \emph{first} and \emph{fourth}
terms, the \indexbf{extremes}, the \emph{second} and \emph{third} terms, the \indexbf{means}.

Thus, in the proportion $a : b = c : d$; $a$ and $c$ are the antecedents, $b$ and
$d$ the consequents, $a$ and $d$ the extremes, $b$ and $c$ the means.
\end{point}

\begin{point}%
The fourth proportional to three given quantities is the
fourth term of the proportion which has for its first three
terms the three given quantities \emph{taken in order.}

Thus, $d$ is the fourth proportional to $a$, $b$, and $c$ in the proportion
\[ a:b = c:d. \]
\end{point}

\begin{point}%
The quantities $a$, $b$, $c$, $d$, $e$, are said to be in \textbf{continued
proportion}\label{continuedprop}, if $a:b = b:c = c:d = d:e$.

If three quantities are in continued proportion, the second
is called the \indexbf{mean proportional} between the other two, and the
third is called the \textbf{third proportional} to the other two.

Thus, in the proportion $a:b = b:c$; $b$ is the mean proportional between
$a$ and $c$; and $c$ is the third proportional to $a$ and $b$.
\end{point}
\scanpage{145.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{In every proportion the product of the extremes is
equal to the product of the means.}

\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{}

\eq[\indent Then]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{§~323}

\eq[\indent Whence]{$ad$}{$= bc$.}{\qed}
\end{proof}


\proposition{Theorem.}

\begin{proof}%
\obs{The mean proportional between two quantities is
  equal to the square root of their product.}

\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= b:c}$.}{}

\eq[\indent Then]{$b^2$}{$= ac$.}{§~327}

Whence, extracting the square root,

\eq{$b$}{$= \sqrt{ac}$.}{\qed}
\end{proof}


\proposition{Theorem.}

\begin{proof}%
\obs{If the product of two quantities is equal to the
  product of two others, either two may be made the extremes of the
  proportion in which the other two are made the means.}

\eq[\indent\textbf{Let}]{$\mathbf{ad}$}{$\mathbf{= bc}$.}{}

\proveq{$a:b$}{$= c:d$}

Divide both members of the given equation by $bd$.

\eq[\indent Then]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{}

\eq[\indent Or]{$a:b$}{$= c:d$.}{\qed}

\end{proof}
\scanpage{146.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If four quantities are in proportion, they are in
proportion by \textnormal{\textbf{alternation}}\label{alternation}; that is, the first term is to
the third as the second is to the fourth.}

\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{}

\proveq{$a:c$}{$=b:d$.}

\eq[\indent Now]{$ad$}{$=bc$.}{§~327}

Divide each member of the equation by $cd$.

\eq[\indent Then]{$\dfrac{a}{c}$}{$= \dfrac{b}{d}$.}{}

\eq[\indent Or]{$a:c$}{$= b:d$.}{\qed}
\end{proof}

\proposition{Theorem.}

\begin{proof}%
\obs{If four quantities are in proportion, they are in
proportion by \textnormal{\indexbf{inversion}}; that is, the second term is to
the first as the fourth is to the third.}

\eq[\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{}

\proveq{$b:a$}{$= d:c$.}

\eq[\indent Now]{$bc$}{$= ad$.}{§~327}

Divide each member of the equation by $ac$.

\eq[\indent Then]{$\dfrac{b}{a}$}{$=\dfrac{d}{c}$.}{}

\eq[\indent Or]{$b:a$}{$= d:c$.}{\qed}
\end{proof}
\scanpage{147.png}%

\proposition{Theorem.}

\begin{proof}
\obs{If four quantities are in proportion, they are in
proportion by \textnormal{\textbf{composition}}\label{composition} that is, the sum of the first
two terms is to the second term as the sum of the last
two terms is to the fourth term.}

\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{}

\proveq{$a+b:b$}{$= c+d:d$.}

\eq[\indent Now]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{}

\eq[\indent Then]{$\dfrac{a}{b}+1$}{$= \dfrac{c}{d}+1$;}{}

\eq[that is,]{$\dfrac{a+b}{b}$}{$=\dfrac{c+d}{d}$.}{}

\eq[\indent Or]{$a+b:b$}{$= c+d:d$.}{}

\eq[\indent In like manner]{$a+b:a$}{$= c+d:c$.}{\qed}
\end{proof}

\proposition{Theorem.}

\begin{proof}%
\obs{If four quantities are in proportion, they are in
proportion by \textnormal{\textbf{division}}\label{division}; that is, the difference of the first
two terms is to the second term as the difference of the
last two terms is to the fourth term.}

\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{}

\proveq{$a-b:b$}{$= c-d:d$.}

\eq[\indent Now]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{}

\eq[\indent Then]{$\dfrac{a}{b}-1$}{$= \dfrac{c}{d}-1$;}{}

\eq[that is,]{$\dfrac{a-b}{b}$}{$= \dfrac{c-d}{d}$.}{}

\eq[\indent Or]{$a-b:b$}{$= c-d:d$.}{}

\eq[\indent In like manner]{$a-b:a$}{$= c-d:c$.}{\qed}

\end{proof}
\scanpage{148.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If four quantities are in proportion, they are in
proportion by \textnormal{\textbf{composition and division}}; that is, the sum
of the first two terms is to their difference as the sum of
the last two terms is to their difference.}

\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$}{}

\eq[\indent Then]{$\dfrac{a+b}{a}$}{$=\dfrac{c+d}{c}$.}{§~332}

\eq[\indent And]{$\dfrac{a-b}{a}$}{$=\dfrac{c-d}{c}$.}{§~333}

\eq[\indent Divide,]{$\dfrac{a+b}{a-b}$}{$=\dfrac{c+d}{c-d}$.}{}

\eq[\indent Or]{$a+b:a-b$}{$ = c+d:c-d$.}{\qed}


\end{proof}

\proposition{Theorem.}

\begin{proof}%
\obs{In a series of equal ratios, the sum of the antecedents
is to the sum of the consequents as any antecedent
is to its consequent.}

\step[\indent\textbf{Let}]{$\mathbf{a:b = c:d = e:f = g:h}$.}{}

\prove{$a+c+e+g : b+d+f+h = a:b$.}

\step[\indent Let]{$r = \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = \dfrac{g}{h}$.}{}

\step[\indent Then]{$a=br$, $c=dr$, $e=fr$, $g=hr$.}{}

\step[\indent And]{$a+c+e+g = ( b+d+f+h )r$.}{}

Divide by $( b+d+f+h )$.

\step[\indent Then]{$\dfrac{a+c+e+g}{b+d+f+h} = r = \dfrac{a}{b}$.}{}

\step[\indent Or]{$a+c+e+g : b+d+f+h = a:b$.}{\qed}

\end{proof}
\scanpage{149.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The products of the corresponding terms of two
or more proportions are in proportion.}

\eq[\indent\textbf{Let}]{$\mathbf{a:b = c:d, \; e:f}$}
 {$\mathbf{= g:h, \; k:l = m:n}$.}{}

\proveq{$aek:bfl$}{$= cgm:dhn$.}

\eq[\indent Now]{$\dfrac{a}{b} = \dfrac{c}{d}$, $\dfrac{e}{f}$}
  {$= \dfrac{g}{h}$, $\dfrac{k}{l} = \dfrac{m}{n}$.}{}

The products of the first members and of the second members
of these equations give

\eq{$\dfrac{aek}{bfl}$}{$= \dfrac{cgm}{dhn}$.}{}

\eq[\indent Or]{$aek:bfl$}{$= cgm:dhn$.}{\qed}

\end{proof}

\pp{\cor{If three quantities are in continued proportion,
the first is to the third as the square of the first is to the
square of the second.}}


\proposition{Theorem.}

\begin{proof}%
\obs{Like powers of the terms of a proportion are in
proportion.}

\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{}

\proveq{$a^n:b^n$}{$= c^n:d^n$.}

\eq[\indent Now]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{}

\eq[\indent Raise to the $n$th power,]
  {$\dfrac{a^n}{b^n}$}{$= \dfrac{c^n}{d^n}$.}{}

\eq[\indent Or]{$a^n:b^n$}{$= c^n:d^n$.}{\qed}

\end{proof}


\pp{\defn{\textbf{Equimultiples}\label{equimultiples} of two quantities are the products
obtained by multiplying each of them by the same number.
Thus, $ma$ and $mb$ are equimultiples of $a$ and $b$.}}
\scanpage{150.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Equimultiples of two quantities are in the same
ratio as the quantities themselves.}

\textbf{Let $a$~and~$b$ be any two quantities.}

\proveq{$ma : mb$}{$= a:b$.}

\eq[\indent Now]{$\dfrac{a}{b}$}{$= \dfrac{a}{b}$.}{}

Multiply both terms of the first fraction by $m$.

\eq[\indent Then]{$\dfrac{ma}{mb}$}{$= \dfrac{a}{b}$.}{}

\eq[\indent Or]{$ma:mb$}{$= a:b$.}{\qed}

\end{proof}



\begin{point}%
\textsc{Scholium.} In the treatment of proportion, it is assumed
that the \emph{quantities} involved are expressed by their
\emph{numerical measures}. It is evident that the ratio of two quantities
of the same kind may be represented by a fraction, if the
two quantities are expressed in \emph{integers} in terms of a \emph{common
unit}. If there is no unit in terms of which both quantities can
be expressed in \emph{integers}, it is still possible by taking the unit
of measure sufficiently small to find a fraction that will represent
the ratio \emph{to any required degree of accuracy.}~\hfill§~269

If we speak of the product of two quantities, it must be
understood that we mean simply \emph{the product of the numbers
which represent them when they are expressed in terms of a
common unit.}

In order that four quantities, $a$, $b$, $c$, $d$, may form a proportion,
$a$ and $b$ must be quantities of the same kind; and $c$ and $d$
must be quantities of the same kind; though $c$ and $d$ need not
be of the same kind as $a$ and $b$. In alternation, however, \emph{the
four quantities must be of the same kind.}
\end{point}
\scanpage{151.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If a line is drawn through two sides of a triangle
parallel to the third side, it divides those sides proportionally.}

\figc{151aa342}{In the triangle $ABC$, let $EF$ be drawn parallel to $BC$.}

\proveq{$EB : AE$}{$= FC: AF$.}

\textsc{Case 1.} \textit{When $AE$ and $EB$ \textup{(Fig.~1)} are commensurable.}

\textbf{Proof.} Find a common measure of $AE$ and $EB$, as $MB$.

Let $MB$ be contained $m$ times in $EB$, and $n$ times in $AE$.

\eq[\indent Then]{$EB:AE$}{$= m:n$.}{}

At the points of division on $BE$ and $AE$ draw lines $\parallel$ to $BC$.
These lines will divide $AC$ into $m + n$ equal parts, of which $FC$
will contain $m$, and $AF$ will contain $n$.~\hfill§~187

\eq{$\therefore FC:AF$}{$= m:n$.}{}

\eq{$\therefore EB:AE$}{$=FC:AF$.}{Ax.~1}


\textsc{Case 2.}  \textit{When $AE$ and $EB$ \textup{(Fig.~2)} are incommensurable.}

\textbf{Proof.} Divide $AE$ into any number of equal parts, and apply
one of these parts to $EB$ as many times as $EB$ will contain it.

Since $AE$ and $EB$ are incommensurable, a certain number
of these parts will extend from $E$ to some point $K$, leaving a
remainder $KB$ less than one of these parts. Draw $KH \parallel BC$.

\eq[\indent Then]{$EK:AE$}{$= FH:AF$}{Case~1}
\scanpage{152.png}%

By increasing the \emph{number} of equal parts into which $AE$ is
divided, we can make the \emph{length} of each part less than any
assigned value, however small, but not zero.

Hence, $KB$, which is less than one of these equal parts, has
zero for a limit.~\hfill§~275

And the corresponding segment $HC$ has zero for a limit.

Therefore, $EK$ approaches $EB$ as a limit,~\hfill§~271

and $FH$ approaches $FC$ as a limit.

\step{$\therefore$ the variable $\dfrac{EK}{AE}$ approaches
  $\dfrac{EB}{AE}$ as a limit,}{§~280}

\step{and the variable $\dfrac{FH}{AF}$ approaches $\dfrac{FC}{AF}$ as a limit.}{}

\step[\indent But]{$\dfrac{EK}{AE}$ is constantly equal to $\dfrac{FH}{AF}$}{Case~1}

\step{$\therefore \dfrac{EB}{AE} = \dfrac{FC}{AF}$.}{§~284}

\hfill\qed

\end{proof}

\begin{point}%
\cor[1]{One side of a triangle is to either part cut
off by a straight line parallel to the base as the other side is
to the corresponding part.}

\eq[\indent For]{$AE:EB$}{$=AF:FC$.}{}

\eq[\indent By composition,]{$AE+EB:AE$}{$= AF+FC:AF$.}{§~332}

\eq[\indent Or]{$AB:AE$}{$= AC:AF$.}{}
\end{point}

\begin{point}%
\cor[2]{If two lines are cut by any number of parallels
the corresponding intercepts are proportional.}

\figc{152aa344}{}
Draw $AN \parallel$ to $CD$. Then

\setlength{\eqalign}{.33\dentwidth}
\eq{$AL=CG$, $LM$}{$=GK$, $MN=KD$.}{§~180}

\eq[\indent Now]{$AH:AM$}{$=AF:AL=FH:LM$}{}

\eq{}{$=HB:MN$.}{§~343}

\eq[\indent Or]{$AF:CG$}{$=FH:GK=HB:KD$.}{}

\setlength{\eqalign}{.5\dentwidth}
\end{point}
\scanpage{153.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If a straight line divides two sides of a triangle proportionally, it is parallel to the third side.}

\figc{153aa345}{In the triangle $ABC$, let $EF$ be drawn so that}

\eq{$\dfrac{AB}{AE}$}{$= \dfrac{AC}{AF}$.}{}

\proveq{$EF$ is}{$\parallel$ to $BC$.}

\step[\indent\textbf{Proof.}]{From $E$ draw $EH \parallel$ to $BC$.}{}

\eq[\indent Then]{$AB:AE$}{$= AC:AH$,}{§~343}

\pnote{(one side of a triangle is to either part cut off by a line parallel to the base
as the other side to the corresponding part).}

\eq[\indent But]{$AB:AE$}{$= AC:AF$.}{Hyp.}

\eq{$\therefore AC:AF$}{$= AC:AH$.}{Ax.~1}

\eq{$\therefore AF$}{$= AH$.}{}

\step{$\therefore EF$ and $EH$ coincide.}{§~47}

\eq[\indent But]{$EH$ is}{$\parallel$ to $BC$.}{Const.}

\step{$\therefore EF$, which coincides with $EH$, is $\parallel$ to $BC$.}{\qed}

\end{proof}


\ex{Find the fourth proportional to $91$,~$65$, and~$133$.}

\ex{Find the mean proportional between $39$~and~$351$.}

\ex{Find the third proportional to $54$~and~$3$.}
\scanpage{154.png}%

\begin{point}%
If a given line $AB$ is divided at $M$, a point between
the extremities $A$ and $B$, it is said to be divided \textbf{internally}
into the segments $MA$ and $MB$; and if it is divided at $M'$,
a point in the prolongation of $AB$, it is said to be divided
\textbf{externally} into the segments $M'A$ and $M'B$.

\figc{154aa346}{}

In either case the segments are the \emph{distances} from the point
of division to the extremities of the line. If the line is divided
internally, the \emph{sum} of the segments is equal to the line; and
if the line is divided externally, the \emph{difference} of the segments
is equal to the line.

Suppose it is required to divide the given line $AB$ \textbf{internally
and externally in the same ratio}; as, for example, the ratio of
the two numbers $3$~and~$5$.

\figc{154bb346}{}

We divide $AB$ into $5 + 3$, or~$8$, equal parts, and take $3$~parts
from $A$; we then have the point $M$, such that

\eq{$MA:MB$}{$= 3:5$.}{(1)}

Secondly, we divide $AB$ into $5-3$, or~$2$, equal parts, and lay
off on the prolongation of $AB$, to the left of $A$, three of these
equal parts; we then have the point $M'$, such that

\eq{$M'A:M'B$}{$= 3:5$.}{(2)}

Comparing (1) and (2),

\eq{$MA:MB$}{$= M'A:M'B$.}{}
\end{point}

\pp{\defn{If a given straight line is divided internally and
externally into segments having the same ratio, the line is
said to be \indexbf{divided harmonically}.}}
\scanpage{155.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The bisector of an angle of a triangle divides the
opposite side into segments which are proportional to
the adjacent sides.}

\figc{155aa348}{Let $CM$ bisect the angle $C$ of the triangle $CAB$.}

\proveq{$MA:MB$}{$= CA:CB$.}

\textbf{Proof.} Draw $AE \parallel$ to $MC$, meeting $BC$ produced at $E$.

\eq[\indent Then]{$MA:MB$}{$= CE:CB$,}{§~342}

\pnote{(if a line is drawn through two sides of a $\triangle$ parallel to the third side, it
divides those sides proportionally).}

\eq[\indent Also,]{$\angle ACM$}{$= \angle CAE$,}{§~110}

\pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines);}

\eq[and]{$\angle BCM$}{$= \angle CEA$,}{§~112}

\pnote{(being ext.-int.~$\angle_s$ of $\parallel$ lines).}

\eq[\indent But]{$\angle ACM$}{$= \angle BCM$.}{Hyp.}

\eq{$\therefore \angle CAE$}{$= \angle CEA$.}{Ax.~1}

\eq{$\therefore CE$}{$=CA$.}{§~147}

Put $CA$ for its equal, $CE$, in the first proportion.

\eq[\indent Then]{$MA:MB$}{$=CA:CB$.}{\qed}


\end{proof}

\ex{In a triangle $ABC$, $AB=12$, $AC=14$, $BC=13$. Find the
segments of $BC$ made by the bisector of the angle $A$.}

\ex{In a triangle $CAB$, $CA=6$, $CB=12$, $AB=15$. Find the
segments of $AB$ made by the bisector of the angle $C$.}
\scanpage{156.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The bisector of an exterior angle of a triangle
divides the opposite side externally into segments which
are proportional to the adjacent sides.}

\figc{156aa349}{Let $CM'$ bisect the exterior angle $ACE$ of the triangle $CAB$, and
meet $BA$ produced at $M'$.}

\proveq{$M'A:M'B$}{$=CA:CB$.}

\step[\indent\textbf{Proof.}]{Draw $AF \parallel$ to $M'C$, meeting $BC$ at $F$.}{}

\eq[\indent Then]{$M'A:M'B$}{$CF:CB$.}{§~343}

\eq[\indent Now]{$\angle M'CE$}{$=\angle AFC$,}{§~112}

\eq[and]{$\angle M'CA$}{$=\angle CAF$,}{§~110}

\pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).}

\eq[\indent But]{$\angle M'CE$}{$=\angle M'CA$.}{Hyp.}

\eq{$\therefore \angle AFC$}{$= \angle CAF$.}{Ax.~1}

\eq{$\therefore CA$}{$=CF$.}{§~147}

Put $CA$ for its equal, $CF$, in the first proportion.

\eq[\indent Then]{$M'A:M'B$}{$=CA:CB$.}{\qed}

\end{proof}

\textbf{Question.} To what case does this theorem not apply? (See
\hyperref[page:69]{Ex.~41, page~\pageref{page:69}}.)

\pp{\cor{The bisectors of an interior angle and an exterior
angle at one vertex of a triangle meeting the opposite side
divide that side harmonically.}~\hfill§~347}
\scanpage{157.png}%


\section{SIMILAR POLYGONS.}

\begin{point}%
\defn{\indexbf{Similar polygons} are polygons that have their
homologous angles equal, and their homologous sides proportional.}

\figc{157aa351}{}

Thus, the polygons $ABCDE$ and $A'B'C'D'E'$ are similar, if
the $\angle_s A$, $B$, $C$, etc., are equal, respectively, to the $\angle_s A'$, $B'$, $C'$,
etc., and if

\step{$AB:A'B' = BC:B'C' = CD:C'D'$, etc.}{}
\end{point}

\pp{\defn{\indexbf{Homologous lines} in similar polygons are lines
similarly situated.}}

\pp{\defn{The ratio of any two homologous lines in similar
polygons, is called the \indexbf{ratio of similitude} of the polygons.}}

The primary idea of similarity is \textbf{likeness of form}. The two
conditions \emph{necessary} to similarity are:

\begin{myenum}
\item \emph{For every angle in one of the figures there must be an
equal angle in the other.}

\item \emph{The homologous sides must be proportional.}
\end{myenum}

Thus, $Q$ and $Q'$ are not similar; the homologous sides are
proportional, but the homologous angles are not equal. Also
$R$ and $R'$ are not similar; the homologous angles are equal,
but the sides are not proportional.

\figc{157bb353}{}

In the case of \emph{triangles}, either condition involves the other
(see §~354 and §~358).
\scanpage{158.png}%

\proposition{Theorem.}

\label{similar triangles}
\begin{proof}%
\obs{Two mutually equiangular triangles are similar.}

\figc{158aa354}{In the triangles $ABC$ and $A'B'C'$, let the angles $A$, $B$, $C$ be equal
to the angles $A'$, $B'$, $C'$, respectively.}

\prove{the $\triangle_s ABC$ and $A'B'C'$ are similar.}

Since the $\triangle_s$ are mutually equiangular, we have only to
prove that

\step{$AB:A'B' = AC:A'C' = BC:B'C'$.}{§~351}

\textbf{Proof.}  Place the $\triangle A'B'C'$ on the $\triangle ABC$ so that $\angle A'$ shall
coincide with its equal, the $\angle A$; and $B'C'$ take the position $EH$.

\eq[\indent Then]{$\angle AEH$}{$= \angle B$}{Hyp.}

\eq{$\therefore EH$ is}{$\parallel$ to $BC$.}{§~114}

\eq{$\therefore AB:AE$}{$= AC:AH$.}{§~343}

\eq[\indent That is,]{$AB:A'B'$}{$= AC:A'C'$.}{}

Similarly, by placing $\triangle A'B'C'$ on $\triangle ABC$, so that $\angle B'$
 shall coincide with its equal, the $\angle B$, we may prove that

\eq{$AB:A'B'$}{$= BC:B'C'$}{\qed}

\end{proof}

\pp{\cor[1]{Two triangles are similar if two angles of the
one are equal, respectively, to two angles of the other.}}

\pp{\cor[2]{Two right triangles are similar if an acute
angle of the one is equal to an acute angle of the other.}}
\scanpage{159.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two triangles have an angle of the one equal
to an angle of the other, and the including sides proportional,
they are similar}.

\figc{159aa357}{In the triangles $ABC$ and $A'B'C'$, let $\angle A$ = $\angle A'$, and let}

\eq{$\mathbf{AB : A'B'}$}{$\mathbf{= AC : A'C'}$.}{}

\prove{the $\triangle_s ABC$ and $A'B'C'$ are similar.}

In this case we prove the $\triangle_s$ similar by proving them mutually
equiangular.


\textbf{Proof.}  Place the $\triangle A'B'C'$ on the $\triangle ABC$, so that the $\angle A'$
shall coincide with its equal, the $\angle A$.

Then the $\triangle A'B'C'$ will take the position of $\triangle AEH$.


\eq[\indent Now]{$\dfrac{AB}{A'B'}$}{$=\dfrac{AC}{A'C'}$.}{Hyp.}

\eq[\indent That is,]{$\dfrac{AB}{AE}$}{$=\dfrac{AC}{AH}$.}{}

\step{$\therefore EH$ is $\parallel$ to $BC$,}{§~345}

\pnote{(if a line divides two sides of a $\triangle$ proportionally, it is $\parallel$ to the third side).}

\step{$\therefore\angle AEH = \angle B$, and $\angle AHE = \angle C$.}{§~112}

\step{$\therefore\triangle AEH$ is similar to $\triangle ABC$.}{§~354}

\step{$\therefore\triangle A'B'C'$ is similar to $\triangle ABC$.}{\qed}

\end{proof}
\scanpage{160.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two triangles have their sides respectively proportional,
they are similar}.

\figc{160aa358}{In the triangles $ABC$ and $A'B'C'$, let}

\step{$\mathbf{AB : A'B' = AC : A'C' = BC : B'C'}$.}{}

\prove{the $\triangle_s ABC$ and $A'B'C'$ are similar.}


\textbf{Proof.}   Upon $AB$ take $AE$ equal to $A'B'$, and upon $AC$ take
$AH$ equal to $A'C'$; and draw $EH$.

\eq[\indent Now]{$AB:A'B'$}{$= AC: A'C'$.}{Hyp.}

\eq[\indent Or, since]{$AE$}{$ = A'B'$ and $AH = A'C'$,}{}

\eq{$AB:AE$}{$= AC:AH$.}{}

\step{$\therefore\triangle_s ABC$ and $AEH$ are similar.}{§~357}

\eq{$\therefore AB:AE$}{$= BC:EH$;}{§~351}

\eq[that is,]{$AB:A'B'$}{$= BC:EH$.}{}

\eq[\indent But]{$AB:A'B'$}{$= BC: B'C'$.}{Hyp.}

\eq{$\therefore BC:EH$}{$= BC:B'C'$.}{Ax.~1}

\eq{$\therefore EH$}{$= B'C'$.}{}

\step{Hence, the $\triangle_s AEH$ and $A'B'C'$ are equal.}{§~150}

\step[\indent But]{$\triangle AEH$ is similar to $\triangle ABC$.}{}

\step{$\therefore\triangle A'B'C'$ is similar to $\triangle ABC$.}{\qed}

\end{proof}
\scanpage{161.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Two triangles which have their sides respectively
parallel, or respectively perpendicular, are similar.}

\figc{161aa359}{Let $ABC$ and $A'B'C'$ have their sides respectively parallel; and
$DEF$ and $D'E'F'$ have their sides respectively perpendicular.}

\prove{the $\triangle_s ABC$ and $A'B'C'$ are similar; and
that the $\triangle_s DEF$ and $D'E'F'$ are similar.}

\textbf{Proof.} 1.~Prolong $BC$ and $AC$ to $B'A'$, forming $\angle_s x$ and $y$.

\step{Then $\angle B' = \angle x$ (§~112), and $\angle B = \angle x$.}{§~110}

\eq[\indent Therefore,]{$\angle B'$}{$= \angle B$}{Ax.~1}

\eq[\indent In like manner,]{$\angle A'$}{$=  \angle A$.}{}

\step{Therefore, $\triangle A'B'C'$ is similar to $\triangle ABC$.}{§~355}

2.~Prolong $DE$ and $FD$ to meet $D'E'$ at $H$ and $D'F'$ at $K$.

The quadrilateral $EHE'O$ has $\angle_s EHE'$ and $E'OE$ right
angles, by hypothesis.

\setlength{\eqalign}{.33\dentwidth}
\eq[\indent Therefore,]{}{$\angle E'$ and $\angle OEH$ are supplementary.}{§~206}

\eq[\indent But]{}{$\angle DEF$ and $\angle OEH$ are supplementary.}{§~86}

\setlength{\eqalign}{.5\dentwidth}

\eq{Therefore, $\angle DEF$}{$= \angle E'$.}{§~85}

\eq[\indent In like manner,]{$\angle EDF$}{$= \angle D'$.}{}

\step{Therefore, $\triangle DEF$ is similar to $\triangle D'E'F'$.}{§~355}

\hfill\qed


\end{proof}


\pp{\cor{The parallel sides and the perpendicular sides
are homologous sides of the triangles.}}
\scanpage{162.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The homologous altitudes of two similar triangles
have the same ratio as any two homologous sides.}

\figc{162aa361}{In the two similar triangles $ABC$ and $A'B'C'$, let $CO$ and $C'O'$ be
homologous altitudes.}

\prove{$\dfrac {CO}{C'O'}=\dfrac {AC}{A'C'}=\dfrac {AB}{A'B'}=\dfrac {BC}{B'C'}$.}

\textbf{Proof.}  In the rt.~$\triangle_s COA$ and $C'O'A'$,

\step{$\angle A = \angle A'$,}{§~351}

\pnote{(being homologous $\triangle_s$ of the similar $\triangle_s ABC$ and $A'B'C'$).}

\step{$\therefore\triangle_s COA$ and $C'O'A'$ are similar,}{§~356}

\pnote{(two rt.~$\triangle_s$ having an acute $\angle$ of the one equal to an acute $\angle$ of the other
are similar).}

\step{$\therefore\dfrac {CO}{C'O'}=\dfrac {AC}{A'C'}$.}{§~351}

In the similar $\triangle_s ABC$ and $A'B'C'$,

\step{$\dfrac {AC}{A'C'}=\dfrac {AB}{A'B'}=\dfrac {BC}{B'C'}$.}{§~351}

\step[Therefore,]{$\dfrac {CO}{C'O'}=\dfrac {AC}{A'C'}=
  \dfrac {AB}{A'B'}=\dfrac {BC}{B'C'}$.}{\qed}


\end{proof}

\ex{The base and altitude of a triangle are $7$~feet $6$~inches and
$5$~feet $6$~inches, respectively. If the homologous base of a similar triangle
is $5$~feet $6$~inches, find its homologous altitude.}
\scanpage{163.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two parallels are cut by three or more transversals
that pass through the same point, the corresponding
segments are proportional}.

\figc{163aa362}{Let the two parallels $AE$ and $A'E'$ be cut by the transversals $OA$,
$OB$, $OC$, $OD$, $OE$ in $A$, $A'$, $B$, $B'$, etc.}

\prove{$\dfrac{AB}{A'B'}= \dfrac{BC}{B'C'}= \dfrac{CD}{C'D'}= \dfrac{DE}{D'E'}$.}

\textbf{Proof.}   Since $A'E'$ is $\parallel$ to $AE$, the pairs of  $\triangle_s OAB$ and $OA'B'$,
$OBC$ and $OB'C'$, etc., are similar.~\hfill§~354

\eq{$\therefore\dfrac{AB}{A'B'} = \dfrac{OB}{OB'}$}
  {and $\dfrac{BC}{B'C'} = \dfrac{OB}{OB'}$.}{§~351}

\pnote{(homologous sides of similar $\triangle_s$ are proportional).}

\eq{$\therefore \dfrac{AB}{A'B'}$}{$= \dfrac{BC}{B'C'}$.}{Ax.~1}

In a similar way it may be shown that

\eq{$\dfrac{BC}{B'C'} = \dfrac{CD}{C'D'}$}
  {and $\dfrac{CD}{C'D'} = \dfrac{DE}{D'E'}$.}{\qed}


\end{proof}

\note{A condensed form of writing the above is
\par
\step{\( \dfrac{AB} {A'B'}=\left(\dfrac{OB} {OB'}\right)=\dfrac{BC} {B'C'}=\left(\dfrac{OC} {OC'}\right)=\dfrac{CD} {C'D'}=\left(\dfrac{OD} {OD'}\right)=\dfrac{DE} {D'E'} \). }{}
\par
A parenthesis about a ratio signifies that this ratio is used to prove the
equality of the ratios immediately preceding and following it.}
\scanpage{164.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{\textsc{Conversely:} If three or more non-parallel straight
lines intercept proportional segments upon two parallels,
they pass through a common point.}

\figc{164aa363}{Let $AB$, $CD$, $EF$ cut the parallels $AE$ and $BF$ so that}

\eq{$\mathbf{AC : BD}$}{$\mathbf{= CE : DF}$.}{}

\prove{$AB$, $CD$, $EF$ prolonged meet in a point.}

\textbf{Proof.}   Prolong $AB$ and $CD$ until they meet in $O$.

\step{Draw $OE$.}{}

\step{Designate by $F'$ the point where $OE$ cuts $BF$.}{}

\eq[\indent Then]{$AC:BD$}{$=CE:DF'$.}{§~362}

\eq[\indent But]{$AC:BD$}{$=CE:DF$.}{Hyp.}

\eq{$\therefore CE:DF'$}{$= CE:DF$.}{Ax.~1}

\eq{$\therefore DF'$}{$= DF$.}{}

\step{$\therefore F'$ coincides with $F$.}{}

\step{$\therefore EF$ coincides with $EF'$.}{§~47}

\step{$\therefore EF$ prolonged passes through $O$.}{}

\step{$\therefore AB$, $CD$, and $EF$ prolonged meet in the point $O$.}{\qed}

\end{proof}
\scanpage{165.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The perimeters of two similar polygons have the
same ratio as any two homologous sides.}

\figc{165aa364}{Let the two similar polygons be $ABCDE$ and $A'B'C'D'E'$, and let
$P$ and $P'$ represent their perimeters.}

\proveq{$P:P'$}{$= AB: A'B'$.}

\step[\indent\textbf{Proof.}]{$AB : A'B' = BC : B'C' = CD : C'D'$, etc.}{§~351}

\step{$\therefore AB + BC + \text{etc.}\ : A'B' + B'C' + \text{etc.}\ = AB : A'B'$,}{§~335}

\pnote{(in a series of equal ratios the sum of the antecedents is to the sum of the
consequents as any antecedent is to its consequent).}

\eq[\indent That is,]{$P : P'$}{$= AB : A'B'$.}{\qed}

\end{proof}

\ex{If the line joining the middle points of the bases of a trapezoid
is produced, and the two legs are also produced, the three lines will
meet in the same point.}

\ex{$AB$ and $AC$ are chords drawn from any point $A$ in the circumference
of a circle, and $AD$ is a diameter. The tangent to the circle
at $D$ intersects $AB$ and $AC$ at $E$ and $F$, respectively. Show that the
triangles $ABC$ and $AEF$ are similar.}

\ex{$AD$ and $BE$ are two altitudes of the triangle $CAB$. Show
that the triangles $CED$ and $CAB$ are similar.}

\ex{If two circles are tangent to each other, the chords formed
by a straight line drawn through the point of contact have the same ratio
as the diameters of the circles.}
\scanpage{166.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two polygons are similar, they are composed
of the same number of triangles, similar each to each,
and similarly placed.}

\figc{166aa365}{Let the polygons $ABCDE$ and $A'B'C'D'E'$ be similar.}

From two homologous vertices, as $E$ and $E'$, draw diagonals
$EB$, $EC$, and $E'B'$, $E'C'$.

\prove{the $\triangle_s EAB$, $EBC$, $ECD$ are similar,
respectively, to the $\triangle_s E'A'B'$, $E'B'C'$, $E'C'D'$.}

\textbf{Proof.} The $\triangle_s EAB$ and $E'A'B'$ are similar.~\hfill§~357

\eq[\indent For]{$\angle A$}{$= \angle A'$,}{§~351}

\eq[and]{$AE:A'E'$}{$=AB:A'B'$.}{§~351}

\eq[\indent Now]{$\angle ABC$}{$= \angle A'B'C'$,}{§~351}

\eq[and]{$\angle ABE$}{$=\angle A'B'E'$.}{§~351}

%proofrule

\eq[\indent By subtracting,]{$\angle EBC$}{$=\angle E'B'C'$.}{Ax.~3}

\eq[\indent Now]{$EB:E'B'$}{$=AB:A'B'$}{§~351}

\eq[and]{$BC:B'C'$}{$=AB:A'B'$}{§~351}

\eq{$\therefore EB:E'B'$}{$=BC:B'C'$.}{Ax.~1}

\step{$\therefore \triangle_s EBC$ and $E'B'C'$ are similar.}{§~357}

In like manner $\triangle_s ECD$ and $E'C'D'$ are similar.~\hfill\qed

\end{proof}
\scanpage{167.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{\textsc{Conversely:} If two polygons are composed of
the same number of triangles, similar each to each, and
similarly placed, the polygons are similar.}

\figc{167aa366}{In the two polygons $ABCDE$ and $A'B'C'D'E'$, let the triangles
$AEB$, $BEC$, $CED$ be similar, respectively, to the triangles $A'E'B'$,
$B'E'C'$, $C'E'D'$; and similarly placed.}

\prove{$ABCDE$ is similar to $A'B'C'D'E'$.}

\eq[\indent\textbf{Proof.}]{$\angle A$}{$= \angle A'$}{§~351}

\eq[\indent Also,]{$\angle ABE$}{$= \angle A'B'E'$,}{}

\eq[and]{$\angle EBC$}{$= \angle E'B'C'$.}{§~351}

%proofrule

\eq[\indent By adding,]{$\angle ABC$}{$= \angle A'B'C'$.}{Ax.~2}

In like manner, $\angle BCD = \angle B'C'D'$, $\angle CDE=\angle C'D'E'$, etc.

\step{Hence, the polygons are mutually equiangular.}{}

Also, \( \dfrac{AB}{A'B'} = \left(\dfrac{EB}{E'B'}\right) =
  \dfrac{BC}{B'C'} = \left(\dfrac{EC}{E'C'}\right) =
  \dfrac{CD}{C'D'} \), etc.~\hfill§~351


Hence, the polygons have their homologous sides proportional.

\step{Therefore, the polygons are similar.}{§~351}

\hfill\qed

\end{proof}
\scanpage{168.png}%


\pagebreak
\section[EXERCISES.]{THEOREMS.}

\ex{If two circles are tangent to each other externally, the corresponding
segments of two lines drawn through the point of contact and
terminated by the circumferences are proportional.}

\ex{In a parallelogram $ABCD$, a line $DE$ is drawn, meeting the }
diagonal $AC$ in $F$, the side $BC$ in $G$, and the side $AB$ produced in $E$.
Prove that $\overline{DF}^2 = FG × FE$.

\ex{Two altitudes of a triangle are inversely proportional to the
corresponding bases.}

\ex{Two circles touch at $P$. Through $P$ three lines are drawn,
meeting one circle in $A$, $B$, $C$, and the other in $A'$, $B'$, $C'$, respectively.
Prove that the triangles $ABC$, $A'B'C'$ are similar.}

\begin{proofex}%
Two chords $AB$, $CD$ intersect at $M$, and $A$ is the middle point
of the arc $CD$.   Prove that the product $AB × AM$ is constant if the chord
$AB$ is made to turn about the fixed point $A$.

Draw the diameter $AE$, and draw $BE$.

\end{proofex}

\begin{proofex}%
If two circles touch each other, their common external tangent
is the mean proportional between their diameters.

Let $AB$ be the common tangent. Draw the diameters $AC$, $BD$. Join
the point of contact $P$ to $A$, $B$, $C$, and $D$. Show that $APD$ and $BPC$ are
straight lines $\perp$ to each other, and that $\triangle_s CAB$, $ABD$ are similar.

\end{proofex}


\begin{proofex}%
If two circles are tangent internally, all chords of the greater
circle drawn from the point of contact are divided proportionally by the
circumference of the smaller circle.

Draw any two of the chords, and join the points where they meet the
circumferences. The $\triangle_s$ thus formed are similar (Ex.~120).

\end{proofex}


\figc{168aa263}{}
\begin{proofex}%
In an inscribed quadrilateral, the product of
the diagonals is equal to the sum of the products of the
opposite sides.

Draw $DE$, making $\angle CDE = \angle ADB$. The $\triangle_s ABD$ and
$ECD$ are similar; and the $\triangle_s BCD$ and $AED$ are similar.

\end{proofex}

\ex{Two isosceles triangles with equal vertical angles are similar.}

\begin{proofex}%
The bisector of the vertical angle $A$ of the triangle $ABC$ intersects
the base at $D$ and the circumference of the circumscribed circle at $E$.

Show that $ AB × AC = AD × AE$.

\end{proofex}
\scanpage{169.png}%

\clearpage
\section{NUMERICAL PROPERTIES OF FIGURES.}

\proposition{Theorem.}
\label{160}

\begin{proof}%
\obs{If in a right triangle a perpendicular is drawn
from the vertex of the right angle to the hypotenuse:}

\begin{myenum}
\item \emph{The triangles thus formed are similar to the given
triangle, and to each other.}

\item \emph{The perpendicular is the mean proportional between
the segments of the hypotenuse.}

\item \emph{Each leg of the right triangle is the mean proportional
between the hypotenuse and its adjacent segment.}
\end{myenum}

\figc{169aa367}{In the right triangle $ABC$, let $CF$ be drawn from the vertex of
the right angle $C$, perpendicular to $AB$.}

\prove[\textup{\textbf{1.~}}To prove that ]{$\triangle$'s $BCA$, $CFA$, $BFC$ are similar.}

\textbf{Proof.} The rt.~$\triangle_s CFA$ and $BCA$ are similar,~\hfill§~356

\step{since the $\angle a'$ is common.}{}

The rt.~$\triangle_s BFC$ and $BCA$ are similar,~\hfill§~356

\step{since the $\angle b$ is common.}{}

Since the $\triangle_s CFA$ and $BFC$ are each similar to $\triangle BCA$,
they are similar to each other.~\hfill§~354

\proveq[\indent\textup{\textbf{2.~}}To prove that]{$AF:CF$}{$=CF:FB$.}

\textbf{Proof.} In the similar $\triangle_s CFA$ and $BFC$,

\eq{$AF:CF$}{$=CF:FB$.}{§~351}
\scanpage{170.png}%

\filbreak
\label{161}

\proveq[\indent\textup{\textbf{3.~}}To prove that]{$AB:AC$}{$= AC:AF$,}

\eq[\emph{and}]{$AB:BC$}{$= BC:BF$.}{}

\textbf{Proof.} In the similar $\triangle_s BCA$ and $CFA$,

\eq{$AB:AC$}{$= AC:AF$}{§~351}

In the similar $\triangle_s BCA$ and $BFC$,

\eq{$AB:BC$}{$=BC:BF$.}{§~351}

\hfill\qed

\end{proof}

\begin{point}%
\cor[1]{The squares of the two legs of a right triangle
are proportional to the adjacent segments of the hypotenuse.}

From the proportions in §~367,3,

\step{$\overline{AC}^2=AB × AF$, and $\overline{BC}^2=AB × BF$.}{§~327}

\step[\indent Hence,]{\( \dfrac{\overline{AC}^2}{\overline{BC}^2} =
  \dfrac{AB × AF}{AB × BF} =
  \dfrac{AF}{BF} \).}{}
\end{point}

\begin{point}%
\cor[2]{The squares of the hypotenuse and either leg
are proportional to the hypotenuse and the adjacent segment.}

\step[\indent For]{\( \dfrac{\overline{AB}^2}{\overline{AC}^2} =
  \dfrac{AB × AB}{AB × AF} =
  \dfrac{AB}{AF} \).}{}
\end{point}

\figc{170aa370}{}
\begin{point}%
\cor[3]{The perpendicular from any point in the
circumference to the diameter of a circle
is the mean proportional between the segments
of the diameter.

The chord drawn from any point in
the circumference to either extremity of
the diameter is the mean proportional between the diameter
and the adjacent segment.}

\step[\indent For]{the $\angle ACB$ is a rt.~$\angle$.}{§~290}
\end{point}
\scanpage{171.png}%

\proposition{Theorem.}
\label{162}

\begin{proof}%
\obs{The sum of the squares of the two legs of a right
triangle is equal to the square of the hypotenuse.}

\figc{171aa371}{Let $ABC$ be a right triangle with its right angle at $C$.}

\proveq{$\overline{AC}^2 + \overline{CB}^2$}{$= \overline{AB}^2$.}

\eq[\indent\textbf{Proof.}]{Draw $CF$}{$\perp$ to $AB$.}{}

\eq[\indent Then]{$\overline{AC}^2$}{$= AB × AF$,}{}

\eq[and]{$\overline{CB}^2$}{$= AB × BF$.}{§~367}

%proofrule

\eq[\indent By adding,]{$\overline{AC}^2 + \overline{CB}^2$}
  {$= AB(AF + BF) = \overline{AB}^2$}{\qed}

\end{proof}

\pp{\cor[1]{The square of either leg of a right triangle is
equal to the difference of the square of the hypotenuse and
the square of the other leg.}}

\figcc{171bb373}{171cc374}
\begin{point}%
\cor[2]{The  diagonal and a side of a
square are incommensurable.}

\step[\indent For]{$\overline{AC}^2 = \overline{AB}^2 + \overline{BC}^2 = 2 \overline{AB}^2$.}{}

\step{$\therefore AC = AB \sqrt{2}$.}{}
\end{point}

\pp{\defn{The \textbf{projection} of any line
upon a second line is the segment of
the second line included between the
perpendiculars drawn to it from the
extremities of the first line. Thus,
$PR$ is the projection of $CD$ upon $AB$.}}
\scanpage{172.png}%

\proposition{Theorem.}
\label{163}

\begin{proof}%
\obs{In any triangle, the square of the side opposite an
acute angle is equal to the sum of the squares of the
other two sides diminished by twice the product of one of
those sides by the projection of the other upon that side.}

\figc{172aa375}{Let $C$ be an acute angle of the triangle $ABC$, and $DC$ the projection
of $AC$ upon $BC$.}

\prove{$\overline{AB}^2 = \overline{BC}^2 + \overline{AC}^2 - 2 BC × DC$.}

\textbf{Proof.} If $D$ falls upon the base (Fig.~1),

\eq{$DB$}{$= BC - DC$.}{}

If $D$ falls upon the base produced (Fig.~2),

\eq{$DB$}{$= DC - BC$.}{}

In either case,

\step{$\overline{DB}^2 = \overline{BC}^2 + \overline{DC}^2 - 2 BC × DC$.}{}

Add $\overline{AD}^2$ to both sides of this equality, and we have

\step{\( \overline{AD}^2 + \overline{DB}^2 =
  \overline{BC}^2 + \overline{AD}^2 + \overline{DC}^2 - 2 BC × DC \).}{}

\eq[\indent But]{$\overline{AD}^2 + \overline{DB}^2$}{$= \overline{AB}^2$}{}

\eq[and]{$\overline{AD}^2 + \overline{DC}^2$}{$= \overline{AC}^2$}{§~371}

Put $\overline{AB}^2$ and $\overline{AC}^2$ for their equals in the above equality.

\step[\indent Then]{$\overline{AB}^2 = \overline{BC}^2 + \overline{AC}^2 - 2 BC × DC$.}{\qed}

\end{proof}
\scanpage{173.png}%

\proposition{Theorem.}
\label{164}

\begin{proof}%
\obs{In any obtuse triangle, the square of the side
opposite the obtuse angle is equal to the sum of the
squares of the other two sides increased by twice the
product of one of those sides by the projection\label{projection} of
the other upon that side.}

\figc{173aa376}{Let $C$ be the obtuse angle of the triangle $ABC$, and $CD$ be the projection
of $AC$ upon $BC$ produced.}

\prove{\( \overline{AB}^2 = \overline{BC}^2 + \overline{AC}^2 + 2 BC × DC \).}

\step[\indent\textbf{Proof.}]{$DB = BC + DC$.}{}

\step[\indent Squaring,]{\( \overline{DB}^2 =
  \overline{BC}^2 + \overline{DC}^2 + 2 BC × DC \).}{}

Add $\overline{AD}^2$ to both sides, and we have

\step{\( \overline{AD}^2 + \overline{DB}^2 =
  \overline{BC}^2 + \overline{AD}^2 + \overline{DC}^2 + 2 BC × DC \).}{}

\step[\indent But]{\( \overline{AD}^2 + \overline{DB}^2 = \overline{AB}^2 \text{, and }
  \overline{AD}^2 + \overline{DC}^2 = \overline{AC}^2 \).}{§~371}

Put $\overline{AB}^2$ and $\overline{AC}^2$ for their equals in the above equality.

\step[\indent Then]{\( \overline{AB}^2 =
  \overline{BC}^2 + \overline{AC}^2 + 2 BC × DC \).}{\qed}

\end{proof}

\note[1]{By the Principle of Continuity the last three theorems may
be included in one theorem. Let the student explain.}

\note[2]{The last three theorems enable us to compute the lengths of
the altitudes of a triangle if the lengths of the three sides are known.}
\scanpage{174.png}%

\proposition{Theorem.}

\begin{proof}%
1.~\obs{The sum of the squares of two sides of a triangle
is equal to twice the square of half the third side
increased by twice the square of the median upon that
side.}

2.~\obs{The difference of the squares of two sides of a triangle
is equal to twice the product of the third side by
the projection of the median upon that side.}

\figc{174aa377}{In the triangle $ABC$, let $AM$ be the median and $MD$ the projection
of $AM$ upon the side $BC$.   Also, let $AB$ be greater than $AC$.}

\proveq{\textup{1.} $\overline{AB}^2+\overline{AC}^2$}{$=
  2\overline{BM}^2+2\overline{AM}^2$.}

\proveq[]{\textup{2.} $\overline{AB}^2-\overline{AC}^2$}{$=
  2BC × MD$.}

\textbf{Proof.}   Since $AB > AC$, the $\angle AMB$ will be obtuse, and the
$\angle AMC$ will be acute.~\hfill§~155

\setlength{\eqalign}{.33\dentwidth}
\eq[\indent Then]{$\overline{AB}^2$}{$=\overline{BM}^2+\overline{AM}^2+2BM × MD$,}{§~376}

\eq[and]{$\overline{AC}^2$}{$=\overline{MC}^2+\overline{AM}^2-2MC × MD$.}{§~375}

\setlength{\eqalign}{.5\dentwidth}

Add these two equalities, and observe that $BM=MC$.


\eq[\indent Then]{$\overline{AB}^2+\overline{AC}^2$}{$=2 \overline{BM}^2+2 \overline{AM}^2$.}{}

Subtract the second equality from the first.

\eq[\indent Then]{$\overline{AB}^2-\overline{AC}^2$}{$=2 BC × MD$.}{\qed}

\end{proof}

\note{This theorem enables us to compute the lengths of the medians
of a triangle if the lengths of the three sides are known.}
\scanpage{175.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If two chords intersect in a circle, the product
of the segments of one is equal to the product of the
segments of the other.}

\figc{175aa378}{Let any two chords $MN$ and $PQ$ intersect at $O$.}

\proveq{$OM × ON$}{$= OQ × OP$.}

\step[\indent\textbf{Proof.}]{Draw $MP$ and $NQ$.}{}

\eq{$\angle a$}{$= \angle a'$,}{§~289}

\pnote{(each being measured by $\frac{1}{2} \arc PN$).}

\eq[\indent And]{}{$\angle c = \angle c'$,}{§~289}

\pnote{(each being measured by $\frac{1}{2} \arc MQ$).}

\step{$\therefore$ the $\triangle_s NOQ$ and $POM$ are similar.}{§~355}

\eq{$\therefore OQ:OM$}{$=ON:OP$.}{§~351}

\eq{$\therefore OM × ON$}{$= OQ × OP$.}{§~327}

\hfill\qed

\end{proof}

\begin{point}\textsc{Scholium.} This proportion may be written

\step{$\dfrac{OM}{OQ} = \dfrac{OP}{ON}$, or
  $\dfrac{OM}{OQ} = \dfrac{1}{\dfrac{ON}{OP}}$;}{}

that is, the ratio of two corresponding segments is equal to the
\emph{reciprocal} of the ratio of the other two segments.

Hence, these segments are said to be \emph{reciprocally proportional}.
\end{point}
\scanpage{176.png}%

\pp{\defn{\textbf{A secant from a point to a circle}\label{secant2} is understood to
mean the segment of the secant lying between the point and
the \emph{second point} of intersection of the secant and circumference.}}

\proposition{Theorem.}

\begin{proof}%
\obs{If from a point without a circle a secant and a
tangent are drawn, the tangent is the mean proportional
between the whole secant and its external segment.}

\figc{176aa381}{Let $AD$ be a tangent and $AC$ a secant drawn from the point $A$ to
the circle $BCD$.}

\prove{$AC : AD = AD : AB$.}

\step[\indent\textbf{Proof.}]{Draw $DC$ and $DB$.}{}

\step{The $\triangle_s ADC$ and $ABD$ are similar.}{§~355}

\step{For $\angle b$ is common; and $\angle a' = \angle a$,}{§§~289,~295}

\pnote{(each being measured by $\frac{1}{2} \arc BD$).}

\step{$\therefore AC : AD = AD : AB$.}{§~351}

\hfill\qed

\end{proof}

\begin{point}%
\cor{If from a fixed point without a circle a secant
is drawn, the product of the secant and its external segment
is constant in whatever direction the secant is drawn.}

\step[\indent For]{$AC × AB = \overline{AD}^2$.}{§~327}
\end{point}
\scanpage{177.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The square of the bisector of an angle of a triangle
is equal to the product of the sides of this angle
diminished by the product of the segments made by
the bisector upon the third side of the triangle.}

\figc{177aa383}{Let $NO$ bisect the angle $MNP$ of the triangle $MNP$.}

\prove{$\overline{NO}^2 = NM × NP - OM × OP$.}

\step[\indent\textbf{Proof.}]{Circumscribe the $\odot MNP$ about the $\triangle MNP$.}{§~314}

Produce $NO$ to meet the circumference in $Q$, and draw $PQ$.

\step{The $\triangle_s NQP$ and $NMO$ are similar.}{§~355}

\eq[\indent For]{$\angle b$}{$= \angle b'$}{Hyp.}

\eq[and]{$\angle a$}{$= \angle a'$}{§~289}

\eq[\indent Whence]{$NQ:NM$}{$= NP:NO$.}{§~351}

\eq{$\therefore NM × NP$}{$= NQ × NO$}{}

\eq{}{$= (NO+OQ)NO$}{}

\eq{}{$= \overline{NO}^2 + NO × OQ$.}{}

\eq[\indent But]{$NO × OQ$}{$= MO × OP$.}{§~378}

\eq{$\therefore MN × NP$}{$= \overline{NO}^2 + MO × OP$.}{}

\step[\indent Whence]{$\overline{NO}^2 = NM × NP = MO × OP$.}{Ax.~3}

\hfill\qed

\end{proof}

\note{This theorem enables us to compute the lengths of the bisectors
of the angles of a triangle if the lengths of the sides are known.}
\scanpage{178.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{In any triangle the product of two sides is equal
to the product of the diameter of the circumscribed circle
by the altitude upon the third side.}

\figc{178aa384}{Let NMQ be a triangle, NO the altitude, and QNMP the circle circumscribed
about the triangle NMQ.}

\step{Draw the diameter $NP$, and draw $PQ$.}{}

\prove{$NM × NQ = NP × NO$.}

\textbf{Proof.} In the $\triangle_s NOM$ and $NQP$,

\step{$\angle NOM$ is a rt.~$\angle$,}{Hyp.}

\step{$\angle NQP$ is a rt.~$\angle$,}{§~290}

\eq{and $\angle a$}{$= \angle a'$,}{§~289}

\pnote{(each being measured by $\frac{1}{2} \arc NQ$).}

\step{$\therefore \triangle_s NOM$ and $NQP$ are similar.}{§~356}

\eq[\indent Whence]{$NM:NP$}{$= NO:NQ$.}{§~351}

\eq{$\therefore NM × NQ$}{$= NP × NO$.}{§~327}

\hfill\qed

\end{proof}

\note{This theorem enables us to compute the length of the radius of
a circle circumscribed about a triangle, if the lengths of the three sides of
the triangle are known. }

\ex{If $OE$, $OF$, $OG$ are the perpendiculars from any point $O$
within the triangle $ABC$ upon the sides $AB$, $BC$, $CA$, respectively, show
that \( \overline{AE}^2 + \overline{BF}^2 +  \overline{CG}^2 =  \overline{EB}^2 +  \overline{FC}^2 +  \overline{GA}^2 \).}
\scanpage{179.png}%


\section[EXERCISES.]{THEOREMS.}

\begin{proofex}%
The sum of the squares of the segments of two perpendicular
chords is equal to the square of the diameter of the circle.

If $AB$, $CD$ are the chords, draw the diameter $BE$, draw $AC$, $ED$, $BD$.
Prove that $AC = ED$, and apply §~371.

\end{proofex}

\ex{The tangents to two intersecting circles drawn from any point
in their common chord produced, are equal. (§~381.)}

\ex{The common chord of two intersecting circles, if produced,
will bisect their common tangents. (§~381.)}

\figc{179aa270}{}
\begin{proofex}%
If three circles intersect one another, the common chords all
pass through the same point.

Let two of the chords $AB$ and $CD$ meet at $O$.   Join
the point of intersection $E$ to $O$, and suppose that $EO$
produced meets the same two circles at two different
points $P$ and $Q$.    Then prove that $OP = OQ$ (§~378);
hence, that the points $P$ and $Q$ coincide.

\end{proofex}

\ex{If two circles are tangent to each other, the common internal
tangent bisects the two common external tangents.}

\begin{proofex}%
If the perpendiculars from the vertices of the triangle $ABC$
upon the opposite sides intersect at $D$, show that

\step{\( \overline{AB}^2-\overline{AC}^2 = \overline{BD}^2-\overline{CD}^2 \).}{}

\end{proofex}

\ex{In an isosceles triangle, the square of a leg is equal to the
square of any line drawn from the vertex to the base, increased by the
product of the segments of the base.}

\ex{The squares of two chords drawn from the same point in a
circumference have the same ratio as the projections of the chords on the
diameter drawn from the same point.}

\ex{The difference of the squares of two sides of a triangle is
equal to the difference of the squares of the segments of the third side,
made by the perpendicular on the third side from the opposite vertex.}

\begin{proofex}%
$E$ is the middle point of $BC$, one of the parallel sides of the
trapezoid $ABCD$; $AE$ and $DE$ produced meet $DC$ and $AB$ produced at
$F$ and $G$, respectively. Show that $FG$ is parallel to $DA$.

$\triangle_s AGD$ and $BGE$ are similar; and $\triangle_s AFD$ and $EFC$ are similar.

\end{proofex}
\scanpage{180.png}%

\ex{If two tangents are drawn to a circle at the extremities of a
diameter, the portion of a third tangent intercepted between them is
divided at its point of contact into segments whose product is equal to the
square of the radius.}

\ex{If two exterior angles of a triangle are bisected, the line
drawn from the point of intersection of the bisectors to the opposite angle
of the triangle bisects that angle.}

\ex{The sum of the squares of the diagonals of a quadrilateral is
equal to twice the sum of the squares of the lines that join the middle
points of the opposite sides.}

\figcc{180aa280}{180bb281}
\begin{proofex}%
The sum of the squares of the four sides of any quadrilateral
is equal to the sum of the squares of the diagonals, increased
by four times the square of the line joining the
middle points of the diagonals.

Apply §~377 to the $\triangle_s$ $ABC$ and $ADC$, add the results,
and eliminate $\overline{BE}^2 + \overline{DE}^2$ by applying §~377 to the $\triangle BDE$.

\end{proofex}

\begin{proofex}%
The square of the bisector of an exterior angle of a triangle is
equal to the product of the external segments determined
by the bisector upon one of the sides, diminished by the
product of the other two sides.

Let $CD$ bisect the exterior $\angle BCH$ of the $\triangle ABC$.
$\triangle_s$ $ACD$ and $FCB$ are similar (§~355).    Apply §~382.

\end{proofex}

\ex{If a point $O$ is joined to the vertices of a triangle $ABC$;
through any point $A'$ in $OA$ a line parallel to $AB$ is drawn, meeting $OB$
at $B'$; through $B'$ a line parallel to $BC$, meeting $OC$ at $C'$; and $C'$ is
joined to $A'$; the triangle $A'B'C'$ is similar to the triangle $ABC$.}

\ex{If the line of centres of two circles meets the circumferences
at the consecutive points $A$, $B$, $C$, $D$, and meets the common external tangent
at $P$, then $PA × PD = PB × PC$.}

\begin{proofex}%
The line of centres of two circles meets the common external
tangent at~$P$, and a secant is drawn from~$P$, cutting the circles at the
consecutive points $E$, $F$, $G$,~$H$. Prove that $PE × PH = PF × PG$.

Draw radii to the points of contact, and to $E$, $F$, $G$, $H$. Let fall $\perp_s$ on
$PH$ from the centres of the $\odot_s$. The various pairs of $\triangle_s$ are similar.

\end{proofex}

\ex{If a line drawn from a vertex of a triangle divides the opposite
side into segments proportional to the adjacent sides, the line bisects
the angle at the vertex.}
\scanpage{181.png}%

\clearpage
\section{PROBLEMS OF CONSTRUCTION.}

\proposition{Problem.}

\begin{proof}%
\obs{To divide a given straight line into parts
  proportional to any number of given lines.}

\figc{181aa385}{Let $AB$, $m$, $n$, and $p$ be given straight lines.}

\prove[To divide ]{$AB$ into parts proportional to $m$, $n$, and $p$.}

\step{Draw $AX$, making any convenient $\angle$ with $AB$.}{}

\step{On $AX$ take $AC$ equal to $m$, $CE$ to $n$, $EF$ to $p$.}{}

\step{Draw $BF$.}{}

\step{From $E$ and $C$ draw $EK$ and $CH \parallel$ to $FB$.}{}

\step{Through $A$ draw a line $\parallel$ to $BF$.}{}

\step{$K$ and $H$ are the division points required.}{}

\step[\indent\textbf{Proof.}]{$\dfrac{AH}{AC} = \dfrac{HK}{CE} = \dfrac{KB}{EF}$,}{§~344}

\pnote{(if two lines are cut by any number of parallels, the corresponding
intercepts are proportional).}

Substitute $m$, $n$, and $p$ for their equals $AC$, $CE$, and $EF$.

\step[\indent Then]{$\dfrac{AH}{m} = \dfrac{HK}{n} = \dfrac{KB}{p}$.}{\qef}
\end{proof}

\ex{Divide a line $12$~inches long into three parts
proportional to the numbers $3$,~$5$,~$7$.}
\scanpage{182.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To find the fourth proportional to three given
  straight lines.}

\figc{182aa386}{Let the three given lines be $m$, $n$, and $p$.}

\prove[To find ]{the fourth proportional to $m$, $n$, and $p$.}


\step {Draw $Ax$ and $Ay$ containing any convenient angle.} {}

\step {On $Ax$ take $AB$ equal to $m$, $BC$ to $n$.} {}

\step {On $Ay$ take $AD$ equal to $p$.} {}

\step {Draw $BD$.} {}

\step {From $C$ draw $CF \parallel$ to $BD$, meeting $Ay$ at $F$.} {}

\step {$DF$ is the fourth proportional required.} {}

\eq [\indent\textbf{Proof.}] {$AB:BC$ } {$ = AD:DF$,} {§~342}

\pnote{(a line drawn through two sides of a $\triangle \parallel$ to the third side divides those sides
    proportionally).}

\step{Substitute $m$, $n$, and $p$ for their equals $AB$, $BC$, and $AD$.} {}

\eq [\indent Then] {$m:n$} {$ = p:DF$} {\qef}



\end{proof}

\ex {The square of the altitude of an equilateral
triangle is equal to three fourths of the square of one side of the
triangle.}
\scanpage{183.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To find the third proportional to two given
straight lines.}

\figc{183aa387}{Let $m$ and $n$ be the two given straight lines.}

\prove[To find ]{the third proportional to $m$ and $n$.}

\step{Construct any convenient angle $A$,}{}

\step{and take $AB$ equal to $m$, $AC$ equal to $n$.}{}

\step{Produce $AB$ to $D$, making $BD$ equal to $AC$.}{}

\step{Draw $BC$.}{}

\step{Through $D$ draw $DE \parallel$ to $BC$, meeting $AC$ produced at $E$.}{}

\step{$CE$ is the third proportional required.}{}

\eq[\indent\textbf{Proof.}]{$AB:BD$}{$= AC:CE$,}{§~342}

\pnote{(a line drawn through two sides of a $\triangle$ parallel to the third side divides
those sides proportionally).}

Substitute, in the above proportion, $AC$ for its equal $BD$.

\eq[\indent Then]{$AB:AC$}{$= AC:CE$,}{}

\eq[that is,]{$m:n$}{$=n:CE$.}{\qef}

\end{proof}

\begin{proofex}%
Construct $x$, if (1) $x=\dfrac{ab}{c}$, (2) $x = \dfrac{a^2}{c}$.

Special cases: (1)~$a = 2$, $b = 8$, $c = 4$; (2)~$a = 3$, $b = 7$, $c = 11$; (3)~$a = 2$,
$c = 3$; (4)~$a = 3$, $c = 5$; (5)~$a = 2c$.

\end{proofex}
\scanpage{184.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To find the mean proportional between two given
straight lines.}

\figc{184aa388}{Let the two given lines be $m$ and $n$.}

\prove[To find ]{the mean proportional between $m$ and $n$.}

\step{On the straight line $AE$}{}

\step{take $AC$ equal to $m$, and $CB$ equal to $n$.}{}

\step{On $AB$ as a diameter describe a semicircumference.}{}

\step{At $C$ erect the $\perp$ $CH$ meeting the circumference at $H$.}{}

\step{$CH$ is the mean proportional between $m$ and $n$.}{}

\eq[\indent\textbf{Proof.}]{$AC:CH$}{$=CH:CB$}{§~370}

\pnote{(the $\perp$ let fall from a point in a circumference to the diameter of a circle is
the mean proportional between the segments of the diameter).}

Substitute for $AC$ and $CB$ their equals $m$ and $n$.

\eq[\indent Then]{$m:CH$}{$=CH:n$.}{\qef}

\end{proof}


\pp{\defn{A straight line is divided \textbf{in extreme and mean
ratio}\label{extrememean}, when one of the segments is the mean proportional
between the whole line and the other segment.}}

\begin{proofex}%
Construct $x$, if $ x=\sqrt{ab} $.

Special cases: (1)~$a = 2$, $b = 3$; (2)~$a = 1$, $b = 6$; (3)~$a = 3$, $b = 7$.

\end{proofex}
\scanpage{185.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To divide a given line in extreme and mean ratio.}

\figc{185aa390}{Let $AB$ be the given line.}

\prove[To divide ]{$AB$ in extreme and mean ratio.}

\step{At $B$ erect a $\perp BE$ equal to half of $AB$.}{}

\step{From $E$ as a centre, with a radius equal to $EB$, describe a $\odot$.}{}

\step{Draw $AE$, meeting the circumference in $F$ and $G$.}{}

\step{On $AB$ take $AC$ equal to $AF$.}{}

\step{On $BA$ produced take $AC'$ equal to $AG$.}{}

Then $AB$ is divided internally at $C$ and externally at $C'$ in
extreme and mean ratio.

\step{$AG:AB = AB:AF$.}{§~381}

\begin{center}
\begin{tabular}{r@{}l@{}l | r@{}l@{}l}
$\overline{AB}^2$& $= AF × AG$ &&
$\overline{AB}^2$& $= AG × AF$ \\

& $= AC(AF+AG)$ &&
& $= C'A(AG-AF)$ \\

& $= AC(AC+AB)$ &&
& $= C'A(C'A-AB)$ \\

& $= \overline{AC}^2 + AB × AC$. &&
& $= \overline{C'A}^2 - AB × C'A$. \\

& $\therefore \overline{AB}^2-AB × AC$ & $=\overline{AC}^2$. &
& $\therefore \overline{AB}^2+AB × C'A$ & $=\overline{C'A}^2$. \\

& $\therefore AB(AB-AC)$ & $=\overline{AC}^2$. &
& $\therefore AB(AB+C'A)$ & $=\overline{C'A}^2$. \\

& $\therefore AB × CB$ & $=\overline{AC}^2$. &
& $\therefore AB × C'B$ & $=\overline{C'A}^2$.
\end{tabular}
\end{center}

\hfill\qef

\end{proof}
\scanpage{186.png}%

\proposition{Problem.}

\begin{proof}%
\obs{Upon a given line homologous to a given side of
a given polygon, to construct a polygon similar to the
given polygon.}

\figc{186aa391}{Let $A'E'$ be the given line homologous to $AE$ of the given polygon
$ABCDE$.}

\prove[To construct ]{on $A'E'$ a polygon similar to the given polygon.}

\step{From $E$ draw the diagonals $EB$ and $EC$.}{}

\step{From $E'$ draw $E'B'$, $E'C'$, and $E'D'$,}{}

\step{making $ \triangle$'s $A'E'B'$, $B'E'C'$, and $C'E'D'$ equal, respectively, to}{}

\step{$ \triangle_s AEB$, $BEC$, and $CED$.}{}

\step{From $A'$ draw $A'B'$, making $\angle E'A'B'$ equal to $\angle EAB$,}{}

\step{and meeting $E'B'$ at $B'$.}{}

\step{From $B'$ draw $B'C'$, making $\angle E'B'C'$ equal to $\angle EBC$,}{}

\step{and meeting $E'C'$ at $C'$.}{}

\step{From $C'$ draw $C'D'$, making $\angle E'C'D'$ equal to $\angle ECD$,}{}

\step{and meeting $E'D'$ at $D'$.}{}

\step{Then $A'B'C'D'E'$ is the required polygon.}{}

\step[\indent\textbf{Proof.}]{The $\triangle_s ABE$, $A'B'E'$, etc., are similar.}{§~354}

\step{Therefore, the two polygons are similar.}{§~366}

\hfill\qef

\end{proof}
\scanpage{187.png}%


\section[EXERCISES.]{PROBLEMS OF CONSTRUCTION.}

\ex{To divide one side of a given triangle into segments proportional
to the adjacent sides (§~348).}

\figcccc{187aa291}{187bb292}{187cc293}{187dd294}

\begin{proofex}%
To find in one side of a given triangle a point whose distances
from the other sides shall be to each other in the given ratio $m : n$.

Take $AG = m \perp$ to $AC$, $GH=n \perp$ to $BC$.   Draw $CD \parallel$ to $OG$.

\end{proofex}

\ex{Given an obtuse triangle; to draw a line from the vertex of
the obtuse angle to the opposite side which shall be the mean proportional
between the segments of that side.}

\begin{proofex}%
Through a given point $P$ within a given circle to draw a chord
$AB$ so that the ratio $AP: BP$ shall equal the given ratio $m : n$.

Draw $OPC$ so that $OP:PC = n:m$.    Draw $CA$ equal to the fourth
proportional to $n$, $m$, and the radius of the circle.

\end{proofex}

\begin{proofex}%
To draw through a given point $P$ in the arc subtended by a
chord $AB$ a chord which shall be bisected by $AB$.

On radius $OP$ take $CD$ equal to $CP$.   Draw $DE \parallel$ to $BA$.
\end{proofex}

\figcccc{187ee295}{187ff296}{187gg297}{187hh298}
\begin{proofex}%
To draw through a given external point $P$ a secant $PAB$ to a
given circle so that the ratio $PA:AB$ shall equal the given ratio $m : n$.
\[ PD:DC = m:n. \quad PD:PA = PA:PC. \]

\end{proofex}

\begin{proofex}%
To draw through a given external point $P$ a secant $PAB$ to a
given circle so that $\overline{AB}^2 = PA × PB$.
\[ PC:CD = CD:PD. \quad PA = CD. \]

\end{proofex}
\scanpage{188.png}%

\ex{To find a point $P$ in the arc subtended by a given chord $AB$
so that the ratio $PA:PB$ shall equal the given ratio $m : n$.}

\ex{To draw through one of the points of intersection of two
circles a secant so that the two chords that are formed shall be in the
given ratio $m:n$.}

\ex{Having given the greater segment of a line divided in extreme
and mean ratio, to construct the line.}

\ex{To construct a circle which shall pass through two given points
and touch a given straight line.}

\ex{To construct a circle which shall pass through a given point
and touch two given straight lines.}

\ex{To inscribe a square in a semicircle.}

\figc{188aa303}{}
\begin{proofex}%
To inscribe a square in a given triangle.

Let $DEFG$ be the required inscribed square. Draw $CM \parallel$ to $AB$, meeting
$AF$ produced in $M$. Draw $CH$ and $MN \perp$ to $AB$, and
produce $AB$ to meet $MN$ at $N$. The $\triangle_s ACM$, $AGF$ are
similar; also, the $\triangle_s AMN$, $AFE$ are similar. By these
triangles show that the figure $CMNH$ is a square. By
constructing this square, the point $F$ can be found.

\end{proofex}

\ex{To inscribe in a given triangle a rectangle similar to a given
rectangle.}

\ex{To inscribe in a circle a triangle similar to a given triangle.}

\ex{To inscribe in a given semicircle a rectangle similar to a given
rectangle.}

\ex{To circumscribe about a circle a triangle similar to a given
triangle.}

\ex{To construct the expression, $x = \dfrac{2abc}{de}$;
  that is, $\dfrac{2ab}{d} × \dfrac{c}{e}$.}

\ex{To construct two straight lines, having given their sum and
their ratio.}

\ex{To construct two straight lines, having given their difference
and their ratio.}

\ex{Given two circles, with centres $O$ and $O'$, and a point $A$ in
their plane, to draw through the point $A$ a straight line, meeting the circumferences
at $B$ and $C$, so that $AB:AC=m:n$.}
\scanpage{189.png}%


\subsection{PROBLEMS OF COMPUTATION.}

\begin{proofex}%
To compute the altitudes of a triangle in terms of its sides.

\figc{189aa312}{}

At least one of the angles $A$ or $B$ is acute. Suppose $B$ is acute.

\eq[\indent In the $\triangle$ $CDB$,]{$h^2$}{$=a^2- \overline{BD}^2$,}{§~372}

\eq[\indent In the $\triangle$ $ABC$,]{$b^2$}{$=a^2+c^2 - 2c ×\ BD$.}{§~376}

\eq[\indent Whence]{$BD$}{$=\dfrac{a^2+c^2-b^2}{2c}$.}{}

\setlength{\eqalign}{0.25\dentwidth}
\eq[\indent Hence,]{$h^2$}{\(=a^2-\dfrac{(a^2+c^2-b^2)^2} {4c^2}=
  \dfrac{4a^2c^2-(a^2+c^2-b^2)^2} {4c^2} \)}{}

\eq{}{\( =\dfrac{(2ac+a^2+c^2-b^2)(2ac-a^2-c^2+b^2)} {4c^2} \)}{}

\eq{}{\( =\dfrac{\{(a+c)^2-b^2\}\{b^2-(a-c)^2\}} {4c^2} \)}{}

\eq{}{\( =\dfrac{(a+b+c)(a+c-b)(b+a-c)(b-a+c)} {4c^2}. \)}{}

\setlength{\eqalign}{0.5\dentwidth}

\eq[\indent Let]{$a+b+c$}{$=2s$.}{}

\eq[\indent Then]{$a+c-b$}{$=2(s-b)$,}{}

\eq{$b+a-c$}{$= 2(s-c)$,}{}

\eq{$b-a+c$}{$=2(s-a)$.}{}

\setlength{\eqalign}{0.25\dentwidth}
\eq[\indent Hence,]{$h^2$}{\( =\dfrac{2s × 2(s-a) × 2(s-b) × 2(2-c)} {4c^2} \).}{}

By simplifying, and extracting the square root,

\label{formtrialtitude}%
\eq{$h$}{\( =\dfrac{2}{c} \sqrt{s(s-a)(s-b)(s-c)} \).}{}

\setlength{\eqalign}{0.5\dentwidth}
\end{proofex}


\figc{189bb313}{}
\begin{proofex}%
To compute the medians of a triangle in terms of its sides.

\setlength{\eqalign}{0.33\dentwidth}
\eq[\indent By §~377,]{}{\( a^2+b^2 = 2m^2+2\left(\dfrac{c}{2}\right)^2 \).}{}

\eq[\indent Whence]{$4m^2$}{$=2(a^2+b^2)-c^2$.}{}

\label{formtrimedian}%
\eq{$\therefore m$}{$=\dfrac{1}{2} \sqrt{2(a^2+b^2)-c^2}$.}{}

\end{proofex}
\setlength{\eqalign}{0.5\dentwidth}
\scanpage{190.png}%

\figc{190aa314}{}
\begin{proofex}%
To compute the bisectors of a triangle in terms of the sides.

\setlength{\eqalign}{.33\dentwidth}
\eq[\indent By §~383,]{$t^2$}{$=ab-AD × BD$.}{}

\eq[\indent By §348,]{$\dfrac{AD}{B}$}{$=\dfrac{BD}{a}=\dfrac{AD+BD}{a+b}=\dfrac{c}{a+b}$.}{}

\eq{$\therefore AD$}{$=\dfrac{bc}{a+b}$, and $BD=\dfrac{ac}{a+b}$.}{}

\eq[\indent Whence]{$t^2$}{$= ab - \dfrac{abc^2}{(a+b)^2}$}{}

\eq{}{$=ab\left[1-\dfrac{c^2}{(a+b)^2}\right]$}{}

\eq{}{$=\dfrac{ab\{(a+b)^2-c^2\}}{(a+b)^2}$}{}

\eq{}{$=\dfrac{ab(a+b+c)(a+b-c)}{(a+b)^2}$}{}

\eq{}{$=\dfrac{ab × 2s × 2(s-c)}{(a+b^2)}$.}{}

\label{formtribisector}%
\eq[\indent Whence]{$t$}{$= \dfrac{2}{a+b} \sqrt{abs(s-c)}$.}{}

\setlength{\eqalign}{.5\dentwidth}


\end{proofex}

\figc{190bb315}{}
\begin{proofex}%
To compute the radius of the circle circumscribed about a triangle in terms of the sides of the triangle.

\eq[\indent By §384,]{$AC × AB$}{$= AE × AD$,}{}

\eq[or,]{$bc$}{$= 2 B × AD$.}{}

\setlength{\eqalign}{0.33\dentwidth}
\eq[\indent But]{$AD$}{$=\dfrac{2}{a}\sqrt{s(s-a)(s-b)(s-c)}$.}{Ex.~312}

\label{formradcircum}%
\eq{$\therefore R$}{$=\dfrac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$.}{}

\setlength{\eqalign}{0.5\dentwidth}

\end{proofex}

\ex{If the sides of a triangle are $3$,~$4$, and~$5$, is the angle
opposite~$5$ right, acute, or obtuse?}

\ex{If the sides of a triangle are $7$,~$9$, and~$12$, is the angle
opposite~$12$ right, acute, or obtuse?}

\ex{If the sides of a triangle are $7$,~$9$, and~$11$, is the angle
opposite~$11$ right, acute, or obtuse?}

\ex{The legs of a right triangle are $8$~inches and $12$~inches; find
the lengths of the projections of these legs upon the hypotenuse, and the
distance of the vertex of the right angle from the hypotenuse.}

\ex{If the sides of a triangle are $6$~inches, $9$~inches, and
$12$~inches, find the lengths (1)~of the altitudes; (2)~of the medians;
(3)~of the bisectors; (4)~of the radius of the circumscribed circle.
}
\scanpage{191.png}%

\ex{A line is drawn parallel to a side $AB$ of a triangle $ABC$,
cutting $AC$ in $D$, $BC$ in $E$. If $AD:DC = 2:3$, and $AB = 20$ inches,
find $DE$.}

\ex{The sides of a triangle are $9$,~$12$,~$15$. Find the segments of
the sides made by bisecting the angles.}

\ex{A tree casts a shadow $90$~feet long, when a post $6$~feet high
casts a shadow~$4$ feet long. How high is the tree?}

\ex{The lower and upper bases of a trapezoid are $a$, $b$, respectively;
and the altitude is~$h$.  Find the altitudes of the two triangles
formed by producing the legs until they meet.}

\ex{The sides of a triangle are $6$,~$7$,~$8$, respectively. In a similar
triangle the side homologous to~$8$ is~$40$.  Find the other two sides.}

\ex{The perimeters of two similar polygons are $200$~feet and $300$~feet.
If a side of the first is $24$~feet, find the homologous side of the
second.}

\ex{How long a ladder is required to reach a window $24$~feet high,
if the lower end of the ladder is $10$~feet from the side of the house?}

\ex{If the side of an equilateral triangle is~$a$, find the altitude.}

\ex{If the altitude of an equilateral triangle is~$h$, find the side.}

\ex{Find the length of the longest chord and of the shortest chord
that can be drawn through a point $6$~inches from the centre of a circle
whose radius is $10$~inches.}

\ex{The distance from the centre of a circle to a chord $10$~feet long
is $12$~feet. Find the distance from the centre to a chord $24$~feet long.}

\ex{The radius of a circle is $5$~inches. Through a point $3$~inches
from the centre a diameter is drawn, and also a chord perpendicular to
the diameter. Find the length of this chord, and the distance from one
end of the chord to the ends of the diameter.}

\ex{The radius of a circle is $6$~inches. Find the lengths of the
tangents drawn from a point $10$~inches from the centre, and also the
length of the chord joining the points of contact.}

\ex{The sides of a triangle are $407$~feet, $368$~feet, and $351$~feet.
Find the three bisectors and the three altitudes.
}
\scanpage{192.png}%

\ex{If a chord $8$~inches long is $8$~inches distant from the centre of
the circle, find the radius, and the chords drawn from the end of the chord
to the ends of the diameter which bisects the chord.}

\ex{From the end of a tangent $20$~inches long a secant is drawn
through the centre of the circle. If the external segment of this secant is
$8$~inches, find the radius of the circle.}

\ex{The radius of a circle is $13$~inches. Through a point $5$~inches
from the centre any chord is drawn. What is the product of the two
segments of the chord? What is the length of the shortest chord that
can be drawn through the point?}

\ex{The radius of a circle is $9$~inches and the length of a tangent
$12$~inches. Find the length of a line drawn from the extremity of the
tangent to the centre of the circle.}

\ex{Two circles have radii of $8$~inches and $3$~inches, respectively,
and the distance between their centres is $15$~inches. Find the lengths of
their common tangents.}

\ex{Find the segments of a line $10$~inches long divided in extreme
and mean ratio.}

\ex{The sides of a triangle are $4$,~$5$,~$5$. Is the largest angle acute,
right, or obtuse?}

\ex{Find the third proportional to two lines whose lengths are
$28$~feet and $42$~feet.}

\ex{If the sides of a triangle are $a$,~$b$,~$c$, respectively, find the
lengths of the three altitudes.}

\ex{The diameter of a circle is $30$~feet and is divided into five
equal parts. Find the lengths of the chords drawn through the points of
division perpendicular to the diameter.}

\ex{The radius of a circle is $2$~inches. From a point $4$~inches
from the centre a secant is drawn so that the internal segment is $1$~inch.
Find the length of the secant.}

\ex{The sides of a triangular pasture are $1551$ yards, $2068$ yards,
$2585$ yards. Find the median to the longest side.}

\ex{The diagonal of a rectangle is $d$, and the perimeter is $p$.
Find the sides.}

\ex{The radius of a circle is $r$. Find the length of a chord whose
distance from the centre is $\frac{1}{2} r$.}

\scanpage{193.png}%


\chapter{BOOK IV\@. AREAS OF POLYGONS.}
\markboth{\Headings{BOOK IV\@. PLANE GEOMETRY.}}
{\Headings{AREAS OF POLYGONS.}}%

\pp{\defn{The \textbf{unit of surface} is a square whose side is a
\emph{unit of length}.}}

\pp{\defn{The \textbf{area of a surface}\label{area} is the \emph{number of units of
surface} it contains.}}

\pp{\defn{Plane figures that \emph{have equal areas but cannot
be made to coincide} are called \textbf{equivalent}\label{equivalent2}.}}

\note{In propositions relating to \emph{areas}, the words ``rectangle,'' ``triangle,''
etc., are often used for ``area of rectangle,'' ``area of triangle,'' etc.}


\proposition{Theorem.}

\begin{proof}%
\obs{Two rectangles having equal altitudes are to
each other as their bases.}

\figc{193aa395}{Let the rectangles $AC$ and $AF$ have the same altitude $AD$.}

\prove{$\rect AC: \rect AF = \base AB: \base AE$.}

\textsc{Case 1.~} \emph{When $AB$ and $AE$ are commensurable.}

\textbf{Proof.} Suppose $AB$ and $AE$ have a common measure, as
$AO$, which is contained $m$ times in $AB$ and $n$ times in $AE$.

\eq[\indent Then]{$AB:AE$}{$ = m:n$.}{}
\scanpage{194.png}%

\filbreak
Apply $AO$ as a unit of measure to $AB$ and $AE$, and at the
several points of division erect $\perp_s$.

\step[\indent The]{$\rect AC$ is divided into $m$~rectangles,}{}

\step[and the]{$\rect AF$ is divided into $n$~rectangles.}{§~107}

\step{These rectangles are all equal.}{§~186}

\eq[\indent Hence,]{}{$\rect AC: \rect AF = m:n$.}{}

\eq[\indent Therefore,]{}{$\rect AC: \rect AF = AB:AE$.}{Ax.~1}

\textsc{Case 2.} \emph{When $AB$ and $AE$ are incommensurable.}

\figc{194aa395}{}

\textbf{Proof.} Divide $AB$ into any number of equal parts, and apply
one of them to $AE$ as many times as $AE$ will contain it.

Since $AB$ and $AE$ are incommensurable, a certain number
of these parts will extend from $A$ to some point $K$, leaving a
remainder $KE$ less than one of the equal parts of $AB$.

\step{Draw $KH \parallel$ to $EF$.}{}

Then $AB$ and $AK$ are commensurable by construction.

\step[\indent Therefore,]{\( \dfrac{\rect AH}{\rect AC} = \dfrac{AK}{AB}. \)}{Case~1}

If the number of equal parts into which $AB$ is divided is
indefinitely increased, the varying values of these ratios will
continue equal, and approach for their respective limits the
ratios

\step{\( \dfrac{\rect AF}{\rect AC} \) and \( \dfrac{AE}{AB} \). (See §~287.)}{}

\step{\( \therefore \dfrac{\rect AF}{\rect AC} = \dfrac{AE}{AB}. \)}{§~284}

\hfill\qed

\end{proof}

\pp{\cor{Two rectangles having equal bases are to each
other as their altitudes.}}
\scanpage{195.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Two rectangles are to each other as the products
of their bases by their altitudes.}

\vspace{1ex}
\figc{195aa397}{Let $R$ and $R'$ be two rectangles, having for their bases $b$ and $b'$,
and for their altitudes $a$ and $a'$, respectively.}

\proveq{$\dfrac{R}{R'}$}{$= \dfrac{a × b}{a' × b'}$.}

\textbf{Proof.} Construct the rectangle $S$, with its base equal to that
of $R$, and its altitude equal to that of $R'$.

\eq[\indent Then]{$\dfrac{R}{S}$}{$=\dfrac{a}{a'}$,}{§~396}

\eq[and]{$\dfrac{S}{R'}$}{$=\dfrac{b}{b'}$.}{§~395}

The products of the corresponding members of these equations
give

\eq{$\dfrac{R}{R'}$}{$=\dfrac{a × b}{a' × b'}$.}{\qed}

\end{proof}

\ex{Find the ratio of a rectangular lawn $72$~yards by $49$~yards to
a grass turf $18$~inches by $14$~inches.}

\ex{Find the ratio of a rectangular courtyard $18\frac{1}{2}$~yards by $15\frac{1}{2}$~yards
to a flagstone $31$~inches by $18$~inches.}

\ex{A square and a rectangle have the same perimeter, $100$~yards.
The length of the rectangle is $4$~times its breadth. Compare their areas.}

\ex{On a certain map the linear scale is $1$~inch to $5$~miles. How
many acres are represented on this map by a square the perimeter of
which is $1$~inch?}
\scanpage{196.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The area of a rectangle is equal to the product
of its base by its altitude.}

\figc{196aa398}{Let $R$ be a rectangle, $b$ its base, and $a$ its altitude.}

\proveq{the area of $R$}{$= a × b$.}

\textbf{Proof.} Let $U$ be the unit of surface.

\eq{\( \dfrac{R}{U} = \dfrac{a × b}{1 × 1} \)}{$= a × b$,}{}

\pnote{(two rectangles are to each other as the products of their bases and altitudes).}

\step[\indent But]{$\dfrac{R}{U} =$ the \emph{number} of units of surface in $R$.}{§~393}

\label{formarearect}%
\eq{$\therefore$ the area of $R$}{$= a × b$.}{\qed}

\end{proof}

\figc{196bb399}{}
\begin{point}\textsc{Scholium.} When the base and altitude each contain
the linear unit an integral number of times, this proposition
is rendered evident by dividing the figure into squares, each
equal to the unit of surface. Thus, if the base contains seven
linear units, and the altitude four, the figure may be divided
into twenty-eight squares, each equal to the unit of surface.
\end{point}
\scanpage{197.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The area of a parallelogram is equal to the
product of its base by its altitude.}

\figc{197aa400}{Let $AEFD$ be a parallelogram, $b$ its base, and $a$ its altitude.}

\prove{the area of the \textnormal{$\Par AEFD = a × b$.}}

\textbf{Proof.} From $A$ draw $AB$ $\parallel$ to $DC$ to meet $FE$ produced.

Then the figure $ABCD$ is a rectangle, with the same base
and the same altitude as the $\Par AEFD$.

\step{The rt.~$\triangle_s ABE$ and $DCF$ are equal.}{§~151}

\step{For $AB = CD$, and $AE = DF$.}{§~178}

From $ABFD$ take the $\triangle DCF$; the $\rect ABCD$ is left.

From $ABFD$ take the $\triangle ABE$; the $\Par AEFD$ is left.

\step{$\therefore \rect ABCD \Bumpeq \Par AEFD$}{Ax.~3}

\step{But the area of the $\rect ABCD = a × b$.}{§~398}

\label{formareapar}%
\step{$\therefore$ the area of the $\Par AEFD = a × b$.}{Ax.~1}

\hfill\qed

\end{proof}

\pp{\cor[1]{Parallelograms having equal bases and equal
altitudes are equivalent.}}

\pp{\cor[2]{Parallelograms having equal bases are to each
other as their altitudes; parallelograms having equal altitudes
are to each other as their bases; any two parallelograms
are to each other as the products of their bases by
their altitudes.}}
\scanpage{198.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The area of a triangle is equal to half the product of its base by its altitude.}

\figc{198aa403}{Let $a$ be the altitude and $b$ the base of the triangle $ABC$.}

\prove{the area of the $\triangle{}ABC=\frac{1}{2}a × b$.}

\textbf{Proof.} Construct on $AB$ and $BC$ the parallelogram $ABCH$.

\step[\indent Then]{$\triangle ABC=\frac{1}{2}\Par ABCH$.}{§~179}

\step{The area of the $\Par ABCH=a × b$.}{§~400}

\label{formareatri}%
\step{Therefore, the area of $\triangle ABC=\frac{1}{2}a × b$.}{Ax.~7}

\hfill\qed

\end{proof}

\pp{\cor[1]{Triangles having equal bases and equal altitudes are equivalent.}}

\pp{\cor[2]{Triangles having equal bases are to each other as their altitudes; triangles having equal altitudes are to each other as their bases; any two triangles are to each other as the products of their bases by their altitudes.}}

\pp{\cor[3]{The product of the legs of a right triangle is equal to the product of the hypotenuse by the altitude from the vertex of the right angle.}}


\ex{The lines which join the middle point of either diagonal of a quadrilateral to the opposite vertices divide the quadrilateral into two equivalent parts.}
\scanpage{199.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The area of a trapezoid is equal to half the sum of its bases multiplied by the altitude.}

\figc{199aa407}{Let $b$ and $b'$ be the bases and $a$ the altitude of the trapezoid $ABCH$.}

\prove{the area of the $ABCH=\frac{1}{2}a(b+b')$.}

\step[\indent\textbf{Proof.}]{Draw the diagonal $AC$.}{}

\eq[\indent Then]{the area of the $\triangle ABC$}{$=\frac{1}{2}a × b$,}{}

\eq[and]{the area of the $\triangle AHC$}{$=\frac{1}{2}a × b'$.}{§~403}

\label{formareatrap}%
\eq{$\therefore$ the area of $ABCH$}{$=\frac{1}{2}a(b+b')$.}{Ax.~2}

\hfill\qed

\end{proof}

\pp{\cor{The area of a trapezoid is equal to the product of the median by the altitude.}~\hfill§~190}

\figc{199bb409}{}
\begin{point}%
\textsc{Scholium.} The area of an irregular polygon may
be found by dividing the polygon into triangles, and by finding the
area of each of these triangles separately.  Or, we may draw the
longest diagonal, and let fall perpendiculars upon this diagonal from
the other vertices of the polygon.

The sum of the areas of the right triangles, rectangles, and
trapezoids thus formed is the area of the polygon.
\end{point}
\scanpage{200.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The areas of two triangles which have an angle of
  the one equal to an angle of the other are to each other as the
  products of the sides including the equal angles.}

\figc{200aa410}{Let the triangles $ABC$ and $ADE$ have the common angle $A$.}

\prove {$\dfrac{\triangle ABC} {\triangle ADE} = \dfrac{AB × AC} {AD × AE}$.}

\step [\indent Proof.] {Draw $BE$.}{}

\step [\indent Now] {$\dfrac{\triangle ABC}{\triangle ABE}= \dfrac{AC}{AE}$,} {}

\step [and] {$\dfrac{\triangle ABE}{\triangle ADE}= \dfrac{AB}{AD}$.} {§~405}


The products of the first members and of the second members of these
equalities give

\step {$\dfrac{\triangle ABC}{\triangle ADE}=\dfrac{AB × AC}{AD × AE}$.} {\qed}
\end {proof}

\ex{The areas of two triangles which have an angle of
the one supplementary to an angle of the other are to each other as
the products of the sides including the supplementary angles.}
\scanpage{201.png}%


\clearpage
\section{COMPARISON OF POLYGONS.}

\proposition{Theorem.}

\begin{proof}%
\obs{The areas of two similar triangles are to each
as the squares of any two homologous sides.}

\figc{201ab411}{Let the two similar triangles be $ACB$ and $A'C'B'$.}

\proveq{$\dfrac{\triangle ACB}{\triangle A'C'B'}$}
  {$= \dfrac{\overline{AB}^2}{\overline{A'B'}^2}$.}

\textbf{Proof.} Draw the altitudes $CO$ and $C'O'$.

\step[\indent Then]{\( \dfrac{\triangle ACB}{\triangle A'C'B'} =
  \dfrac{AB × CO}{A'B' × C'O'} =
  \dfrac{AB}{A'B'} × \dfrac{CO}{C'O'} \),}{§~405}

\pnote{(two $\triangle_s$ are to each other as the products of their bases by their altitudes).}

\eq[\indent But]{$\dfrac{AB}{A'B'}$}{$= \dfrac{CO}{C'O'}$.}{§~361}

\pnote{(the homologous altitudes of two similar $\triangle_s$ have the same ratio as any two
homologous sides).}

Substitute, in the above equality, for $\dfrac{CO}{C'O'}$ its equal $\dfrac{AB}{A'B'}$;

\step[then]{\( \dfrac{\triangle ACB}{\triangle A'C'B'} =
  \dfrac{AB }{A'B'} × \dfrac{AB }{A'B'} =
  \dfrac{\overline{AB}^2 }{\overline{A'B'}^2} \).}{\qed}

\end{proof}

\ex{Prove this proposition by §~410.}
\scanpage{202.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The areas of two similar polygons are to each
  other as the squares of any two homologous sides.}

\figc{202ab412}{Let $S$ and $S'$ denote the areas of the two similar polygons
  $ABC$ etc.\ and $A'B'C'$ etc.}

\proveq{$S:S'$}{$=\overline{AB}^2:\overline{A'B'^2}$.}

\textbf{Proof.} By drawing all the diagonals from any homologous
vertices $E$ and $E'$, the two similar polygons are divided into
similar triangles.~\hfill§~365

\step{\( \displaystyle \therefore \frac{\overline{AB}^2}{\overline{A'B'^2}}=
 \frac{\triangle ABE}{\triangle A'B'E'}=
 \left(\frac{\overline{BE}^2}{\overline{B'E'^2}}\right)=
 \frac{\triangle BCE}{\triangle B'C'E'}=\text{etc.} \)}{§~411}

\step[\indent That is,]{\( \displaystyle \frac{\triangle ABE}{\triangle A'B'E'}=
 \frac{\triangle BCE}{\triangle B'C'E'}=
 \frac{\triangle CDE}{\triangle C'D'E'} \).}{}

\( \displaystyle \therefore
  \frac{\triangle ABE + \triangle BCE + \triangle CDE}
     {\triangle A'B'E' + \triangle B'C'E' + \triangle C'D'E'}=
  \frac{\triangle ABE}{\triangle A'B'E'}=
  \frac{\overline{AB}^2}{\overline{A'B'^2}} \).\hsp§~335

\step{\( \displaystyle \therefore
  S:S'=\overline{AB}^2:\overline{A'B'^2} \)}{\qed}

\end{proof}

\pp{\cor[1]{The areas of two similar polygons
  are to each other as the squares of any two homologous lines.}}

\pp{\cor[2]{The homologous sides of two
  similar polygons have the same ratio as the square roots of their areas.}}
\scanpage{203.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The square on the hypotenuse of a right triangle
  is equivalent to the sum of the squares on the two legs.}

\figc{203aa415}{Let $BE$, $CH$, $AF$, be squares on the three sides of the
  right triangle $ABC$.}

\prove{$BE \Bumpeq CH + AF$.}

\textbf{Proof.} Through $A$ draw $AL \parallel$ to $CE$, and draw $AD$
and $CF$.

Since $\angle_s{} BAC$, $BAG$, and $CAH$ are rt.\ $\angle_s$, $CAG$ and
$BAH$ are straight lines.~\hfill§~90

\eq[\indent The]{$\triangle ABD$}{$=\triangle FBC$.}{§~143}

\eq[\indent For]{$BD$}{$=BC$,}{}

\eq{$BA$}{$=BF$,}{§~168}

\eq[and]{$\angle ABD$}{$=\angle FBC$,}{Ax.~2}

\pnote{(each being the sum of a rt.\ $\angle$ and the $\angle ABC$).}

\step{Now the rectangle $BL$ is double the $\triangle ABD$,}{}

\pnote{(having the same base $BD$, and the same altitude, the distance
  between the $\parallel_s AL$ and $BD$),}

\step{and the square $AF$ is double the $\triangle FBC$,}{}

\pnote{(having the same base $FB$, and the same altitude $AB$).}

$\therefore$ the rectangle $BL$ is equivalent to the square
  $AF$.\hfill~Ax.~6

In like manner, by drawing $AE$ and $BK$, it may be proved that the
rectangle $CL$ is equivalent to the square $CH$.

Hence, the square $BE$, the sum of the rectangles $BL$ and $CL$, is
equivalent to the sum of the squares $CH$ and $AF$.~\hfill\qed

\end{proof}

\pp{\cor{The square on either leg of a right
  triangle is equivalent to the difference of the square on the
  hypotenuse and the square on the other leg.}}
\scanpage{204.png}%


\section[EXERCISES.]{THEOREMS.}

\figccc{204aa356}{204bb357}{204cc358}
\begin{proofex}%
The square constructed upon the sum of two straight
lines is equivalent to the sum of the squares constructed upon these
two lines, increased by twice the rectangle of these lines:

Let $AB$ and $BC$ be the two straight lines, and $AC$ their sum.
Construct the squares $ACGK$ and $ABED$ upon $AC$ and $AB$,
respectively.  Prolong $BE$ and $DE$ until they meet $KG$ and $CG$,
respectively.  Then we have the square $EFGH$, with sides each equal
to $BC$.  Hence, the square $ACGK$ is the sum of the squares $ABED$
and $EFGH$, and the rectangles $DEHK$ and $BCFE$, the dimensions of
which are equal to $AB$ and $BC$.

\end{proofex}

\begin{proofex}%
The square constructed upon the difference of two
straight lines is equivalent to the sum of the squares constructed
upon these two lines, diminished by twice the rectangle of these
lines.

Let $AB$ and $AC$ be the two straight lines, and $BC$ their
difference.  Construct the square $ABFG$ upon $AB$, the square $ACKH$
upon $AC$, and the square $BEDC$ upon $BC$ (as shown in the figure).
Prolong $ED$ to meet $AG$ in $L$.

The dimensions of the rectangles $LEFG$ and $HKDL$ are $AB$ and $AC$,
and the square $BCDE$ is evidently the difference between the whole
figure and the sum of these rectangles; that is, the square
constructed upon $BC$ is equivalent to the sum of the squares
constructed upon $AB$ and $AC$, diminished by twice the rectangle of
$AB$ and $AC$.

\end{proofex}

\begin{proofex}%
The difference between the squares constructed upon
two straight lines is equivalent to the rectangle of the sum and
difference of these lines.

Let $ABDE$ and $BCFG$ be the squares constructed upon the two straight
lines $AB$ and $BC$.  The difference between these squares is the
polygon $ACGFDE$, which is composed of the rectangles $ACHE$ and
$GFDH$.  Prolong $AE$ and $CH$ to $I$ and $K$, respectively, making
$EI$ and $HK$ each equal to $BC$, and draw $IK$.  The rectangles
$GFDH$ and $EHKI$ are equal.  The difference between the squares
$ABDE$ and $BCGF$ is then equivalent to the rectangle $ACKI$, which
has for dimensions $AI$, equal to $AB + BC$, and $EH$, equal to $AB - BC$.

\end{proofex}
\scanpage{205.png}%

\ex{The area of a rhombus is equal to half the product of its
diagonals.}

\ex{Two isosceles triangles are equivalent if their legs are equal
each to each, and the altitude of one is equal to half the base of the other.}

\ex{The area of a circumscribed polygon is equal to half the
product of its perimeter by the radius of the inscribed circle.}

\ex{Two parallelograms are equal if two adjacent sides of the one
are equal, respectively, to two adjacent sides of the other, and the included
angles are supplementary.}

\ex{If $ABC$ is a right triangle, $C$ the vertex of the right angle,
$BD$ a line cutting $AC$ in $D$, then \( \overline{BD}^2 + \overline{AC}^2 =
\overline{AB}^2 + \overline{DC}^2 \).}

\ex{Upon the sides of a right triangle as homologous sides three
similar polygons are constructed. Prove that the polygon upon the
hypotenuse is equivalent to the sum of the polygons upon the legs.}

\ex{If the middle points of two adjacent sides of a parallelogram
are joined, a triangle is formed which is equivalent to one eighth of the
parallelogram.}

\ex{If any point within a parallelogram is joined to the four vertices,
the sum of either pair of triangles having parallel bases is equivalent
to half the parallelogram.}

\ex{Every straight line drawn through the intersection of the
diagonals of a parallelogram divides the parallelogram into two equal
parts.}

\ex{The line which joins the middle points of the bases of a trapezoid
divides the trapezoid into two equivalent parts.}

\ex{Every straight line drawn through the middle point of the
median of a trapezoid cutting both bases divides the trapezoid into two
equivalent parts.}

\ex{If two straight lines are drawn from the middle point of either
leg of a trapezoid to the opposite vertices, the triangle thus formed is
equivalent to half the trapezoid.}

\ex{The area of a trapezoid is equal to the product of one of the
legs by the distance from this leg to the middle point of the other leg.}

\ex{The figure whose vertices are the middle points of the sides
of any quadrilateral is equivalent to half the quadrilateral.}
\scanpage{206.png}%


\clearpage
\section{PROBLEMS OF CONSTRUCTION.}

\proposition{Problem.}

\begin{proof}%
\obs{To construct a square equivalent to the sum of two given squares.}

\figc{206aa417}{Let $R$ and $R'$ be two given squares.}

\prove[To construct ]{a square equivalent to $R'+R$.}

\step {Construct the rt.\ $\angle A$.} {}

\step {Take $AC$ equal to a side of $R'$,} {}

\step {and $AB$ equal to a side of $R$; and draw $BC$.} {}

\step {Construct the square $S$, having each of its sides equal to $BC$.} {}

\step [\indent Then] {$S$ is the square required.} {}

\step [\indent Proof.] {$\overline{BC}^2 \Bumpeq \overline{AC}^2 + \overline{AB}^2$,} {§~415}

\pnote {(the square on the hypotenuse of a rt.\ $\triangle$ is equivalent to the sum of the squares on the two legs).}

\step {$\therefore S \Bumpeq R'+R$.} {}

\hfill\qef

\end{proof}

\ex{If the perimeter of a rectangle is $72$~feet, and the
length is equal to twice the width, find the area.}

\ex{How many tiles $9$~inches long and $4$~inches wide will
be required to pave a path $8$~feet wide surrounding a rectangular court
$120$~feet long and $36$~feet wide?}

\ex{The bases of a trapezoid are $16$~feet and $10$~feet;
each leg is equal to $5$~feet.  Find the area of the trapezoid.}
\scanpage{207.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To construct a square equivalent to the difference
  of two given squares.}

\figc{207aa418}{Let $R$ be the smaller square and $R'$ the larger.}

\prove[To construct ]{a square equivalent to $R'-R$.}

\step{Construct the rt.\ $\angle A$.}{}

\step{Take $AB$ equal to a side of $R$.}{}

\step{From $B$ as a centre, with a radius equal to a side of $R'$,}{}

\step{describe an arc cutting the line $AX$ at $C$.}{}

\step{Construct the square $S$, having each of its sides equal to $AC$.}{}

\step[\indent Then]{$S$ is the square required.}{}

\step[\indent\textbf{Proof.}]{\( \overline{AC}^2 \Bumpeq
  \overline{BC}^2-\overline{AB}^2 \),}{§~416}

\pnote{(the square on either leg of a rt.\ $\triangle$ is equivalent to
  the difference of the square on the hypotenuse and the square on the
  other leg).}

\step{$\therefore S \Bumpeq R'-R$.}{}

\hfill\qef

\end{proof}


\ex{Construct a square equivalent to the sum of two squares
whose sides are $3$~inches and $4$~inches.}

\ex{Construct a square equivalent to the difference of two squares
whose sides are $2\frac{1}{2}$~inches and $2$~inches.}

\ex{Find the side of a square equivalent to the sum of
two squares whose sides are $24$~feet and $32$~feet.}

\ex{Find the side of a square equivalent to the
difference of two squares whose sides are $24$~feet and $40$~feet.}

\ex{A rhombus contains $100$~square feet, and the length
of one diagonal is $10$~feet.  Find the length of the other diagonal.
}
\scanpage{208.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To construct a polygon similar to two given
  similar polygons and equivalent to their sum.}

\figc{208aa419}{Let $R$ and $R'$ be two similar polygons, and $AB$ and $A'B'$
  two homologous sides.}

\prove[To construct ]{a similar polygon equivalent to $R+R'$.}

\step{Construct the rt.\ $\angle P$.}{}

\step{Take $PH$ equal to $A'B'$, and $PO$ equal to $AB$.}{}

\step{Draw $OH$, and take $A''B''$ equal to $OH$.}{}

\step{Upon $A''B''$, homologous to $AB$, construct $R''$ similar to $R$.}{}

\step{Then $R''$ is the polygon required.}{}

\eq[\indent\textbf{Proof.}]{$\overline{PO}^2 + \overline{PH}^2$}{$= \overline{OH}^2$.}{§~415}

Put for $PO$, $PH$, and $OH$ their equals $AB$, $A'B'$, and $A''B''$.

\eq[\indent Then]{$\overline{AB}^2 + \overline{A'B'^2}$}{$= \overline{A''B''^2}$.}{}

\step[\indent Now]{$\dfrac{R}{R''} = \dfrac{\overline{AB}^2}{\overline{A''B''^2}}$,
  and $\dfrac{R'}{R''} = \dfrac{\overline{A'B'^2}}{\overline{A''B''^2}}$.}
  {§~412}

\step[\indent By addition,]{$\dfrac{R+R'}{R''} =
   \dfrac{\overline{AB}^2 + \overline{A'B'^2}}{\overline{A''B''^2}} = 1$.}
   {Ax.~2}

\step{$\therefore R'' \Bumpeq R+R'$.}{\qef}

\end{proof}
\scanpage{209.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To construct a triangle equivalent to a given polygon.}

\figc{209aa420}{Let $ABCDHE$ be the given polygon.}

\prove[To construct ]{a triangle equivalent to the given polygon.}

Let $D$, $H$, and $E$ be any three consecutive vertices of the
polygon.  Draw the diagonal $DE$.

\step{From $H$ draw $HF \parallel$ to $DE$.}{}

\step{Produce $AE$ to meet $HF$ at $F$, and draw $DF$.}{}

Again, draw $CF$, and draw $DK \parallel$ to $CF$ to meet $AF$
produced at $K$, and draw $CK$.

In like manner continue to reduce the number of sides of the polygon
until we obtain the $\triangle CIK$.

\step{Then $\triangle CIK$ is the triangle required.}{}

\textbf{Proof.}  The polygon $ABCDF$ has one side less than the
polygon \newline$ACBDHE$, but the two polygons are equivalent.

\step{For the part $ACBDE$ is common,}{}

\step{and the $\triangle DEF \Bumpeq \triangle DEH$,}{§~404}

\pnote{(for the base $DE$ is common, and their vertices $F$ and $H$ are
  in the line $FH \parallel$ to the base).}

In like manner it may be proved that

\step{$ABCK \Bumpeq ABCDF$, and $CIK \Bumpeq ABCK$.}{\qef}

\end{proof}
\scanpage{210.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To construct a square equivalent to a given parallelogram.}

\figc{210aa421}{Let $ABCD$ be the parallelogram, $b$ its base, and $a$ its altitude.}

\prove[To construct ]{a square equivalent to the $\Par ABCD$.}

\step{Upon a line $MX$ take $MN$ equal to $a$, $NO$ equal to $b$.}{}

\step{Upon $MO$ as a diameter, describe a semicircle.}{}

\step{At $N$ erect $NP \perp$ to $MO$, meeting the circumference at $P$.}{}

Then the square $R$, constructed upon a line equal to $NP$, is
equivalent to the $\Par ACBD$.

\eq[\indent\textbf{Proof.}]{$MN:NP$}{$= NP:NO$,}{§~370}

\pnote{(a $\perp$ let fall from any point of a
  circumference to the diameter is the mean proportional between the
  segments of the diameter).}

\step{$\therefore \overline{NP}^2 = MN × NO = a × b$.}{§~327}

\step[\indent Therefore,]{$R \Bumpeq \Par ABCD$.}{\qef}

\end{proof}

\pp{\cor[1]{A square may be constructed
  equivalent to a given triangle, by taking for its side the mean
  proportional between the base and half the altitude of the triangle.}}

\pp{\cor[2]{A square may be constructed
  equivalent to a given polygon, by first reducing the polygon to an
  equivalent triangle, and then constructing a square equivalent to
  the triangle.}}
\scanpage{211.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To construct a parallelogram equivalent to a given
  square, and having the sum of its base and altitude equal to a given line.}

\figc{211aa424}{Let $R$ be the given square, and let the sum of the base and
  altitude of the required parallelogram be equal to the given line $MN$.}

\prove[To construct ]{a $\Par$ equivalent to $R$, with the sum of its
  base and altitude equal to $MN$.}

\step{Upon $MN$ as a diameter, describe a semicircle.}{}

At $M$ erect $MP$, a $\perp$ to $MN$, equal to a side of the given
square $R$.

\step{Draw $PQ \parallel$ to $MN$, cutting the circumference at $S$.}{}

\step{Draw $SC \perp$ to $MN$.}{}

Any $\Par$ having $CM$ for its altitude and $CN$ for its base is
equivalent to~$R$.

\eq[\indent\textbf{Proof.}]{$SC$}{$=PM$.}{§§~104, 180}

\eq{$\therefore \overline{SC}^2$}{$= \overline{PM}^2 = R$.}{}

\eq{$MC:SC$}{$= SC:CN$,}{§~370}

\pnote{(a $\perp$ let fall from any point of a circumference to the
  diameter is the mean proportional between the segments of the
  diameter).}

\step[\indent Then]{$\overline{SC}^2 \Bumpeq MC × CN$.}{§~327}

\hfill\qef

\end{proof}

\note{This problem may be stated as follows:}

\emph{To construct two straight lines the sum and product of which are
  known.}
\scanpage{212.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To construct a parallelogram equivalent to a given
  square, and having the difference of its base and altitude equal to
  a given line.}

\figc{212aa425}{Let $R$ be the given square, and let the difference of the
  base and altitude of the required parallelogram be equal to the
  given line $MN$.}

\textit{To construct a $\Par$ equivalent to $R$, with the difference of its
  base and altitude equal to $MN$.}

Upon the given line $MN$ as a diameter, describe a circle.

From $M$ draw $MS$, tangent to the $\odot$, and equal to a side of
the given square $R$.

Through the centre of the $\odot$ draw $SB$ intersecting the
circumference at $C$ and $B$.

Then any $\Par$, as $R'$, having $SB$ for its base and $SC$ for its
altitude, is equivalent to $R$.

\step[\indent\textbf{Proof.}]{$SB:SM=SM:SC$,}{§~381}

\pnote{(if from a point without a $\odot$ a secant and a tangent are
  drawn, the tangent is the mean proportional between the whole secant
  and the external segment).}

\step[\indent Then]{$\overline{SM}^2 \Bumpeq SB × SC$,}{§~327}

\noindent and the difference between $SB$ and $SC$ is the diameter of the
$\odot$, that is,~$MN$.

\hfill\qef

\end{proof}

\note{This problem may be stated: \textit{To construct two
  straight lines the difference and product of which are known.}}
\scanpage{213.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To construct a polygon similar to a given polygon
  $P$ and equivalent to a given polygon $Q$.}

\figc{213aa426}{Let $P$ and $Q$ be the two given polygons, and $AB$ a side of $P$.}

\prove[To construct ]{a polygon similar to $P$ and equivalent to $Q$.}

\step{Find squares equivalent to $P$ and $Q$,}{§~423}

\step{and let $m$ and $n$ respectively denote their sides.}{}

Find $A'B'$, the fourth proportional to $m$, $n$, and $AB$.~\hfill§~386

Upon $A'B'$, homologous to $AB$, construct $P'$ similar to $P$.

\eq[\indent Then]{$P'$}{$\Bumpeq Q$.}{}

\eq[\indent\textbf{Proof.}]{$m:n$}{$= AB:A'B'$.}{Const.}

\eq{$\therefore m^2:n^2$}{$= \overline{AB}^2:\overline{A'B'}^2$.}{§~338}

\eq[\indent But]{$P \Bumpeq m^2$,}{and $Q \Bumpeq n^2$. }{Const.}

\eq{$\therefore P:Q = m^2$}{$:n^2 = \overline{AB}^2:\overline{A'B'}^2$.}{}

\eq[\indent But]{$P:P'$}{$= \overline{AB}^2:\overline{A'B'}^2$.}{§~412}

\eq{$\therefore P:Q$}{$= P:P'$.}{Ax.~1}

\eq{$\therefore P'$}{$\Bumpeq Q$.}{\qef}

\end{proof}

\ex{To construct a square equivalent to the sum of any
number of given squares.}

\ex{To construct a polygon similar to two given similar
polygons and equivalent to their difference.
}
\scanpage{214.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To construct a square which shall have a given
  ratio to a given square.}

\figc{214aa427}{Let R be the given square, and $\dfrac{n}{m}$ the given ratio.}

\textit{To construct a square which shall be to $R$ as $n$ is to $m$.}

Take $AB$ equal to a side of $R$, and draw $Ay$, making any convenient
angle with $AB$.

On $Ay$ take $AE$ equal to $m$, $EF$ equal to $n$, and draw $EB$.

\step{Draw $FC \parallel$ to $EB$ meeting $AB$ produced at $C$.}{}

\step{On $AC$ as a diameter, describe a semicircle.}{}

\step{At $B$ erect the $\perp BD$, meeting the semicircumference at $D$.}{}

\step{Then $BD$ is a side of the square required.}{}

\step[\indent\textbf{Proof.}]{Denote $AB$ by $a$, $BC$ by $b$, and $BD$ by $x$.}{}


\eq[\indent Now]{$a:x$}{$= x:b$.}{§~370}

\eq[\indent Therefore,]{$a^2:x^2$}{$= a:b$.}{§~337}

\eq[\indent But]{$a:b$}{$= m:n$.}{§~342}

\eq[\indent Therefore,]{$a^2:x^2$}{$= m:n$.}{Ax.~1}

\eq[\indent By inversion,]{$x^2:a^2$}{$= n:m$.}{§~331}

Hence, the square on $BD$ will have the same ratio to $R$ as $n$ has
to $m$.

\hfill\qef

\end{proof}
\scanpage{215.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To construct a polygon similar to a given polygon
  and having a given ratio to it.}

\figc{215aa428}{Let $R$ be the given polygon, and $\dfrac{n}{m}$ the given
  ratio.}

\prove[To construct ]{a polygon similar to $R$, which shall be to $R$ as
  $n$ is to $m$.}

Construct a line $A'B'$, such that the square on $A'B'$ shall be to
the square on $AB$ as $n$ is to $m$.~\hfill§~427

Upon $A'B'$, as a side homologous to $AB$, construct the polygon $S$
similar to $R$.~\hfill§~391

\step{Then $S$ is the polygon required.}{}

\eq[\indent\textbf{Proof.}]{$S:R$}{$= \overline{A'B'}^2 : \overline{AB}^2$.}{§~412}

\eq[\indent But]{$\overline{A'B'}^2 : \overline{AB}^2$}{$= n:m$.}{Const.}

\eq[\indent Therefore,]{$S:R$}{$= n:m$.}{Ax.~1}

\hfill\qef

\end{proof}


\ex{To construct a triangle equivalent to a given
triangle, and having one side equal to a given length $l$.}

\ex{To transform a triangle into an equivalent right
triangle.}

\ex{To transform a given triangle into an equivalent
right triangle, having one leg equal to a given length.}

\ex{To transform a given triangle into an equivalent
right triangle, having the hypotenuse equal to a given length.}
\scanpage{216.png}%


\section[EXERCISES.]{PROBLEMS OF CONSTRUCTION.}

\begin{proofex}%
To transform a triangle $ABC$ into an equivalent triangle,
having a side equal to a given length $l$, and an angle equal to angle $BAC$.

Upon $AB$ (produced if necessary), take $AD$ equal to $l$, draw $BE \parallel$ to
$CD$, meeting $AC$ (produced if necessary) at $E$.

$\triangle BED \Bumpeq \triangle BEC$.

\end{proofex}

\ex{To transform a given triangle into an equivalent isosceles triangle,
having the base equal to a given length.}


\exheader{To construct a triangle equivalent to:}

\ex{The sum of two given triangles.}

\ex{The difference of two given triangles.}

\ex{To transform a given triangle into an equivalent equilateral
triangle.}


\exheader{To transform a parallelogram into an equivalent:}

\ex{Parallelogram having one side equal to a given length.}

\ex{Parallelogram having one angle equal to a given angle.}

\ex{Rectangle having a given altitude.}


\exheader{To transform a square into an equivalent:}

\ex{Equilateral triangle.}

\ex{Right triangle having one leg equal to a given length.}

\ex{Rectangle having one side equal to a given length.}


\exheader{To construct a square equivalent to:}

\ex{Five eighths of a given square.}

\ex{Three fifths of a given pentagon.}

\ex{To divide a given triangle into two equivalent parts by a line
through a given point $P$ in one of the sides.}

\ex{To find a point within a triangle, such that the lines joining
this point to the vertices shall divide the triangle into three equivalent
parts.}

\ex{To divide a given triangle into two equivalent parts by a line parallel to one of the sides.}

\ex{To divide a given triangle into two equivalent parts by a line
perpendicular to one of the sides.}
\scanpage{217.png}%

\subsection{PROBLEMS OF COMPUTATION.}

\figccc{217aa404}{217bb405}{217cc406}

\begin{proofex}%
To find the area of an equilateral triangle in terms of its side.

Denote the side by $a$, the altitude by $h$, and the area by $S$.

\setlength{\eqalign}{0.33\dentwidth}
\eq[\indent Then]{$h^2$}{$a^2 - \dfrac{a^2}{4} = \dfrac{3a^2}{4} =
   \dfrac{a^2}{4} × 3$.}{§~372}

\eq{$\therefore h$}{$= \dfrac{a}{2}\sqrt{3}$.}{}


\eq[\indent But]{$S$}{$= \dfrac{a × h}{2}$.}{§~403}

\label{formareaequitri}%
\eq{$\therefore S$}
  {$=\dfrac{a}{2} × \dfrac{a\sqrt{3}}{2} = \dfrac{a^2\sqrt{3}}{4}$.}{}

\setlength{\eqalign}{0.5\dentwidth}

\end{proofex}


\begin{proofex}%
To find the area of a triangle in terms of its sides.


\label{formareatri2}%
\setlength{\eqalign}{0.33\dentwidth}
\eq[\indent By Ex.~312,]{$h$}{$= \dfrac{2}{b} \sqrt{s(s - a)(s - b)(s - c)}$.}{}

\eq[\indent Hence,]{$S$}{$= \dfrac{b}{2} × \dfrac{2}{b}
  \sqrt{s(s - a)(s - b)(s - c)}$}{§~403}

\eq{}{$= \sqrt{s(s - a)(s - b)(s - c)}$.}{}

\setlength{\eqalign}{0.5\dentwidth}

\end{proofex}

\begin{proofex}%
To find the area of a triangle in terms of the
radius of the circumscribed circle.

If $R$ denotes the radius of the circumscribed circle, and $h$ the
altitude of the triangle, we have, by §~384,

\eq{$b × c$}{$= 2 R × h$.}{}

Multiply by $a$, and we have,

\eq{$a × b × c$}{$= 2 R × a × h$.}{}

\eq[\indent But]{$a × h$}{$= 2 S$.}{§~403}

\eq{$\therefore a × b × c$}{$= 4 R × S$.}{}

\label{formareatri3}%
\eq{$\therefore S$}{$= \dfrac{abc}{4R}$.}{}

Show that the radius of the circumscribed circle is equal to
$\dfrac{abc}{4S}$.

\end{proofex}
\scanpage{218.png}%

\ex{Find the area of a right triangle, if the length of the hypotenuse
is $17$~feet and the length of one leg is $8$~feet.}

\ex{Find the ratio of the altitudes of two equivalent triangles, if
the base of one is three times that of the other.}

\ex{The bases of a trapezoid are $8$~feet and $10$~feet, and the altitude
is $6$~feet.  Find the base of the equivalent rectangle that has an
equal altitude.}

\ex{Find the area of a rhombus, if the sum of its diagonals is
$12$~feet, and their ratio is~$3:5$.}

\ex{Find the area of an isosceles right triangle, if the hypotenuse
is $20$~feet.}

\ex{In a right triangle the hypotenuse is $13$~feet, one leg is $5$~feet.
Find the area.}

\ex{Find the area of an isosceles triangle, if base $=b$, and leg $=c$.}

\ex{Find the area of an equilateral triangle, if one side $=8$~feet.}

\ex{Find the area of an equilateral triangle, if the altitude $=h$.}

\ex{A house is $40$~feet long, $30$~feet wide, $25$~feet high to the roof,
and $35$~feet high to the ridge-pole. Find the number of square feet in its
entire exterior surface.}

\ex{The sides of a right triangle are as $3:4:5$. The altitude upon
the hypotenuse is $12$~feet. Find the area.}

\ex{Find the area of a right triangle, if one leg $=a$, and the altitude
upon the hypotenuse $=h$.}

\ex{Find the area of a triangle, if the lengths of the sides are
$104$~feet, $111$~feet, and $175$~feet.}

\ex{The area of a trapezoid is $700$~square feet. The bases are
$30$~feet and $40$~feet, respectively. Find the altitude.}

\ex{$ABCD$ is a trapezium; $AB = 87$ feet, $BC = 119$ feet, $CD = 41$ feet, $DA = 169$ feet, $AC = 200$ feet. Find the area.}

\ex{What is the area of a quadrilateral circumscribed about a
circle whose radius is $25$~feet, if the perimeter of the quadrilateral is $400$
feet?  What is the area of a hexagon that has a perimeter of $400$ feet and
is circumscribed about the same circle of $25$~feet radius (Ex.~361)?}

\ex{The base of a triangle is $15$~feet, and its altitude is $8$~feet.
Find the perimeter of an equivalent rhombus, if the altitude is $6$~feet.}
\scanpage{219.png}%

\ex{Upon the diagonal of a rectangle $24$~feet by $10$~feet a triangle
equivalent to the rectangle is constructed.  What is its altitude?}

\ex{Find the side of a square equivalent to a trapezoid whose bases
are $56$~feet and $44$~feet, and each leg is $10$~feet.}

\ex{Through a point $P$ in the side $AB$ of a triangle $ABC$, a line
is drawn parallel to $BC$ so as to divide the triangle into two equivalent
parts. Find the value of $AP$ in terms of $AB$.}

\ex{What part of a parallelogram is the triangle cut off by a line
from one vertex to the middle point of one of the opposite sides?}

\ex{In two similar polygons, two homologous sides are $15$~feet
and $25$~feet. The area of the first polygon is $450$~square feet.  Find the
area of the second polygon.}

\ex{The base of a triangle is $32$~feet, its altitude $20$~feet. What is
the area of the triangle cut off by a line parallel to the base at a distance
of $15$~feet from the base?}

\ex{The sides of two equilateral triangles are $3$~feet and $4$~feet.
Find the side of an equilateral triangle equivalent to their sum.}

\ex{If the side of one equilateral triangle is equal to the altitude
of another, what is the ratio of their areas?}

\ex{The sides of a triangle are $10$~feet, $17$~feet, and $21$~feet.  Find
the areas of the parts into which the triangle is divided by the bisector of
the angle formed by the first two sides.}

\ex{In a trapezoid, one base is $10$~feet, the altitude is $4$~feet, the
area is $32$~square feet.  Find the length of a line drawn between the legs
parallel to the bases and distant $1$~foot from the lower base.}

\ex{The diagonals of a rhombus are $90$~yards and $120$~yards,
respectively. Find the area, the length of one side, and the perpendicular
distance between two parallel sides.}

\ex{Find the number of square feet of carpet that are required to
cover a triangular floor whose sides are, respectively, $26$~feet, $35$~feet, and
$51$~feet.}

\ex{If the altitude $h$ of a triangle is increased by a length $m$, how
much must be taken from the base $a$ that the area may remain the same?}

\ex{Find the area of a right triangle, having given the segments
$p$, $q$, into which the hypotenuse is divided by a perpendicular drawn to
the hypotenuse from the vertex of the right angle.}
\scanpage{220.png}%


\chapter{BOOK V\@. REGULAR POLYGONS AND CIRCLES.}
\markboth{\Headings{BOOK V\@. PLANE GEOMETRY.}}
{\Headings{REGULAR POLYGONS AND CIRCLES.}}%

\pp{\defn{A \indexbf{regular polygon} is a polygon
which is both equilateral and equiangular.  The equilateral triangle
and the square are examples.}}


\proposition{Theorem.}

\begin{proof}%
\obs{An equilateral polygon inscribed in a circle is a
  regular polygon.}

\figc{220aa430}{Let $ABC$ etc.\ be an equilateral polygon inscribed in a
  circle.}

\prove{the polygon $ABC$ etc.\ is a regular polygon.}

\step[\indent\textbf{Proof.}]{The arcs $AB$, $BC$, $CD$, etc., are equal.}{§~243}

\step{Hence, arcs $ABC$, $BCD$, etc., are equal.}{Ax.~2}

\step{Therefore, arcs $CFA$, $DFB$, etc., are equal.}{Ax.~3}

\step{Therefore, $\angle_s A$, $B$, $C$, etc., are equal.}{§~289}

Therefore, the polygon $ABC$ etc.\ is a regular polygon, being
equilateral and equiangular.~\hfill§~429

\hfill\qed

\end{proof}
\scanpage{221.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{A circle may be circumscribed about, and a circle
  may be inscribed in, any regular polygon.}

\figc{221aa431}{Let $ABCDE$ be a regular polygon.}

\prove[\textup{1.} To prove that ]{a circle may be circumscribed about $ABCDE$.}

\textbf{Proof.} Let $O$ be the centre of the circle which may be passed
through $A$, $B$, and $C$.~\hfill§~258

\step{Draw $OA$, $OB$, $OC$, and $OD$.}{}

\eq[\indent Then]{$\angle ABC$}{$= \angle BCD$,}{§~429}

\eq[and]{$\angle OBC$}{$= \angle OCB$.}{§~145}

%proofrule
\eq[\indent By subtraction,]{$\angle OBA$}{$= \angle OCD$.}{Ax.~3}

\step{The $\triangle_s OBA$ and $OCD$ are equal.}{§~143}

\eq[\indent For]{$\angle OBA$}{$= \angle OCD$,}{}

\eq{$OB$}{$= OC$,}{§~217}

\eq[and]{$AB$}{$= CD$.}{§~429}

\eq{$\therefore OA$}{$= OD$.}{§~128}

$\therefore$ the circle passing through $A$, $B$, $C$, passes through
$D$.

In like manner it may be proved that the circle passing through $B$,
$C$, and $D$ also passes through $E$; and so on.
\scanpage{222.png}%

Therefore, the circle described from $O$ as a centre, with a
radius $OA$, will be circumscribed about the polygon.~\hfill§~231

\prove[\textup{2.} To prove that ]{a circle may be inscribed in $ABCDE$.}

\textbf{Proof.}  Since the sides of the regular polygon are equal
chords of the circumscribed circle, they are equally distant
from the centre.~\hfill§~249

Therefore, the circle described from $O$ as a centre, with the
distance from $O$ to a side of the polygon as a radius, will be
inscribed in the polygon (§~232).~\hfill\qed

\end{proof}

\pp{\defn{The radius of the circumscribed circle, $OA$, is
called the \textbf{radius} of the polygon\label{polyradius}.}}

\pp{\defn{The radius of the inscribed circle, $OF$, is called
the \textbf{apothem}\label{apothem} of the polygon.}}

\pp{\defn{The common centre, $O$, of the circumscribed and
inscribed circles is called the \textbf{centre} of the polygon\label{centrepoly}.}}

\pp{\defn{The angle between radii drawn to the extremities
of any side is called the \textbf{angle at the centre} of the polygon.}}

By joining the centre to the vertices of a regular polygon,
the polygon can be decomposed into as many equal isosceles
triangles as it has sides.

\pp{\cor[1]{The angle at the centre of a regular polygon\label{anglecentreregpoly}
is equal to four right angles divided by the number of sides
of the polygon.  Hence, the angles at the centre of any regular
polygon are all equal.}}

\pp{\cor[2]{The radius drawn to any vertex of a regular
polygon bisects the angle at the vertex.}}

\pp{\cor[3]{The angle at the centre of a regular polygon
and an interior angle of the polygon are supplementary.}}

\step[\indent For]{$\angle_s FOB$ and $FBO$ are complementary.}{§~135}

$\therefore$ their doubles $AOB$ and $FBC$ are supplementary.\hfill~Ax.~6
\scanpage{223.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If the circumference of a circle is divided into
any number of equal arcs, the chords joining the successive
points of division form a regular inscribed polygon;
and the tangents drawn at the points of division
form a regular circumscribed polygon.}

\figc{223aa439}{Suppose the circumference divided into equal arcs $AB$, $BC$, etc.
Let $AB$, $BC$, etc., be the chords, $FBG$, $GCH$, etc., the tangents.}

1.~\prove{$ABCDE$ is a regular polygon.}

\step[\indent\textbf{Proof.}]{The sides $AB$, $BC$, $CD$, etc., are equal.}{§~241}

\step{Therefore, the polygon is regular.}{§~430}

2.~\prove{To prove that $FGHIK$ is a regular polygon.}

\textbf{Proof.} The $\triangle_s AFB$, $BGC$, $CHD$, etc., are all equal isosceles
triangles.\hfill\allowbreak\null\nobreak\hfill\nobreak§§~295,139

\step{$\therefore \angle_s F$, $G$, $H$, etc., are equal, and $FB$, $BG$, $GC$, etc., are equal.}{}

\step{$\therefore FG=GH=HI$, etc.}{Ax.~6}

\step{$\therefore FGHIK$ is a regular polygon.}{§~429}

\hfill\qed

\end{proof}

\pp{\cor[1]{Tangents to a circle at the vertices of a regular
inscribed polygon form a regular circumscribed polygon of
the same number of sides as the inscribed polygon.}}
\scanpage{224.png}%

\figccc{224aa441}{224bb442}{224cc443}
\begin{point}%
\cor[2]{Tangents to a circle at the middle points of
the arcs subtended by the sides of a regular inscribed polygon
form a circumscribed regular polygon,
whose sides are parallel to the sides of
the inscribed polygon and whose vertices
lie on the radii (prolonged) of the inscribed
polygon.}

For two corresponding sides, $AB$ and
$A'B'$, are perpendicular to $OM$ (§§~248,
254), and are parallel (§~104); and the tangents $MB'$ and $NB'$,
intersecting at a point equidistant from $OM$ and $ON$ (§~261),
intersect upon the bisector of the $\angle MON$ (§~162); that is,
upon the radius $OB$.
\end{point}

\pp{\cor[3]{If the vertices of a regular inscribed polygon
are joined to the middle points of the arcs subtended
by the sides of the polygon, the joining
lines form a regular inscribed polygon of
double the number of sides.}}

\pp{\cor[4]{Tangents at the middle points
the arcs between adjacent points of contact
of the sides of a regular circumscribed polygon
form a regular circumscribed polygon of
double the number of sides.}}

\begin{point}%
\cor[5]{The perimeter of an inscribed polygon is
less than the perimeter of an inscribed polygon of double
the number of sides; and the perimeter of a circumscribed
polygon is greater than the perimeter of a circumscribed
polygon of double the number of sides.}

For two sides of a triangle are together greater than the
third side.~\hfill§~138
\end{point}
\scanpage{225.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Two regular polygons of the same number of sides
are similar.}

\figc{225aa445}{Let $Q$ and $Q'$ be two regular polygons, each having $n$
sides.}

\prove{$Q$ and $Q'$ are similar.}

\textbf{Proof.}  The sum of the interior $\angle_s$ of each polygon is
equal to

\step{$(n-2)2$ rt.~$\angle_s$,}{§~205}

\pnote{(the sum of the interior $\angle_s$ of a polygon is equal to 2
  rt.\ $\angle_s$ taken as many times less two as the polygon has
  sides).}

\step{Each angle of either polygon $=
  \dfrac{(n-2) 2 \text{ rt.\ } \angle_s}{n}$,}{§~206}

\pnote{(for the $\angle_s$ of a regular polygon are all equal, and
  hence each $\angle$ is equal to the sum of the $\angle_s$ divided by
  their number).}

Hence, the two polygons $Q$ and $Q'$ are mutually equiangular.

\step{Since $AB = BC$, etc., and $A'B' = B'C'$, etc.,}{§~429}

\step{\( AB:A'B' = BC:B'C' \), etc.}{}

Hence, the two polygons have their homologous sides proportional.

\step{Therefore the two polygons are similar.}{§~351}

\hfill\qed

\end{proof}

\pp{\cor{The areas of two regular polygons
  of the same number of sides are to each other as the squares of any
  two homologous sides.}~\hfill§~412}
\scanpage{226.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The perimeters of two regular polygons of the same
  number of sides are to each other as the radii of their
  circumscribed circles, and also as the radii of their inscribed circles.}

\figc{226aa447}{Let $P$ and $P'$ denote the perimeters, $O$ and $O'$ the
  centres, of the two regular polygons.}

From $O$, $O'$ draw $OA$, $O'A'$, $OB$, $O'B'$, and the $\perp_s OM$,
$O'M'$.

\prove{$P:P' = OA:O'A' = OM:O'M'$.}

\step[\indent\textbf{Proof.}]{Since the polygons are similar,}{§~445}

\eq{$P:P'$}{$= AB:A'B'$.}{§~364}

The $\triangle_s OAB$ and $O'A'B'$ are isosceles.~\hfill§~431

\eq[\indent Now]{$\angle O$}{$= \angle O'$,}{§~436}

\eq[and]{$OA:OB$}{$= O'A':O'B'$.}{}

\step{$\therefore$ the $\triangle_s OAB$ and $O'A'B'$ are similar.}{§~357}

\eq{$\therefore AB:A'B'$}{$= OA:O'A'$.}{§~351}

\eq[\indent Also,]{$AB:A'B'$}{$= OM:O'M'$.}{§~361}

\step{$\therefore P:P' = OA:O'A' = OM:O'M'$.}{Ax.~1}

\hfill\qed

\end{proof}


\pp{\cor{The areas of two regular polygons
  of the same number of sides are to each other as the squares of the
  radii of the circumscribed circles, and of the inscribed
  circles.}~\hfill§~413}
\scanpage{227.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{If the number of sides of a regular inscribed
polygon is indefinitely increased, the apothem of the
polygon approaches the radius of the circle as its limit.}

\figc{227aa449}{Let $AB$ be a side and $OP$ the apothem of a regular polygon of $n$
sides inscribed in the circle whose radius is $OA$.}

\prove{$OP$ approaches $OA$ as a limit, when $n$ increases
indefinitely.}

\eq[\indent\textbf{Proof.}]{$OP$}{$<OA$,}{§~97}

\eq[and]{$OA-OP$}{$< AP$.}{§~138}

\eq{$\therefore OA-OP$}{$<AB$, which is twice $AP$.}{§~245 }

Now, if $n$ is taken sufficiently great, $AB$, and consequently
$OA-OP$, can be made less than any assigned value, however
small, but cannot be made zero.

Since $OA-OP$ can be made less than any assigned value
by increasing $n$, but cannot be made zero, $OA$ is the limit of
$OP$ by the test for a limit.~\hfill§~275

\hfill\qed

\end{proof}

\begin{point}%
\cor{If the number of sides of a regular inscribed
polygon is indefinitely increased, the square of the apothem
approaches the square of the radius of the circle as a limit.}

\step[\indent For]{$\overline{OA}^2 - \overline{OP}^2 = \overline{AP}^2$.}{§~372}

But by taking $n$ sufficiently great, $AB$ and consequently $AP$,
the half of $AB$, can be made less than any assigned value.
\end{point}
\scanpage{228.png}%

Therefore, $\overline{AP}^2$, the product of $AP$ by $AP$, can be made
less than any assigned value; for the product of two finite factors
approaches zero as a limit, if \emph{either} factor approaches zero as
a limit (§~276); and for a still stronger reason, the product
approaches zero as a limit, if \emph{each} of the factors approaches
zero as a limit.

\proposition{Theorem.}

\begin{proof}%
\obs{An arc of a circle is less than any line which
envelops it and has the same extremities.}

\figc{228aa451}{Let $ACB$ be an arc of a circle, and $AB$ its chord.}

\prove{the arc $ACB$ is less than any other line which
  envelops this arc and terminates at $A$ and $B$.}

% the following are steps in a proof, but do not agree with the formatting.
\textbf{Proof.} Of all the lines that can be drawn, each to include
the area $ACB$ between itself and the chord $AB$, there must be at
least one shortest line; for all the lines are not equal.

Now the enveloping line $ADB$ cannot be the shortest; for drawing
$ECF$ tangent to the arc $ACB$ at $C$, the line $AECFB < AEDFB$, since
$ECF < EDF$.~\hfill§~49

In like manner it can be shown that no other enveloping line can be
the shortest.  Therefore, $ACB$ is the shortest.

\step {} {\qed}


\end{proof}

\pp{\cor[1]{The circumference of a circle
  is less than the perimeter of any polygon circumscribed about it.}}

\pp{\cor[2]{Any convex curve\label{convexcurve} is less than
  the perimeter of a polygon circumscribed about it.}}
\scanpage{229.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The circumference of a circle is the limit which
  the perimeters of regular inscribed polygons and of similar
  circumscribed polygons approach, if the number of sides of the
  polygons is indefinitely increased; and the area of a circle is the
  limit which the areas of these polygons approach.}

\figc{229aa454}{Let $P$ and $P'$ denote the lengths of the perimeters, $AB$
  and $A'B'$ two homologous sides, $R$ and $R'$ the radii, of the
  polygons, and $C$ the circumference of the circle.}

\prove{$C$ is the limit of $P$ and of $P'$, if
  the number of sides of the polygons is indefinitely increased.}

\step[\indent\textbf{Proof.}]{Since the polygons are similar by hypothesis,}{}

\eq {$P':P $} {$= R':R$.} {§~447}

\eq [\indent Therefore,] {$P'-P:P $} {$= R'-R:R$.} {§~333}

\eq [\indent Whence,] {$R(P'-P) $} {$ = P(R'-R)$.} {§~327}

\eq [\indent Therefore,] {$P'-P $} {$= \frac{P}{R}(R' - R)$.} {}

\step {Now $P$ is always less than $C$.} {§~273}

\eq {$\therefore P'-P $} {$< \frac{C}{R} (R'-R)$.}{}
\scanpage{230.png}%

But $R'-R$, which is less than $A'C$ (§~138), can be made less than
any assigned quantity by increasing the number of sides of the
polygons; and therefore $\dfrac{C}{R}(R'-R)$ can be made less than any
assigned quantity.~\hfill§~276

Hence, $P'-P$ can be made less than any assigned quantity.

Since $P'$ is always greater than $C$ (§~452), and $P$ is always less
than $C$ (§~273), the difference between $C$ and either $P'$ or $P$
is less than the difference $P'-P$, and consequently can be made less
than any assigned quantity, but cannot be made zero.

Therefore, $C$ is the common limit of $P'$ and $P$.~\hfill§~275

\lett{Let $K$ denote the area of the circle, $S$ the area of the
  inscribed polygon, and $S'$ the area of the circumscribed polygon.}

2. \prove{$K$ is the limit of $S$ and $S'$.}

\eq[\indent\textbf{Proof.}]{$S':S$}{$= R'^2:R^2$.}{§~448}

\eq[\indent By division,]{$S'-S:S$}{$= R'^2-R^2:R^2$.}{§~333}

\eq[\indent Whence]{$S'-S$}{$= \dfrac{S}{R^2}(R'^2-R^2)$.}{}

\step{Now $K$ is always greater than $S$.}{Ax.~8}

\eq[\indent Therefore,]{$S'-S$}{$< \dfrac{K}{R^2}(R'^2-R^2)$.}{}

But $R'^2 - R^2$, which is equal to $(R'+R)(R'-R)$, can be made less
than any assigned quantity; and therefore $\dfrac{K}{R^2}(R'^2-R^2)$ can
be made less than any assigned quantity.~\hfill§~276

Hence, $S'-S$ can be made less than any assigned quantity.

Since $S' > K$ always, and $S < K$ always (Ax.~8), the difference
between $K$ and either $S'$ or $S$ is less than the difference $S'-S$,
and consequently can be made less than any assigned quantity, but
cannot be made zero.

Therefore, $K$ is the common limit of $S'$ and $S$.~\hfill§~275

\hfill\qed

\end{proof}
\scanpage{231.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Two circumferences have the same ratio as their
radii.}

\figc{231aa455}{Let $C$ and $C'$ be the circumferences, $R$ and $R'$ the radii, of the two
circles $Q$ and $Q'$.}

\proveq{$C:C'$}{$= R:R'$.}

\textbf{Proof.}   Inscribe in the $\odot_s$ two similar regular polygons, and
denote their perimeters by $P$ and $P'$.

\eq[\indent Then]{$P:P'$}{$= R:R'$.}{§~447}

Conceive the number of sides of these regular polygons to
be indefinitely increased, the polygons continuing similar.

Then $P$ and $P'$ will have $C$ and $C'$ as limits.~\hfill§~454

But $P:P'$ will always be equal to $R:R'$.~\hfill§~447

\eq{$\therefore C:C'$}{$= R:R'$.}{§~285}

\hfill\qed

\end{proof}

\begin{point}%
\cor{The ratio of the circumference of a circle to its
diameter is constant.}

\eq[\indent For]{$C:C'$}{$= R: R'$.}{§~455}

\eq{$\therefore C:C'$}{$= 2R:2R'$.}{§~340}

\eq[\indent By alternation,]{$C: 2R$}{$= C' : 2R'$.}{§~330}
\end{point}

\pp{\defn{The constant ratio of the circumference of a
circle to its diameter is represented by the Greek letter $\pi$\label{pi}.}}

\label{formcircum}%
\pp{\cor{$\pi = \dfrac{C}{2R}$. $\therefore C=2\pi R$.}}
\scanpage{232.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The area of a regular polygon is equal to half
the product of its apothem by its perimeter.}

\figc{232aa459}{Let $P$ represent the perimeter, $R$ the apothem, and $S$ the area of
the regular polygon $ABC$ etc.}

\prove{$S = \frac{1}{2}R × P$.}

\step[\indent\textbf{Proof.}]{Draw the radii $OA$, $OB$, $OC$, etc.}{}

\step{The polygon is divided into as many $\triangle_s$ as it has sides.}{}

\step{The apothem is the common altitude of these $\triangle_s$,}{}

\step{and the area of each $\triangle=\frac{1}{2}R$ multiplied by the base.}{§~403}

Hence, the area of all the $\triangle_s$ is equal to $\frac{1}{2}R$ multiplied by
the sum of all the bases.

But the sum of the areas of all the $\triangle_s$ is equal to the area of
the polygon.\hfill~Ax.~9

And the sum of all the bases of the $\triangle_s$ is equal to the perimeter
of the polygon.\hfill~Ax.~9

\label{formareapoly}%
\step{$\therefore S = \frac{1}{2}R × P$.}{\qed}

\end{proof}

\pp{\defn{In different circles \indexbf{similar arcs}, \indexbf{similar sectors},
and \indexbf{similar segments} are such as correspond to \emph{equal angles at
the centre}.}}
\scanpage{233.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{The area of a circle is equal to half the product
of its radius by its circumference.}

\figc{233aa461}{Let $R$ represent the radius, $C$ the circumference, and $S$ the area,
of the circle whose centre is $O$.}

\proveq{$S$}{$= \frac{1}{2}R × C$.}

\textbf{Proof.} Circumscribe any regular polygon about the circle,
and denote its perimeter by $P$, and its area by $S'$.

\eq[\indent Then]{$S'$}{$= \frac{1}{2} R × P$.}{§~459}

Conceive the number of sides of the polygon to be indefinitely
increased.

\step{Then $P$ approaches $C$ as its limit,}{§~454}

\step{$\frac{1}{2}R × P$ approaches $\frac{1}{2}R× C$ as its limit,}{§~279}

\step{and $S'$ approaches $S$ as its limit.}{§~454}

\step[\indent But]{$S' = \frac{1}{2} R × P$, always.}{§~459}

\step{$\therefore S = \frac{1}{2}R× C$.}{§~284}

\hfill\qed

\end{proof}

\label{formareasector}%
\begin{point}%
\cor[1]{The area of a sector is equal to half the
product of its radius by its arc.}

For the sector and its arc are like parts of the circle and
its circumference, respectively.
\end{point}

\begin{point}%
\cor[2]{The area of a circle is equal to $\pi$ times the
square of its radius.}

\label{formareacircle}%
For the area of the \( \odot = \frac{1}{2} R × C =
\frac{1}{2} R × 2\pi R = \pi R^2 \).

\end{point}
\scanpage{234.png}%

\begin{point}%
\cor[3]{The areas of two circles are to
  each other as the squares of their radii.}

For, if $S$ and $S'$ denote the areas, and $R$ and $R'$ the radii,

\[ S:S' = \pi R^2:\pi R'^2 = R^2:R'^2. \]
\end{point}

\pp{\cor[4]{Similar arcs are to each other
  as their radii; similar sectors are to each other as the squares of
  their radii.}}

\proposition{Theorem.}

\begin{proof}%
\obs{The areas of two similar segments are to each
  other as the squares of their radii.}

\figc{234aa466}{Let $AC$ and $A'C'$ be the radii of the two similar sectors
  $ACB$ and $A'C'B'$, and let $ABP$ and $A'B'P'$ be the corresponding
  segments.}

\proveq{$ABP:A'B'P'$}{$= \overline{AC}^2:\overline{A'C'}^2$.}

\step[\indent\textbf{Proof.}]{Sector $ACB :$ Sector $A'C'B' =
 \overline{AC}^2:\overline{A'C'}^2$.}{§~465}

\step{The $\triangle_s ACB$ and $A'C'B'$ are similar.}{§~357}

\eq{$\therefore \triangle ACB:\triangle A'C'B'$}
  {$=\overline{AC}^2:\overline{A'C'}^2$.}{§~411}

\eq{$\therefore$ sector $ACB :$ sector $A'C'B'$}
  {$=\triangle ACB : \triangle A'C'B'$.}{Ax.~1}

\eq{$\therefore$ sector $ACB : \triangle ACB$}
  {$=$ sector $A'C'B' : \triangle A'C'B'$.}{§~330}

\step{\( \therefore
  \dfrac{\text{sector } ACB-\triangle ACB}
       {\text{sector } A'C'B'-\triangle A'C'B'} =
  \dfrac{\triangle ACB}{\triangle A'C'B'} =
  \dfrac{\overline{AC}^2}{\overline{A'C'}^2} \).}{§~333}

\eq[\indent That is,]{$ABP:A'B'P'$}{$= \overline{AC}^2:\overline{A'C'}^2$.}{\qed}


\end{proof}
\scanpage{235.png}%


\clearpage
\section{PROBLEMS OF CONSTRUCTION.}

\proposition{Problem.}

\begin{proof}%
\obs{To inscribe a square in a given circle.}

\figc{235aa467}{Let $O$ be the centre of the given circle.}

\prove[]{To inscribe a square in the given circle.}

\step{Draw two diameters $AC$ and $BD \perp$ to each other.}{}

\step{Draw $AB$, $BC$, $CD$, and $DA$.}{}

\step{Then $ABCD$ is the square required.}{}

\step[\indent\textbf{Proof.}]{The $\angle_s ABC$, $BCD$, etc., are rt.\ $\angle_s$,}{§~290}

\pnote{(each being inscribed in a semicircle),}

\step{and the sides $AB$, $BC$, etc., are equal,}{§~241}

\pnote{(in the same $\odot$ equal arcs are subtended by equal chords).}

\step{Hence the quadrilateral $ABCD$ is a square.}{§~168}

\hfill\qef

\end{proof}

\pp{\cor{By bisecting the arcs $AB$, $BC$,
  etc., a regular polygon of eight sides may be inscribed in the
  circle; and, by continuing the process, regular polygons of sixteen,
  thirty-two, sixty-four, etc., sides may be inscribed.}}


\ex{The area of a circumscribed square is equal to
twice the area of the inscribed square.}

\ex{The area of a circular ring is equal to that of a
circle whose diameter is a chord of the outer circle tangent to the
inner circle.}
\scanpage{236.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To inscribe a regular hexagon in a given circle.}

\figc{236aa469}{Let $O$ be the centre of the given circle.}

\prove[To inscribe ]{a regular hexagon in the given circle.}

\step{From $O$ draw any radius, as $OC$.}{}

\step{From $C$ as a centre, with a radius equal to $OC$,}{}

\step{describe an arc intersecting the circumference at $F$.}{}

\step{Draw $OF$ and $CF$.}{}

\step{Then $CF$ is a side of the regular hexagon required.}{}

\step[\indent\textbf{Proof.}]{The $\triangle OFC$ is equiangular,}{§~146}

\pnote{(since it is equilateral by construction).}

Hence, the $\angle FOC$ is $\frac{1}{3}$ of $2$~rt.~$\angle_s$, or
$\frac{1}{6}$ of $4$~rt.~$\angle_s$.~\hfill§~136

\step{$\therefore$ the arc $FC$ is $\frac{1}{6}$ of the circumference,}{}

and the chord $FC$ is a side of a regular inscribed hexagon.

Hence, to inscribe a regular hexagon apply the radius six times as a
chord.~\hfill\qef

\end{proof}

\pp{\cor[1]{By joining the alternate
  vertices $A$, $C$, $D$, an equilateral triangle is inscribed in the
  circle.}}

\pp{\cor[2]{By bisecting the arcs $AB$,
  $BC$, etc., a regular polygon of twelve sides may be inscribed in
  the circle; and, by continuing the process, regular polygons of
  twenty-four, forty-eight, etc., sides may be inscribed.}}
\scanpage{237.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To inscribe a regular decagon in a given circle.}

\figc{237aa472}{Let $O$ be the centre of the given circle.}

\prove[To inscribe ]{a regular decagon in the given circle.}

\step{Draw any radius $OC$,}{}

\step{and divide it in extreme and mean ratio, so that $OC$ shall}{}

\step{be to $OS$ as $OS$ is to $SC$.}{§~389}

\step{From $C$ as a centre, with a radius equal to $OS$,}{}

\step{describe an arc intersecting the circumference at $B$.}{}

\step{Draw $BC$.}{}

\step{Then $BC$ is a side of the regular decagon required.}{}

\step[\indent\textbf{Proof.}]{Draw $BS$ and $BO$.}{}

\eq[\indent Now]{$OC:OS$}{$= OS:SC$,}{Const.}

\eq[and]{$BC$}{$= OS$.}{Const.}

\eq{$\therefore OC:BC$}{$= BC:SC$.}{}

\eq[\indent Moreover,]{$\angle OCB$}{$= \angle SCB$.}{Iden.}

\step{Hence, the $\triangle_s OCB$ and $BCS$ are similar.}{§~357}

\step{But the $\triangle OCB$ is isosceles.}{§~217}

\step{$\therefore \triangle BCS$, which is similar to the $\triangle OCB$,
  is isosceles,}{}

\step{and $CB = BS = SO$.}{§~120}
\scanpage{238.png}%

\step{$\therefore$ the $\triangle SOB$ is isosceles, and the $\angle O =
  \angle SBO$.}{§~145}

\step{But the ext.~$\angle CSB = \angle O + \angle SBO = 2 \angle O$.}{§~137}

\eq[\indent Hence,]{$\angle SCB$}{$= 2\angle O$,}{}

\eq[and]{$\angle OBC$}{$= 2\angle O$.}{}

\step{$\therefore$ the sum of the $\angle_s$ of the $\triangle OCB = 5
\angle O = 2 \text{rt.\ } \angle_s$,}{}

\step[and]{$\angle O = \frac{1}{5}$ of $2$~rt.~$\angle_s$, or $\frac{1}{10}$ of $4$~rt.~$\angle_s$.}{}

\step{Therefore, the arc $BC$ is $\frac{1}{10}$ of the circumference,}{}

and the chord $BC$ is a side of a regular inscribed decagon.

Therefore, to inscribe a regular decagon, divide the radius internally
in extreme and mean ratio, and apply the greater segment ten times as
a chord.~\hfill\qef

\end{proof}

\pp{\cor[1]{By joining the alternate vertices
  of a regular inscribed decagon, a regular pentagon is inscribed.}}

\pp{\cor[2]{By bisecting the arcs $BC$, $CF$,
  etc., a regular polygon of twenty sides may be inscribed in the
  circle; and, by continuing the process, regular polygons of forty,
  eighty, etc., sides may be inscribed.}}


If $R$ denotes the radius of a regular inscribed polygon, $r$ the
apothem, $a$ one side, $A$ an interior angle, and $C$ the angle at the
centre, show that

\ex{In a regular inscribed triangle $a = R\sqrt{3}$,
$r =\frac{1}{2}R$, $A = 60°$, $C = 120°$.}

\ex{In an inscribed square $a = R\sqrt{2}$,
$r = \frac{1}{2}R\sqrt{2}$, $A=90°$, $C=90°$.}

\ex{In a regular inscribed hexagon $a = R$,
$r = \frac{1}{2}R\sqrt{3}$, $A=120°$, $C=60°$.}

\ex{In a regular inscribed decagon
\[ a = \frac{R(\sqrt{5}-1)}{2},\
   r = \frac{1}{4} R \sqrt{10+2\sqrt{5}},\
   A = 144°,\
   C = 36°. \]}
\scanpage{239.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To inscribe in a given circle a regular
  pentedecagon, or polygon of fifteen sides.}

\figc{239aa475}{Let $Q$ be the given circle.}

\prove[To inscribe ]{in $Q$ a regular pentedecagon.}

\step{Draw $EH$ equal to the radius of the circle,}{}

\step{and $EF$ equal to a side of the regular inscribed decagon.}{§~472}

\step{Draw $FH$.}{}

Then $FH$ is a side of the regular pentedecagon required.

\step[\indent\textbf{Proof.}]{The arc $EH$ is $\frac{1}{6}$ of the circumference,}{§~469}

\step{and the arc $EF$ is $\frac{1}{10}$ of the circumference.}{Const.}

Hence, the arc $FH$ is $\frac{1}{6} - \frac{1}{10}$, or
$\frac{1}{15}$, of the circumference.

And the chord $FH$ is a side of a regular inscribed pentedecagon.

By applying $FH$ fifteen times as a chord, we have the polygon
required.~\hfill\qef

\end{proof}

\pp{\cor{By bisecting the arcs $FH$, $HA$,
  etc., a regular polygon of thirty sides may be inscribed; and, by
  continuing the process, regular polygons of sixty, one hundred
  twenty, etc., sides may be inscribed.}}
\scanpage{240.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To inscribe in a given circle a regular
  polygon similar to a given regular polygon.}

\figc{240aa477}{Let $ABC$ etc.\ be the given regular polygon, and $O'$ the
  centre of the given circle.}

\prove[To inscribe ]{in the circle a regular polygon similar to $ABC$
  etc.}

\step{From $O$, the centre of the given polygon,}{}

\step{draw $OD$ and $OC$.}{}

\step{From $O'$, the centre of the given circle,}{}

\step{draw $O'C'$ and $O'D'$,}{}

\step{making the $\angle O'$ equal to the $\angle O$.}{}

\step{Draw $C'D'$.}{}

\step{Then $C'D'$ is a side of the regular polygon required.}{}

\textbf{Proof.} Each polygon has as many sides as the $\angle O$, or
$\angle O'$, is contained times in $4$~rt.~$\angle_s$.

Therefore, the polygon $C'D'E'$ etc.\ is similar to the polygon $CDE$
etc.,~\hfill§~445

\pnote{(two regular polygons of the same number of sides are similar).}

\hfill\qef

\end{proof}


\ex{The area of an inscribed regular octagon is equal to
that of the rectangle whose sides are equal to the sides of the
inscribed and the circumscribed squares.
}
\scanpage{241.png}%

\proposition{Problem.}

\begin{proof}%
\obs{Given the side and the radius of a regular
  inscribed polygon, to find the side of the regular inscribed polygon
  of double the number of sides.}

\figc{241aa478}{Let $AB$ be a side of the regular inscribed polygon.}

\prove[To find ]{$AD$, a side of the regular inscribed polygon of double
  the number of sides.}

\step{Denote the radius by $R$, and $AB$ by $a$.}{}

\step{From $D$ draw $DH$ through the centre $O$, and draw $OA$, $AH$.}{}

\step{$DH$ is $\perp$ to $AB$ at its middle point $C$.}{§~161}

\eq[\indent In the rt.~$\triangle OCA$,]{$\overline{OC}^2$}
  {$= R^2 - \frac{1}{4}a^2$.}{§~372}

\eq[\indent Therefore,]{$OC$}{$= \sqrt{R^2 - \frac{1}{4}a^2}$,}{}

\eq[and]{$DC$}{$= R - \sqrt{R^2 - \frac{1}{4}a^2}$.}{}

\step{The $\angle DAH$ is a rt.~$\angle$.}{§~290}

In the rt.~$\triangle DAH$, \( \overline{AD}^2 = DH × DC \).~\hfill§~367

But $DH = 2R$, and \( DC = R - \sqrt{R^2 - \frac{1}{4}a^2} \).

\eq{$\therefore AD$}{$= \sqrt{2R(R - \sqrt{R^2 - \frac{1}{4}a^2})}$}{}

\eq{}{$= \sqrt{R(2R - \sqrt{4R^2 - a^2})}$.}{\qef}

\end{proof}

\pp{\cor{If $R=1$, $AD =
  \sqrt{2-\sqrt{4-a^2}}$.}}
\scanpage{242.png}%

\proposition{Problem.}

\begin{proof}%
\obs{To find the numerical value of the ratio of the
  circumference of a circle to its diameter.}

\figc{242aa480}{Let $C$ be the circumference, when the radius is unity.}

\prove[To find ]{the numerical value of $\pi$.}

By §~458, $2\pi R = C$. \qquad $\therefore \pi = \frac{1}{2}C$ when
$R = 1$.

Let $S_6$ be the length of a side of a regular polygon of $6$~sides,
$S_{12}$ of $12$~sides, and so on.

If $R=1$, by §~469, $S_6=1$ and by §~479 we have
\[
\begin{array}{lcc}
\multicolumn{1}{c}{\text{\footnotesize Form of Computation.}} &
\text{\footnotesize Length of Side.} &
\text{\footnotesize Length of Perimeter.} \\
%
S_{12} = \sqrt{2 - \sqrt{4 - 1^2}} &
0.51763809 &
6.21165708 \\
%
S_{24} = \sqrt{2 - \sqrt{4 - (0.51763809)^2}} &
0.26105238 &
6.26525722 \\
%
S_{48} = \sqrt{2 - \sqrt{4 - (0.26105238)^2}} &
0.13080626 &
6.27870041 \\
%
S_{96} = \sqrt{2 - \sqrt{4 - (0.13080626)^2}} &
0.06543817 &
6.28206396 \\
%
S_{192} = \sqrt{2 - \sqrt{4 - (0.06543817)^2}} &
0.03272346 &
6.28290510 \\
%
S_{384} = \sqrt{2 - \sqrt{4 - (0.03272346)^2}} &
0.01636228 &
6.28311544 \\
%
S_{768} = \sqrt{2 - \sqrt{4 - (0.01636228)^2}} &
0.00818121 &
6.28316941 \\
\end{array}
\]
$\therefore C = 6.28317$ approximately; that is, $\pi = 3.14159$
nearly.
\hfill\qef

\end{proof}

\begin{point}%
\textsc{Scholium.~}$\pi$ is incommensurable.  We
generally take
\[ \pi = 3.1416, \text{ and } \frac{1}{\pi} = 0.31831. \]
\end{point}
\scanpage{243.png}%


\section{MAXIMA AND MINIMA.}

\begin{point}%
\defn{Among geometrical magnitudes which satisfy
given conditions, the \emph{greatest} is called the \indexbf{maximum}; and
the \emph{smallest} is called the \indexbf{minimum}.}

Thus, the diameter of a circle is the maximum among all chords; and
the perpendicular is the minimum among all lines drawn to a given line
from a given external point.
\end{point}

\pp{\defn{\indexbf{Isoperimetric} polygons are polygons which have
equal \newline perimeters.}}

\proposition{Theorem.}

\begin{proof}%
\obs{Of all triangles having two given sides, that in
which these sides include a right angle is the maximum.}

\figc{243aa484}{Let the triangles $ABC$ and $EBC$ have the sides $AB$ and $BC$ equal
to $EB$ and $BC$, respectively; and let the angle $ABC$ be a right angle.}

\proveq{$\triangle ABC$}{$> \triangle EBC$.}

\step[\indent\textbf{Proof.}]{From $E$ draw the altitude $ED$.}{}

The $\triangle_s ABC$ and $EBC$, having the same base, $BC$, are to
each other as their altitudes $AB$ and $ED$.~\hfill§~405

\eq[\indent Now]{$EB$}{$> ED$.}{§~97}

\eq[\indent But]{$EB$}{$= AB$.}{Hyp.}

\eq{$\therefore AB$}{$> ED$.}{}

\eq{$\therefore \triangle ABC$}{$> \triangle EBC$.}{§~405}

\hfill\qed

\end{proof}
\scanpage{244.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Of all isoperimetric triangles having the same
  base the isosceles triangle is the maximum.}

\figc{244aa485}{Let the $\triangle_s ACB$ and $ADB$ have equal perimeters, and
  let $AC$ and $CB$ be equal, and $AD$ and $DB$ be unequal.}

\prove {$\triangle ACB > \triangle ADB$.}

\step [\indent Proof.] {Produce $AC$ to $H$, making $CH = AC$; and draw $HB$.} {}

\step {Produce $HB$, take $DP$ equal to $DB$, and draw $AP$.} {}

\step {Draw $CE$ and $DF \perp$ to $AB$, and $CK$ and $DM \parallel$ to $AB$.} {}

\step {The $\angle ABH$ is a right $\angle$, for it may be inscribed in the
semicircle whose centre is $C$ and radius $CA$.} {§~290}

$ADP$ is not a straight line, for then the $\angle_s DBA$ and $DAB$
would be equal, being complements of the equal $\angle_s DBM$ and
$DPM$, respectively; and $DA$ and $DB$ would be equal (§~147), which
is contrary to the hypothesis.  Hence,

\step {$AP < AD + DP$, $\therefore < AD+DB$, $\therefore < AC+CB$,
$\therefore < AH$.} {}

\eq {$\therefore BH $} {$> BP$.} {§~102}

\eq {$\therefore CE(=\frac{1}{2} BH) $} {$> DF (=\frac{1}{2}BP)$.} {Ax.~7}

\eq [\indent Therefore,] {$\triangle ACB $} {$> \triangle ADB$.} {§~405}

\step {} {\qed}

\end {proof}
\scanpage{245.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Of all polygons with sides all given but one, the
  maximum can be inscribed in a semicircle which has the undetermined
  side for its diameter.}

\figc{245aa486}{Let $ABCDE$ be the maximum of polygons with sides $AB$, $BC$,
  $CD$, $DE$, and the extremities $A$ and $E$ on the straight line
  $MN$.}

\prove{$ABCDE$ can be inscribed in a semicircle.}

\textbf{Proof.} From \emph{any} vertex, as $C$, draw $CA$ and $CE$.

The $\triangle ACE$ must be the maximum of all $\triangle_s$ having
the
sides $CA$ and $CE$, and the third side on $MN$; otherwise by
increasing or diminishing the $\angle ACE$, keeping the lengths of
the sides $CA$ and $CE$ unchanged, but sliding the extremities
$A$ and $E$ along the line $MN$, we could increase the $\triangle ACE$,
while the rest of the polygon would remain unchanged; and
therefore increase the polygon.  But this is contrary to the hypothesis that the polygon is the maximum polygon.

Hence, the $\triangle ACE$ is the maximum of $\triangle_s$ that have
the sides $CA$ and $CE$.


\step{Therefore, the $\angle ACE$ is a right angle.}{§~484}

\step{Therefore, $C$ lies on the semicircumference.}{§~290}

Hence, \emph{every} vertex lies on the circumference; that is, the
maximum polygon can be inscribed in a semicircle having the
undetermined side for a diameter.~\hfill\qed

\end{proof}
\scanpage{246.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Of all polygons with given sides, that which can
  be inscribed in a circle is the maximum.}

\figc{246aa487}{Let $ABCDE$ be a polygon inscribed in a circle, and
  $A'B'C'D'E'$ be a polygon, equilateral with respect to $ABCDE$,
  which cannot be inscribed in a circle.}

\prove{that $ABCDE > A'B'C'D'E'$.}

\step [\indent Proof.] {Draw the diameter $AH$, and draw $CH$ and $DH$.} {}

\step {Upon $C'D'$ construct the $\triangle C'H'D' = \triangle CHD$, and draw
$A'H'$.} {}

Since, by hypothesis, a $\odot$ cannot pass through \emph{all} the
vertices of $A'B'C'D'E'$, \emph{one} or \emph{both} of the parts
$ABCH$, $AEDH$ must be greater than the corresponding part of
$A'B'C'H'D'E'$.\hfill§~486

If either of these parts is \emph{not greater than} its corresponding
part, it is equal to it,\hfill§~486

\pnote{(for $ABCH$ and $AEDH$ are the maxima of polygons that have
  sides equal to $AB$, $BC$, $CH$, and $AE$, $ED$, $DH$, respectively,
  and the remaining side undetermined).}

\eq {$\therefore ABCHDE $} {$> A'B'C'H'D'E'$.} {Ax.~4}

\step {Take away from the two figures the equal $\triangle_s CHD$ and
$C'H'D'$.}{}

\eq [\indent Then] {$ABCDE $} {$> A'B'C'D'E'$.} {Ax.~5}

\step {} {\qed}

\end {proof}
\scanpage{247.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Of isoperimetric polygons of the same number of
  sides, the maximum is equilateral.}

\figc{247aa488}{Let $ABCD$ etc.\ be the maximum of isoperimetric polygons of
  any given number of sides.}

\prove{$AB$, $BC$, $CD$, etc., are equal.}

\step[\indent\textbf{Proof.}]{Draw $AC$.}{}

The $\triangle ABC$ must be the maximum of all the $\triangle_s$ which
are formed upon $AC$ with a perimeter equal to that of $\triangle
ABC$.

Otherwise a greater $\triangle AKC$ could be substituted for
$\triangle ABC$, without changing the perimeter of the polygon.

But this is inconsistent with the hypothesis that the polygon $ABCD$
etc.\ is the maximum polygon.


\step{$\therefore$ the $\triangle ABC$ is isosceles.}{§~485}

\step{$\therefore AB = BC$.}{}

In like manner it may be proved that $BC=CD$, etc.~\hfill\qed

\end{proof}

\begin{point}%
\cor{The maximum of isoperimetric
  polygons of the same number of sides is a regular polygon.}

For the maximum polygon is equilateral (§~488), and can be inscribed
in a circle (§~487), and is, therefore, regular.~\hfill§~430
\end{point}
\scanpage{248.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Of isoperimetric regular polygons, that which has
  the greatest number of sides is the maximum.}

\figc{248aa490}{Let $Q$ be a regular polygon of three sides, and $Q'$ a
  regular polygon of four sides, and let the two polygons have equal
  perimeters.}

\prove{$Q'$ is greater than $Q$.}

\textbf{Proof.} Draw $CD$ from $C$ to any point in $AB$.

Invert the $\triangle CDA$ and place it in the position $DCE$, letting
$D$ fall at $C$, $C$ at $D$, and $A$ at $E$.

The polygon $DBCE$ is an irregular polygon of four sides, which by
construction has the same perimeter as $Q'$, and the same area as $Q$.

Then the irregular polygon $DBCE$ of four sides is less than the
isoperimetric regular polygon $Q'$ of four sides.~\hfill§~489

In like manner it may be shown that $Q'$ is less than an isoperimetric
regular polygon of five sides, and so on.~\hfill\qed

\end{proof}

\ex{Of all equivalent parallelograms that have equal
bases, the rectangle has the minimum perimeter.}

\ex{Of all equivalent rectangles, the square has the
minimum perimeter.}

\ex{Of all triangles that have the same base and the
same altitude, the isosceles has the minimum perimeter.}

\ex{Of all triangles that can be inscribed in a given
circle, the equilateral is the maximum and has the maximum perimeter.}
\scanpage{249.png}%

\proposition{Theorem.}

\begin{proof}%
\obs{Of regular polygons having a given area, that
  which has the greatest number of sides has the least perimeter.}

\figc{249aa491}{Let $Q$ and $Q'$ be regular polygons having the same area, and
  let $Q'$ have the greater number of sides.}

\proveq[\indent To prove ]{the perimeter of $Q$}{$>$ the perimeter of $Q'$.}

\textbf{Proof.} Let $Q''$ be a regular polygon having the same
perimeter as $Q'$, and the same number of sides as $Q$.

\eq[\indent Then]{$Q'$}{$> Q''$}{§~490}

\pnote{(of isoperimetric regular polygons, that which has the greatest
number of sides is the maximum).}

\eq[\indent But]{$Q$}{$\Bumpeq Q'$.}{Hyp.}

\eq{$\therefore Q$}{$> Q''$.}{}

\eq{$\therefore$ the perimeter of $Q$}{$>$ the perimeter of $Q''$.}{}

\eq{But the perimeter of $Q'$}{$=$ the perimeter of $Q''$.}{Hyp.}

\eq{$\therefore$ the perimeter of $Q$}{$>$ the perimeter of $Q'$.}{\qed}


\end{proof}



\ex{To inscribe in a semicircle the maximum rectangle.}

\ex{Of all polygons of a given number of sides which may
be inscribed in a given circle, that which is regular has the maximum
area and the maximum perimeter.}

\ex{Of all polygons of a given number of sides which may
be circumscribed about a given circle, that which is regular has the
minimum area and the minimum perimeter.
}
\scanpage{250.png}%


\section[EXERCISES.]{THEOREMS.}

\ex{Every equilateral polygon circumscribed about a circle is
regular if it has an \emph{odd} number of sides.}

\ex{Every equiangular polygon inscribed in a circle is regular if
it has an \emph{odd} number of sides.}

\ex{Every equiangular polygon circumscribed about a circle is
regular.}

\ex{The side of a circumscribed equilateral triangle is equal to
twice the side of the similar inscribed triangle.}

\ex{The apothem of an inscribed regular hexagon is equal to half
the side of the inscribed equilateral triangle.}

\ex{The area of an inscribed regular hexagon is three fourths of
the area of the circumscribed regular hexagon.}

\ex{The area of an inscribed regular hexagon is the mean proportional
between the areas of the inscribed and the circumscribed equilateral
triangles.}

\ex{The square of the side of an inscribed equilateral triangle is
equal to three times the square of a side of the inscribed regular hexagon.}

\ex{The area of an inscribed equilateral triangle is equal to half
the area of the inscribed regular hexagon.}

\ex{The square of the side of an inscribed equilateral triangle is
equal to the sum of the squares of the sides of the inscribed square and of
the inscribed regular hexagon.}

\ex{The square of the side of an inscribed regular pentagon is
equal to the sum of the squares of the radius of the circle and the side of
the inscribed regular decagon.}

\exheader{If $R$ denotes the radius of a circle, and $a$ one side of an inscribed regular
polygon, show that:}

\ex{In a regular pentagon, $a = \frac{1}{2} R \sqrt{10 - 2 \sqrt{5}}$.}

\ex{In a regular octagon, $a = R \sqrt{2 - \sqrt{2}}$.}

\ex{In a regular dodecagon, $a = R \sqrt{2 - \sqrt{3}}$.}

\ex{If two diagonals of a regular pentagon intersect, the longer
segment of each is equal to a side of the pentagon.}
\scanpage{251.png}%

\ex{The apothem of an inscribed regular pentagon is equal to half
the sum of the radius of the circle and the side of the inscribed regular
decagon.}

\ex{The side of an inscribed regular pentagon is equal to the
hypotenuse of the right triangle which has for legs the radius of the
circle and the side of the inscribed regular decagon.}

\ex{The radius of an inscribed regular polygon is the mean proportional
between its apothem and the radius of the similar circumscribed
regular polygon.}

\ex{If squares are constructed outwardly upon the six sides of a
regular hexagon, the exterior vertices of these squares are the vertices of
a regular dodecagon.}

\ex{If the alternate vertices of a regular hexagon are joined by
straight lines, show that another regular hexagon is thereby formed.
Find the ratio of the areas of these two hexagons.}

\ex{If on the legs of a right triangle as diameters semicircles are
described external to the triangle, and from the whole figure a semicircle
on the hypotenuse is subtracted, the remaining figure is equivalent to the
given right triangle.}

\ex{The star-shaped polygon, formed by producing the sides of a
regular hexagon, is equivalent to twice the given hexagon.}

\ex{The sum of the perpendiculars drawn to the sides of a regular
polygon from any point within the polygon is equal to the apothem multiplied
by the number of sides.}

\ex{If two chords of a circle are perpendicular to each other, the
sum of the four circles described on the four segments as diameters is
equivalent to the given circle.}

\ex{If the diameter of a circle is divided into any two segments,
and upon these segments as diameters semicircumferences are described
upon opposite sides of the diameter, these semicircumferences divide the
circle into two parts which have the same ratio as the two segments of
the diameter.}

\ex{The diagonals that join any vertex of a regular polygon to
all the vertices not adjacent divide the angle at that vertex into as many
equal parts less two as the polygon has sides.}
\scanpage{252.png}%


\subsection{PROBLEMS OF CONSTRUCTION.}

\ex{To circumscribe an equilateral triangle about a given circle.}

\ex{To circumscribe a square about a given circle.}

\ex{To circumscribe a regular hexagon about a given circle.}

\ex{To circumscribe a regular octagon about a given circle.}

\ex{To circumscribe a regular pentagon about a given circle.}

\ex{To draw through a given point a line so as to divide a given
circumference into two parts having the ratio $3:7$.}

\ex{To construct a circumference equal to the sum of two given
circumferences.}

\ex{To construct a circumference equal to the difference of two
given circumferences.}

\ex{To construct a circle equivalent to the sum of two given
circles.}

\ex{To construct a circle equivalent to the difference of two given
circles.}

\ex{To construct a circle equivalent to three times a given circle.}

\ex{To construct a circle equivalent to three fourths of a given
circle.}

\ex{To construct a circle whose ratio to a given circle shall be
equal to the given ratio $m:n$.}

\ex{To divide a given circle by a concentric circumference into
two equivalent parts.}

\ex{To divide a given circle by concentric circumferences into five
equivalent parts.}

\ex{To construct an angle of~$18°$; of~$36°$; of~$9°$.}

\ex{To construct an angle of$12°$; of~$24°$; of~$6°$.}

\exheader{To construct with a side of a given length:}

\ex{An equilateral triangle.}

\ex{A square.}

\ex{A regular hexagon.}

\ex{A regular octagon.}

\ex{A regular pentagon.}

\ex{A regular decagon.}

\ex{A regular dodecagon.}

\ex{A regular pentedecagon.}
\scanpage{253.png}%


\subsection{PROBLEMS OF COMPUTATION.}

\ex{Find the area of a circle whose radius is $12$ inches.}

\ex{Find the circumference and the area of a circle whose diameter
is $8$ feet.}

\ex{A regular pentagon is inscribed in a circle whose radius is $R$.
If the length of a side is $a$, find the apothem.}

\ex{A regular polygon is inscribed in a circle whose radius is $R$.
If the length of a side is $a$, show that the apothem is $\frac{1}{2} \sqrt{R^2 - a^2}$.}

\ex{Find the area of a regular decagon inscribed in a circle whose
radius is $16$ inches.}

\ex{Find the side of a regular dodecagon inscribed in a circle
whose radius is $20$ inches.}

\ex{Find the perimeter of a regular pentagon inscribed in a circle
whose radius is $25$ feet.}

\ex{The length of each side of a park in the shape of a regular
decagon is $100$ yards. Find the area of the park.}

\ex{Find the cost, at $\$2$ per yard, of building a wall around a
cemetery in the shape of a regular hexagon, that contains $16,627.84$ square
yards.}

\ex{The side of an inscribed regular polygon of $n$ sides is $16$ feet.
Find the side of an inscribed regular polygon of $2n$ sides.}

\ex{If the radius of a circle is $R$, and the side of an inscribed
regular polygon is $a$, show that the side of the similar circumscribed regular polygon is
$\dfrac{2aR}{\sqrt{4R^2-a^2}}$.}

\ex{What is the width of the circular ring between two concentric
circumferences whose lengths are $650$ feet and $425$ feet?}

\ex{Find the angle subtended at the centre by an arc $5$ feet $10$
inches long, if the radius of the circle is $9$ feet $4$ inches.}

\ex{The chord of a segment is $10$ feet, and the radius of the circle
is $16$ feet. Find the area of the segment.}

\ex{Find the area of a sector, if the angle at the centre is $20°$, and
the radius of the circle is $20$ inches.}
\scanpage{254.png}%

\ex{The chord of half an arc is $12$~feet, and the radius of the
circle is $18$~feet. Find the height of the segment subtended by the whole
arc.}

\ex{Find the side of a square which is equivalent to a circle whose
diameter is $35$~feet.}

\ex{The diameter of a circle is $15$~feet. Find the diameter of a
circle twice as large. Three times as large.}

\ex{Find the radii of the concentric circumferences that divide a
circle $11$~inches in diameter into five equivalent parts.}

\ex{The perimeter of a regular hexagon is $840$ feet, and that of a
regular octagon is the same. By how many square feet is the octagon
larger than the hexagon?}

\ex{The diameter of a bicycle wheel is $28$~inches. How many
revolutions does the wheel make in going $10$~miles?}

\ex{Find the diameter of a carriage wheel that makes $264$~revolutions
in going half a mile.}

\ex{The sides of three regular octagons are $6$~feet, $7$~feet, $8$~feet,
respectively. Find the side of a regular octagon equivalent to the sum of
the three given octagons.}

\ex{A circular pond $100$ yards in diameter is surrounded by a walk
$10$~feet wide. Find the area of the walk.}

\ex{The span (chord) of a bridge in the form of a circular arc is
$120$~feet, and the highest point of the arch is $15$~feet above the piers. Find
the radius of the arc.}

\ex{Three equal circles are described each tangent to the other
two. If the common radius is~$R$, find the area contained between the
circles.}

\ex{Given $p$, $P$, the perimeters of regular polygons of $n$~sides
inscribed in and circumscribed about a given circle. Find $p'$, $P'$, the
perimeters of regular polygons of $2n$~sides inscribed in and circumscribed
about the given circle.}

\ex{Given the radius $R$, and the apothem $r$ of an inscribed regular
polygon of $n$ sides. Find the radius $R'$ and the apothem $r'$ of an isoperimetrical
regular polygon of $2n$ sides.}
\scanpage{255.png}%


\subsection{MISCELLANEOUS EXERCISES.}

\subsection{THEOREMS.}

\ex{If two adjacent angles of a quadrilateral are right angles, the
bisectors of the other two angles are perpendicular.}

\ex{If two opposite angles of a quadrilateral are right angles, the
bisectors of the other two angles are parallel.}

\ex{The two lines that join the middle points of the opposite sides
of a quadrilateral bisect each other.}

\ex{The line that joins the feet of the perpendiculars dropped from
the extremities of the base of an isosceles triangle to the opposite sides is
parallel to the base.}

\ex{If $AD$ bisects the angle $A$ of a triangle $ABC$, and $BD$ bisects
the exterior angle $CBF$, then angle $ADB$ equals one half angle $ACB$.}

\ex{The sum of the acute angles at the vertices of a pentagram\label{pentagram}
(five-pointed star) is equal to two right angles.}

\begin{proofex}%
The altitudes $AD$, $BE$, $CF$ of the triangle $ABC$ bisect the
angles of the triangle $DEF$.

Circles with $AB$, $BC$, $AC$ as diameters will pass through $E$ and $D$, $E$
and $F$, $D$ and $F$, respectively.

\end{proofex}

\ex{The segments of any straight line intercepted between the
circumferences of two concentric circles are equal.}

\ex{If a circle is circumscribed about any triangle, the feet of the
perpendiculars dropped from any point in the circumference to the sides
of the triangle lie in one straight line.}

\ex{Two circles are tangent internally at $P$, and a chord $AB$ of
the larger circle touches the smaller circle at $C$. Prove that $PC$ bisects
the angle $APB$.}

\ex{The diagonals of a trapezoid divide each other into segments
which are proportional.}

\ex{If through a point $P$ in the circumference of a circle two
chords are drawn, the chords and the segments between $P$ and a chord
parallel to the tangent at $P$ are reciprocally proportional.}
\scanpage{256.png}%

\ex{The perpendiculars from two vertices of a triangle upon the
opposite sides divide each other into segments reciprocally proportional.}

\ex{The perpendicular from any point of a circumference upon a
chord is the mean proportional between the perpendiculars from the same
point upon the tangents drawn at the extremities of the chord.}

\ex{In an isosceles right triangle either leg is the mean proportional
between the hypotenuse and the perpendicular upon it from the
vertex of the right angle.}

\ex{If two circles intersect in the points $A$ and $B$, and through $A$
any secant $CAD$ is drawn limited by the circumferences at $C$ and $D$, the
straight lines $BC$, $BD$ are to each other as the diameters of the circles.}

\ex{The area of a triangle is equal to half the product of its perimeter
by the radius of the inscribed circle.}

\ex{The perimeter of a triangle is to one side as the perpendicular
from the opposite vertex is to the radius of the inscribed circle.}

\begin{proofex}%
If three straight lines $AA'$, $BB'$, $CC'$, drawn from the vertices
of a triangle $ABC$ to the opposite sides, pass through a common point $O$
within the triangle, then

\step{\( \dfrac{OA'}{AA'} + \dfrac{OB'}{BB'} + \dfrac{OC'}{CC'} = 1 \).}{}

\end{proofex}

\ex{$ABC$ is a triangle, $M$ the middle point of $AB$, $P$ any point
in $AB$ between $A$ and $M$. If $MD$ is drawn parallel to $PC$, meeting $BC$
at $D$, the triangle $BPD$ is equivalent to half the triangle $ABC$.}

\ex{Two diagonals of a regular pentagon, not drawn from a common
vertex, divide each other in extreme and mean ratio.}

\ex{If all the diagonals of a regular pentagon are drawn, another
regular pentagon is thereby formed.}

\ex{The area of an inscribed regular dodecagon is equal to three
times the square of the radius.}

\ex{The area of a square inscribed in a semicircle is equal to two
fifths the area of the square inscribed in the circle.}

\ex{The area of a circle is greater than the area of any polygon
of equal perimeter.}

\ex{The circumference of a circle is less than the perimeter of any
polygon of equal area.}
\scanpage{257.png}%


\subsection{PROBLEMS OF LOCI.}

\ex{Find the locus of the centre of the circle inscribed in a triangle
that has a given base and a given angle at the vertex.}

\ex{Find the locus of the intersection of the altitudes of a triangle
that has a given base and a given angle at the vertex.}

\ex{Find the locus of the extremity of a tangent to a given circle,
if the length of the tangent is equal to a given line.}

\ex{Find the locus of a point, tangents drawn from which to a
given circle form a given angle.}

\ex{Find the locus of the middle point of a line drawn from a
given point to a given straight line.}

\ex{Find the locus of the vertex of a triangle that has a given
base and a given altitude.}

\ex{Find the locus of a point the sum of whose distances from
two given parallel lines is equal to a given length.}

\ex{Find the locus of a point the difference of whose distances
from two given parallel lines is equal to a given length.}

\ex{Find the locus of a point the sum of whose distances from two
given intersecting lines is equal to a given length.}

\ex{Find the locus of a point the difference of whose distances
from two given intersecting lines is equal to a given length.}

\ex{Find the locus of a point whose distances from two given
points are in the given ratio $m:n$.}

\ex{Find the locus of a point whose distances from two given
parallel lines are in the given ratio $m:n$.}

\ex{Find the locus of a point whose distances from two given
intersecting lines are in the given ratio $m:n$.}

\ex{Find the locus of a point the sum of the squares of whose
distances from two given points is constant.}

\ex{Find the locus of a point the difference of the squares of whose
distances from two given points is constant.}

\ex{Find the locus of the vertex of a triangle that has a given base
and the other two sides in the given ratio $m:n$.
}
\scanpage{258.png}%


\subsection{PROBLEMS OF CONSTRUCTION.}

\ex{To divide a given trapezoid into two equivalent parts by a
line parallel to the bases.}

\ex{To divide a given trapezoid into two equivalent parts by a
line through a given point in one of the bases.}

\ex{To construct a regular pentagon, given one of the diagonals.}

\ex{To divide a given straight line into two segments such that
their product shall be the maximum.}

\ex{To find a point in a semicircumference such that the sum of
its distances from the extremities of the diameter shall be the maximum.}

\ex{To draw a common secant to two given circles exterior to
each other such that the intercepted chords shall have the given lengths
$a$, $b$.}

\ex{To draw through one of the points of intersection of two
intersecting circles a common secant which shall have a given length.}

\ex{To construct an isosceles triangle, given the altitude and one
of the equal base angles.}

\ex{To construct an equilateral triangle, given the altitude.}

\ex{To construct a right triangle, given the radius of the inscribed
circle and the difference of the acute angles.}

\ex{To construct an equilateral triangle so that its vertices shall
lie in three given parallel lines.}

\ex{To draw a line from a given point to a given straight line
which shall be to the perpendicular from the given point as $m : n$.}

\ex{To find a point within a given triangle such that the perpendiculars
from the point to the three sides shall be as the numbers $m$, $n$, $p$.}

\ex{To draw a straight line equidistant from three given points.}

\ex{To draw a tangent to a given circle such that the segment
intercepted between the point of contact and a given straight line shall
have a given length.}

\ex{To inscribe a straight line of a given length between two given
circumferences and parallel to a given straight line.
}
\scanpage{259.png}%

\ex{To draw through a given point a straight line so that its distances
from two other given points shall be in a given ratio.}

\ex{To construct a square equivalent to the sum of a given triangle
and a given parallelogram.}

\ex{To construct a rectangle having the difference of its base and
altitude equal to a given line, and its area equivalent to the sum of a given
triangle and a given pentagon.}

\ex{To construct a pentagon similar to a given pentagon and
equivalent to a given trapezoid.}

\ex{To find a point whose distances from three given straight
lines shall be as the numbers $m$, $n$, $p$.}

\ex{Given an angle and two points $P$ and $P'$ between the sides of
the angle. To find the shortest path from $P$ to $P'$ that shall touch both
sides of the angle.}

\ex{To construct a triangle, given its angles and its area.}

\ex{To transform a given triangle into a triangle similar to
another given triangle.}

\ex{Given three points $A$, $B$, $C$. To find a fourth point $P$ such
that the areas of the triangles $APB$, $APC$, $BPC$ shall be equal.}

\ex{To construct a triangle, given its base, the ratio of the other
sides, and the angle included by them.}

\ex{To divide a given circle into $n$ equivalent parts by concentric
circumferences.}

\ex{In a given equilateral triangle to inscribe three equal circles
tangent to each other, each circle tangent to two sides of the triangle.}

\ex{Given an angle and a point $P$ between the sides of the angle.
To draw through $P$ a straight line that shall form with the sides of the
angle a triangle with the perimeter equal to a given length $a$.}

\ex{In a given square to inscribe four equal circles, so that each
circle shall be tangent to two of the others and also tangent to two sides
of the square.}

\ex{In a given square to inscribe four equal circles, so that each
circle shall be tangent to two of the others and also tangent to one side
of the square.}
\scanpage{260.png}%


\chapter{TABLE OF FORMULAS.}
\markboth{\Headings{PLANE GEOMETRY.}}{\Headings{TABLE OF FORMULAS.}}%

\subsection{PLANE FIGURES.}

\subsection{NOTATION.}

\begin{tabular}{r@{~}c@{~}l}
$P$ &=& perimeter. \\
$h$ &=& altitude. \\
$b$ &=& lower base. \\
$b'$ &=& upper base. \\
$R$ &=& radius of circle. \\
$D$ &=& diameter of circle. \\
$C$ &=& circumference of circle. \\
$r$ &=& apothem of regular polygon. \\
$a$, $b$, $c$ &=& sides of triangle. \\
$s$ &=& \( \frac{1}{2}(a+b+c) \). \\
$p$ &=& perpendicular of triangle. \\
$m,n$ &=& segments of third side of triangle adjacent to \\
&& sides $b$ and $a$, respectively. \\
$S$ &=& area. \\
$\pi$ &=& 3.1416.
\end{tabular}


\newpage
\subsection{FORMULAS.}

\noindent\begin{supertabular}{lr@{~}c@{~}l@{\qquad}r}

\multicolumn{5}{l}{\hspace{-2ex}\textbf{Line Values.}} \\

\multicolumn{5}{r}{\tiny PAGE}\\
Right triangle, &
  $b^2$ &=& $c × m$; $a^2 = c × n$ & \pageref{160} \\
& $p^2$ &=& $m × n$ & \pageref{160} \\
& $b^2:a^2$ &=& $m:n$ & \pageref{161} \\
& $b^2:c^2$ &::& $m:c$ & \pageref{161} \\
& $a^2+b^2$ &=& $c^2$ & \pageref{162} \\
\scanpage{261.png}%

Any triangle, &
$a^2$ &=& $b^2+c^2 \pm 2c × m$ & \llap{\pageref{163},}\pageref{164} \\

\multicolumn{4}{l}{Altitude of triangle on side $a$,} \\
& $h$ &=& \( \displaystyle \frac{2}{a}
             \sqrt{s(s-a)(s-b)(s-c)} \) & \pageref{formtrialtitude} \\

\multicolumn{4}{l}{Median of triangle on side $a$,} \\
& $m$ &=& \( \frac{1}{2} \sqrt{2(b^2+c^2) - a^2} \) & \pageref{formtrimedian} \\

\multicolumn{4}{l}{Bisector of triangle on side $a$,} \\
& $t$ &=& \( \displaystyle \frac{2}{b+c}
             \sqrt{bcs(s-a)} \) & \pageref{formtribisector} \\

\multicolumn{4}{l}{Radius of circumscribed circle,} \\
& $R$ &=& \( \displaystyle \frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}} \) & \pageref{formradcircum} \\

Circumference of circle, & $C$ &=& $2\pi R$ & \pageref{formcircum} \\
\qquad\DittoMark\qquad\qquad\DittoMark & $C$ &=& $\pi D$ & \pageref{formcircum} \\

\multicolumn{5}{l}{\hspace{-2ex}\textbf{Areas.}} \\

Rectangle,     & $S$ &=& $b × h$ & \pageref{formarearect} \\
Square,        & $S$ &=& $b^2$ & \pageref{formarearect} \\
Parallelogram, & $S$ &=& $b × h$ & \pageref{formareapar} \\
Triangle,      & $S$ &=& $\frac{1}{2}b × h$ & \pageref{formareatri} \\
\qquad\DittoMark       & $S$ &=& $\sqrt{s(s-a)(s-b)(s-c)}$ & \pageref{formareatri2} \\
\qquad\DittoMark       & $S$ &=& \( \displaystyle \frac{abc}{4R} \) & \pageref{formareatri3} \\
Equilateral triangle, & $S$ &=&
   \( \displaystyle \frac{a^2}{4}\sqrt{3} \) & \pageref{formareaequitri} \\
Trapezoid,     & $S$ &=& $\frac{1}{2}h(b+b')$ & \pageref{formareatrap} \\
Regular polygon, & $S$ &=& $\frac{1}{2}r × P$ & \pageref{formareapoly} \\
Circle,        & $S$ &=& $\frac{1}{2}R × C$ & \pageref{formareacircle} \\
\qquad\DittoMark       & $S$ &=& $\pi R^2$ & \pageref{formareacircle} \\
Sector,        & $S$ &=& $\frac{1}{2}R × \arc$ & \pageref{formareasector} \\
\end{supertabular}
\scanpage{262.png}%


\twocolumn
\chapter{INDEX.}
\markboth{INDEX.}{INDEX.}

\small
\noindent\begin{supertabular}{lr}

\multicolumn{2}{r}{\tiny PAGE} \\
\tablehead{\multicolumn{2}{r}{\tiny PAGE} \\}%

Abbreviations & \pageref{abbr} \\
Alternation & \pageref{alternation} \\
Altitude of parallelogram & \pageref{altpar} \\
\Ditto of trapezoid & \pageref{alttrap} \\
\Ditto of triangle & \pageref{alttri} \\
Analysis & \pageref{analysis} \\
Angle & \pageref{angle} \\
\Ditto acute & \pageref{acute} \\
\multicolumn{2}{l}{\Ditto at centre of}\\
\quad\quad\quad regular polygon & \pageref{anglecentreregpoly} \\
\Ditto central & \pageref{central} \\
\Ditto exterior of triangle & \pageref{exteriortri} \\
\Ditto inscribed in circle & \pageref{inscribedcirc} \\
\Ditto inscribed in segment & \pageref{inscribedseg} \\
\Ditto oblique & \pageref{oblique} \\
\Ditto obtuse & \pageref{obtuse} \\
\Ditto reflex & \pageref{reflex} \\
\Ditto right & \pageref{right} \\
\Ditto salient & \pageref{salient} \\
\Ditto straight & \pageref{straight} \\
\Ditto vertical & \pageref{vertical angle} \\
Angles, adjacent & \pageref{adjacent1},\pageref{adjacent2} \\
\Ditto alternate-exterior & \pageref{altext} \\
\Ditto alternate-interior & \pageref{altint} \\
\Ditto complementary & \pageref{complementary} \\
\Ditto conjugate & \pageref{conjugate angles} \\
\Ditto exterior & \pageref{exterior} \\
\Ditto exterior-interior & \pageref{extint} \\
\Ditto interior & \pageref{interior} \\
\Ditto supplementary & \pageref{supplementary} \\
\Ditto supplementary-adjacent & \pageref{suppladj} \\
\Ditto vertical & \pageref{vertical angles} \\
Antecedents & \pageref{antecedents} \\
Apothem & \pageref{apothem} \\
Arc & \pageref{arc} \\
Area & \pageref{area} \\
Axiom & \pageref{axiom} \\
\Ditto of parallel lines & \pageref{axiomparallel} \\
Axioms of straight lines & \pageref{axiomstraight} \\
\Ditto general & \pageref{generalaxioms} \\
Axis of symmetry & \pageref{axissym} \\
\\
\textbf{B}ase of isosceles triangle & \pageref{baseiso} \\
\Ditto of parallelogram & \pageref{basepar} \\
\Ditto of triangle & \pageref{basetri} \\
Bases of trapezoid & \pageref{basetrap} \\
Bisector & \pageref{bisector} \\
\\
\textbf{C}entre of circle & \pageref{centrecirc} \\
\Ditto of regular polygon & \pageref{centrepoly} \\
\Ditto of symmetry & \pageref{centresym} \\
Chord & \pageref{chord} \\
Circle & \pageref{circle} \\
\Ditto circumscribed & \pageref{circcircumscribed} \\
\Ditto inscribed & \pageref{circinscribed} \\
Circles, concentric & \pageref{concentric} \\
\Ditto escribed & \pageref{escribed} \\
Circum-centre of triangle & \pageref{circum-centre} \\
Circumference & \pageref{circumference} \\
Commensurable & \pageref{commensurable} \\
Complement & \pageref{complement} \\
Composition & \pageref{composition} \\
Conclusion & \pageref{conclusion} \\
Concurrent lines & \pageref{concurrent} \\
Congruent figures & \pageref{congruent} \\
Consequents & \pageref{consequents} \\
Constant & \pageref{constant} \\
Construction & \pageref{construction} \\
Continued proportion & \pageref{continuedprop} \\
Continuity, Principle of & \pageref{princcont} \\
Contradictory of a theorem & \pageref{contradictory} \\
Converse of a theorem & \pageref{converse1},\pageref{converse2} \\
Convex curve & \pageref{convexcurve} \\
Curved surface & \pageref{curvedsurf} \\
\\
\textbf{D}ecagon & \pageref{decagon} \\
Diagonal & \pageref{diagonal1},\pageref{diagonal2} \\
Diameter & \pageref{diameter} \\
Dimensions & \pageref{dimensions} \\
Distance & \pageref{distance1},\pageref{distance2} \\
Division & \pageref{division} \\
Dodecagon & \pageref{dodecagon} \\
Duality, Principle of & \pageref{princduality} \\
\\
\textbf{E}qual figures & \pageref{equal} \\
Equimultiples & \pageref{equimultiples} \\
Equivalent figures & \pageref{equivalent1},\pageref{equivalent2} \\
Ex-centres of triangle & \pageref{ex-centres} \\
Extreme and mean ratio & \pageref{extrememean} \\
Extremes & \pageref{extremes} \\
\\
\textbf{F}igure, curvilinear & \pageref{curvilinear} \\
\Ditto geometrical & \pageref{geometrical figure} \\
\Ditto plane & \pageref{plane figure} \\
\Ditto rectilinear & \pageref{rectilinear} \\
Foot of perpendicular & \pageref{foot} \\
Fourth proportional & \pageref{fourth} \\
\\
\textbf{G}eometrical solid & \pageref{geometrical1},\pageref{geometrical2} \\
Geometry & \pageref{Geometry} \\
Geometry, Plane & \pageref{Plane Geometry} \\
\Ditto Solid & \pageref{Solid Geometry} \\
\\
\textbf{H}armonic division & \pageref{divided harmonically} \\
Heptagon & \pageref{heptagon} \\
Hexagon & \pageref{hexagon} \\
Homologous angles & \pageref{homologous angles},\pageref{homangles} \\
\Ditto lines & \pageref{Homologous lines} \\
\Ditto sides & \pageref{homologous sides},\pageref{homsides} \\
Hypotenuse & \pageref{hypotenuse} \\
Hypothesis & \pageref{hypothesis} \\
\\
\textbf{I}n-centre of triangle & \pageref{in-centre} \\
Incommensurable ratio & \pageref{incommensurable ratio} \\
Intersection & \pageref{intersection} \\
Inversion & \pageref{inversion} \\
Isoperimetric figures & \pageref{Isoperimetric} \\
\\
\textbf{L}egs of right triangle & \pageref{legs} \\
\Ditto of trapezoid & \pageref{legstrap} \\
Limit & \pageref{limit} \\
Line & \pageref{line},\pageref{line2},\pageref{line3} \\
\Ditto curved & \pageref{curved line} \\
\Ditto of centres & \pageref{line of centres} \\
\Ditto straight & \pageref{straight line} \\
Lines, oblique & \pageref{oblique lines} \\
\Ditto parallel & \pageref{parallel lines} \\
\Ditto perpendicular & \pageref{perpendicular} \\
\\
\textbf{M}ajor arc & \pageref{major} \\
Maximum & \pageref{maximum} \\
Mean proportional & \pageref{mean proportional} \\
Means & \pageref{means} \\
Median of trapezoid & \pageref{mediantrap} \\
Minimum & \pageref{minimum} \\
Minor arc & \pageref{minor} \\
\\
\textbf{N}egative quantities & \pageref{negative} \\
Numerical measure & \pageref{numerical measure} \\
\\
\textbf{O}ctagon & \pageref{octagon} \\
Opposite of a theorem & \pageref{opposite} \\
Origin & \pageref{origin} \\
\\
\textbf{P}arallel lines & \pageref{parallel lines} \\
Parallelogram & \pageref{parallelogram} \\
Pentagon & \pageref{pentagon} \\
Pentagram & \pageref{pentagram} \\
Perigon & \pageref{perigon} \\
Perimeter & \pageref{perimeter},\pageref{perimeter2} \\
Perpendicular bisector & \pageref{perpbisector} \\
Perpendicular lines & \pageref{perpendicular} \\
Pi ($\pi$) & \pageref{pi} \\
Plane & \pageref{planes},\pageref{plane} \\
Point & \pageref{point},\pageref{point2} \\
\Ditto of contact & \pageref{point of contact} \\
\Ditto of tangency & \pageref{point of tangency} \\
Polygon & \pageref{polygon} \\
\Ditto angles of & \pageref{polyangles} \\
\Ditto circumscribed & \pageref{polycircumscribed} \\
\Ditto concave & \pageref{concave polygon} \\
\Ditto convex & \pageref{convex polygon} \\
\Ditto equiangular & \pageref{equiangular polygon} \\
\Ditto equilateral & \pageref{equilateral polygon} \\
\Ditto inscribed & \pageref{polyinscribed} \\
\Ditto regular & \pageref{regular polygon} \\
Polygons mut.\ equiangular & \pageref{mutually equiangular} \\
\Ditto mutually equilateral & \pageref{mutually equilateral} \\
Positive quantities & \pageref{positive} \\
Postulate & \pageref{postulate} \\
Projection & \pageref{projection} \\
Proof & \pageref{proof} \\
Proportion & \pageref{proportion} \\
Proposition & \pageref{proposition} \\
\\
\textbf{Q}uadrant & \pageref{quadrant} \\
Quadrilateral & \pageref{quadrilateral},\pageref{quadrilateral2} \\
Radius of regular polygon & \pageref{polyradius} \\
Ratio & \pageref{ratio} \\
Ratio of similitude & \pageref{ratio of similitude} \\
Reciprocity, Principle of & \pageref{princreciprocity} \\
Rectangle & \pageref{rectangle} \\
Rhomboid & \pageref{rhomboid} \\
Rhombus & \pageref{rhombus} \\
\\
\textbf{S}cholium & \pageref{scholium} \\
Secant & \pageref{secant},\pageref{secant2} \\
Sector & \pageref{sector} \\
Segment of circle & \pageref{segment} \\
\Ditto of line & \pageref{lineseg} \\
Semicircle & \pageref{semicircle} \\
Semicircumference & \pageref{semicircumference} \\
Sides of an angle & \pageref{anglesides} \\
\Ditto of polygon & \pageref{polysides} \\
\Ditto of triangle & \pageref{trisides} \\
Similar arcs & \pageref{similar arcs} \\
\Ditto figures & \pageref{similar} \\
\Ditto polygons & \pageref{Similar polygons} \\
\Ditto sectors & \pageref{similar sectors} \\
\Ditto segments & \pageref{similar segments} \\
\Ditto triangles & \pageref{similar triangles} \\% not sure about this ref
Square & \pageref{square} \\
Superposition & \pageref{superposition} \\
Supplement & \pageref{supplement} \\
Surface & \pageref{surface},\pageref{surface2},\pageref{surface3} \\
Symbols & \pageref{symbols} \\
Symmetry & \pageref{symmetry} \\
\\
\textbf{T}angent & \pageref{tangent},\pageref{tangent2} \\
\Ditto common external & \pageref{common external tangent} \\
\Ditto common internal & \pageref{common internal tangent} \\
Terms of a proportion & \pageref{terms} \\
Theorem & \pageref{theorem} \\
Third proportional & \pageref{third} \\
Transversal & \pageref{transversal} \\
Trapezium & \pageref{trapezium} \\
Trapezoid & \pageref{trapezoid} \\
\Ditto isosceles & \pageref{isosceles trapezoid} \\
Triangle & \pageref{triangle},\pageref{triangle2} \\
\Ditto equiangular & \pageref{equiangular triangle} \\
\Ditto equilateral & \pageref{equilateral triangle} \\
\Ditto isosceles & \pageref{isosceles triangle} \\
\Ditto obtuse & \pageref{obtuse triangle} \\
\Ditto right & \pageref{right triangle} \\
\Ditto scalene & \pageref{scalene triangle} \\
\Ditto altitudes of & \pageref{alttri} \\
\Ditto angles of & \pageref{anglestri} \\
\Ditto bisectors of & \pageref{tribisectors} \\
\Ditto medians of & \pageref{trimedians} \\
\Ditto vertices of & \pageref{trivertices} \\
\\
\textbf{V}ariable & \pageref{variable} \\
Vertex of angle & \pageref{vertex} \\
\Ditto of triangle & \pageref{trivertex} \\
Vertices of polygon & \pageref{polyvertices} \\

\end{supertabular}

\onecolumn

%%%%%%%%%%%%%%%%%%%%%%%%% GUTENBERG LICENSE %%%%%%%%%%%%%%%%%%%%%%%%%%

\cleardoublepage

\backmatter
\phantomsection
\pdfbookmark[-1]{BACK MATTER.}{BACK MATTER}
\phantomsection
\pdfbookmark[0]{PG LICENSE.}{LICENSE}
\fancyhead[C]{\Headings{LICENSE.}}

\begin{PGtext}
End of Project Gutenberg's Plane Geometry, by George Albert Wentworth

*** END OF THIS PROJECT GUTENBERG EBOOK PLANE GEOMETRY ***

***** This file should be named 33063-pdf.pdf or 33063-pdf.zip *****
This and all associated files of various formats will be found in:
        http://www.gutenberg.org/3/3/0/6/33063/

Produced by Jeremy Weatherford, Stan Goodman, Kevin Handy
and the Online Distributed Proofreading Team at
http://www.pgdp.net


Updated editions will replace the previous one--the old editions
will be renamed.

Creating the works from public domain print editions means that no
one owns a United States copyright in these works, so the Foundation
(and you!) can copy and distribute it in the United States without
permission and without paying copyright royalties.  Special rules,
set forth in the General Terms of Use part of this license, apply to
copying and distributing Project Gutenberg-tm electronic works to
protect the PROJECT GUTENBERG-tm concept and trademark.  Project
Gutenberg is a registered trademark, and may not be used if you
charge for the eBooks, unless you receive specific permission.  If you
do not charge anything for copies of this eBook, complying with the
rules is very easy.  You may use this eBook for nearly any purpose
such as creation of derivative works, reports, performances and
research.  They may be modified and printed and given away--you may do
practically ANYTHING with public domain eBooks.  Redistribution is
subject to the trademark license, especially commercial
redistribution.



*** START: FULL LICENSE ***

THE FULL PROJECT GUTENBERG LICENSE
PLEASE READ THIS BEFORE YOU DISTRIBUTE OR USE THIS WORK

To protect the Project Gutenberg-tm mission of promoting the free
distribution of electronic works, by using or distributing this work
(or any other work associated in any way with the phrase "Project
Gutenberg"), you agree to comply with all the terms of the Full Project
Gutenberg-tm License (available with this file or online at
http://gutenberg.org/license).


Section 1.  General Terms of Use and Redistributing Project Gutenberg-tm
electronic works

1.A.  By reading or using any part of this Project Gutenberg-tm
electronic work, you indicate that you have read, understand, agree to
and accept all the terms of this license and intellectual property
(trademark/copyright) agreement.  If you do not agree to abide by all
the terms of this agreement, you must cease using and return or destroy
all copies of Project Gutenberg-tm electronic works in your possession.
If you paid a fee for obtaining a copy of or access to a Project
Gutenberg-tm electronic work and you do not agree to be bound by the
terms of this agreement, you may obtain a refund from the person or
entity to whom you paid the fee as set forth in paragraph 1.E.8.

1.B.  "Project Gutenberg" is a registered trademark.  It may only be
used on or associated in any way with an electronic work by people who
agree to be bound by the terms of this agreement.  There are a few
things that you can do with most Project Gutenberg-tm electronic works
even without complying with the full terms of this agreement.  See
paragraph 1.C below.  There are a lot of things you can do with Project
Gutenberg-tm electronic works if you follow the terms of this agreement
and help preserve free future access to Project Gutenberg-tm electronic
works.  See paragraph 1.E below.

1.C.  The Project Gutenberg Literary Archive Foundation ("the Foundation"
or PGLAF), owns a compilation copyright in the collection of Project
Gutenberg-tm electronic works.  Nearly all the individual works in the
collection are in the public domain in the United States.  If an
individual work is in the public domain in the United States and you are
located in the United States, we do not claim a right to prevent you from
copying, distributing, performing, displaying or creating derivative
works based on the work as long as all references to Project Gutenberg
are removed.  Of course, we hope that you will support the Project
Gutenberg-tm mission of promoting free access to electronic works by
freely sharing Project Gutenberg-tm works in compliance with the terms of
this agreement for keeping the Project Gutenberg-tm name associated with
the work.  You can easily comply with the terms of this agreement by
keeping this work in the same format with its attached full Project
Gutenberg-tm License when you share it without charge with others.

1.D.  The copyright laws of the place where you are located also govern
what you can do with this work.  Copyright laws in most countries are in
a constant state of change.  If you are outside the United States, check
the laws of your country in addition to the terms of this agreement
before downloading, copying, displaying, performing, distributing or
creating derivative works based on this work or any other Project
Gutenberg-tm work.  The Foundation makes no representations concerning
the copyright status of any work in any country outside the United
States.

1.E.  Unless you have removed all references to Project Gutenberg:

1.E.1.  The following sentence, with active links to, or other immediate
access to, the full Project Gutenberg-tm License must appear prominently
whenever any copy of a Project Gutenberg-tm work (any work on which the
phrase "Project Gutenberg" appears, or with which the phrase "Project
Gutenberg" is associated) is accessed, displayed, performed, viewed,
copied or distributed:

This eBook is for the use of anyone anywhere at no cost and with
almost no restrictions whatsoever.  You may copy it, give it away or
re-use it under the terms of the Project Gutenberg License included
with this eBook or online at www.gutenberg.org

1.E.2.  If an individual Project Gutenberg-tm electronic work is derived
from the public domain (does not contain a notice indicating that it is
posted with permission of the copyright holder), the work can be copied
and distributed to anyone in the United States without paying any fees
or charges.  If you are redistributing or providing access to a work
with the phrase "Project Gutenberg" associated with or appearing on the
work, you must comply either with the requirements of paragraphs 1.E.1
through 1.E.7 or obtain permission for the use of the work and the
Project Gutenberg-tm trademark as set forth in paragraphs 1.E.8 or
1.E.9.

1.E.3.  If an individual Project Gutenberg-tm electronic work is posted
with the permission of the copyright holder, your use and distribution
must comply with both paragraphs 1.E.1 through 1.E.7 and any additional
terms imposed by the copyright holder.  Additional terms will be linked
to the Project Gutenberg-tm License for all works posted with the
permission of the copyright holder found at the beginning of this work.

1.E.4.  Do not unlink or detach or remove the full Project Gutenberg-tm
License terms from this work, or any files containing a part of this
work or any other work associated with Project Gutenberg-tm.

1.E.5.  Do not copy, display, perform, distribute or redistribute this
electronic work, or any part of this electronic work, without
prominently displaying the sentence set forth in paragraph 1.E.1 with
active links or immediate access to the full terms of the Project
Gutenberg-tm License.

1.E.6.  You may convert to and distribute this work in any binary,
compressed, marked up, nonproprietary or proprietary form, including any
word processing or hypertext form.  However, if you provide access to or
distribute copies of a Project Gutenberg-tm work in a format other than
"Plain Vanilla ASCII" or other format used in the official version
posted on the official Project Gutenberg-tm web site (www.gutenberg.org),
you must, at no additional cost, fee or expense to the user, provide a
copy, a means of exporting a copy, or a means of obtaining a copy upon
request, of the work in its original "Plain Vanilla ASCII" or other
form.  Any alternate format must include the full Project Gutenberg-tm
License as specified in paragraph 1.E.1.

1.E.7.  Do not charge a fee for access to, viewing, displaying,
performing, copying or distributing any Project Gutenberg-tm works
unless you comply with paragraph 1.E.8 or 1.E.9.

1.E.8.  You may charge a reasonable fee for copies of or providing
access to or distributing Project Gutenberg-tm electronic works provided
that

- You pay a royalty fee of 20% of the gross profits you derive from
     the use of Project Gutenberg-tm works calculated using the method
     you already use to calculate your applicable taxes.  The fee is
     owed to the owner of the Project Gutenberg-tm trademark, but he
     has agreed to donate royalties under this paragraph to the
     Project Gutenberg Literary Archive Foundation.  Royalty payments
     must be paid within 60 days following each date on which you
     prepare (or are legally required to prepare) your periodic tax
     returns.  Royalty payments should be clearly marked as such and
     sent to the Project Gutenberg Literary Archive Foundation at the
     address specified in Section 4, "Information about donations to
     the Project Gutenberg Literary Archive Foundation."

- You provide a full refund of any money paid by a user who notifies
     you in writing (or by e-mail) within 30 days of receipt that s/he
     does not agree to the terms of the full Project Gutenberg-tm
     License.  You must require such a user to return or
     destroy all copies of the works possessed in a physical medium
     and discontinue all use of and all access to other copies of
     Project Gutenberg-tm works.

- You provide, in accordance with paragraph 1.F.3, a full refund of any
     money paid for a work or a replacement copy, if a defect in the
     electronic work is discovered and reported to you within 90 days
     of receipt of the work.

- You comply with all other terms of this agreement for free
     distribution of Project Gutenberg-tm works.

1.E.9.  If you wish to charge a fee or distribute a Project Gutenberg-tm
electronic work or group of works on different terms than are set
forth in this agreement, you must obtain permission in writing from
both the Project Gutenberg Literary Archive Foundation and Michael
Hart, the owner of the Project Gutenberg-tm trademark.  Contact the
Foundation as set forth in Section 3 below.

1.F.

1.F.1.  Project Gutenberg volunteers and employees expend considerable
effort to identify, do copyright research on, transcribe and proofread
public domain works in creating the Project Gutenberg-tm
collection.  Despite these efforts, Project Gutenberg-tm electronic
works, and the medium on which they may be stored, may contain
"Defects," such as, but not limited to, incomplete, inaccurate or
corrupt data, transcription errors, a copyright or other intellectual
property infringement, a defective or damaged disk or other medium, a
computer virus, or computer codes that damage or cannot be read by
your equipment.

1.F.2.  LIMITED WARRANTY, DISCLAIMER OF DAMAGES - Except for the "Right
of Replacement or Refund" described in paragraph 1.F.3, the Project
Gutenberg Literary Archive Foundation, the owner of the Project
Gutenberg-tm trademark, and any other party distributing a Project
Gutenberg-tm electronic work under this agreement, disclaim all
liability to you for damages, costs and expenses, including legal
fees.  YOU AGREE THAT YOU HAVE NO REMEDIES FOR NEGLIGENCE, STRICT
LIABILITY, BREACH OF WARRANTY OR BREACH OF CONTRACT EXCEPT THOSE
PROVIDED IN PARAGRAPH F3.  YOU AGREE THAT THE FOUNDATION, THE
TRADEMARK OWNER, AND ANY DISTRIBUTOR UNDER THIS AGREEMENT WILL NOT BE
LIABLE TO YOU FOR ACTUAL, DIRECT, INDIRECT, CONSEQUENTIAL, PUNITIVE OR
INCIDENTAL DAMAGES EVEN IF YOU GIVE NOTICE OF THE POSSIBILITY OF SUCH
DAMAGE.

1.F.3.  LIMITED RIGHT OF REPLACEMENT OR REFUND - If you discover a
defect in this electronic work within 90 days of receiving it, you can
receive a refund of the money (if any) you paid for it by sending a
written explanation to the person you received the work from.  If you
received the work on a physical medium, you must return the medium with
your written explanation.  The person or entity that provided you with
the defective work may elect to provide a replacement copy in lieu of a
refund.  If you received the work electronically, the person or entity
providing it to you may choose to give you a second opportunity to
receive the work electronically in lieu of a refund.  If the second copy
is also defective, you may demand a refund in writing without further
opportunities to fix the problem.

1.F.4.  Except for the limited right of replacement or refund set forth
in paragraph 1.F.3, this work is provided to you 'AS-IS' WITH NO OTHER
WARRANTIES OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO
WARRANTIES OF MERCHANTIBILITY OR FITNESS FOR ANY PURPOSE.

1.F.5.  Some states do not allow disclaimers of certain implied
warranties or the exclusion or limitation of certain types of damages.
If any disclaimer or limitation set forth in this agreement violates the
law of the state applicable to this agreement, the agreement shall be
interpreted to make the maximum disclaimer or limitation permitted by
the applicable state law.  The invalidity or unenforceability of any
provision of this agreement shall not void the remaining provisions.

1.F.6.  INDEMNITY - You agree to indemnify and hold the Foundation, the
trademark owner, any agent or employee of the Foundation, anyone
providing copies of Project Gutenberg-tm electronic works in accordance
with this agreement, and any volunteers associated with the production,
promotion and distribution of Project Gutenberg-tm electronic works,
harmless from all liability, costs and expenses, including legal fees,
that arise directly or indirectly from any of the following which you do
or cause to occur: (a) distribution of this or any Project Gutenberg-tm
work, (b) alteration, modification, or additions or deletions to any
Project Gutenberg-tm work, and (c) any Defect you cause.


Section  2.  Information about the Mission of Project Gutenberg-tm

Project Gutenberg-tm is synonymous with the free distribution of
electronic works in formats readable by the widest variety of computers
including obsolete, old, middle-aged and new computers.  It exists
because of the efforts of hundreds of volunteers and donations from
people in all walks of life.

Volunteers and financial support to provide volunteers with the
assistance they need, are critical to reaching Project Gutenberg-tm's
goals and ensuring that the Project Gutenberg-tm collection will
remain freely available for generations to come.  In 2001, the Project
Gutenberg Literary Archive Foundation was created to provide a secure
and permanent future for Project Gutenberg-tm and future generations.
To learn more about the Project Gutenberg Literary Archive Foundation
and how your efforts and donations can help, see Sections 3 and 4
and the Foundation web page at http://www.pglaf.org.


Section 3.  Information about the Project Gutenberg Literary Archive
Foundation

The Project Gutenberg Literary Archive Foundation is a non profit
501(c)(3) educational corporation organized under the laws of the
state of Mississippi and granted tax exempt status by the Internal
Revenue Service.  The Foundation's EIN or federal tax identification
number is 64-6221541.  Its 501(c)(3) letter is posted at
http://pglaf.org/fundraising.  Contributions to the Project Gutenberg
Literary Archive Foundation are tax deductible to the full extent
permitted by U.S. federal laws and your state's laws.

The Foundation's principal office is located at 4557 Melan Dr. S.
Fairbanks, AK, 99712., but its volunteers and employees are scattered
throughout numerous locations.  Its business office is located at
809 North 1500 West, Salt Lake City, UT 84116, (801) 596-1887, email
business@pglaf.org.  Email contact links and up to date contact
information can be found at the Foundation's web site and official
page at http://pglaf.org

For additional contact information:
     Dr. Gregory B. Newby
     Chief Executive and Director
     gbnewby@pglaf.org


Section 4.  Information about Donations to the Project Gutenberg
Literary Archive Foundation

Project Gutenberg-tm depends upon and cannot survive without wide
spread public support and donations to carry out its mission of
increasing the number of public domain and licensed works that can be
freely distributed in machine readable form accessible by the widest
array of equipment including outdated equipment.  Many small donations
($1 to $5,000) are particularly important to maintaining tax exempt
status with the IRS.

The Foundation is committed to complying with the laws regulating
charities and charitable donations in all 50 states of the United
States.  Compliance requirements are not uniform and it takes a
considerable effort, much paperwork and many fees to meet and keep up
with these requirements.  We do not solicit donations in locations
where we have not received written confirmation of compliance.  To
SEND DONATIONS or determine the status of compliance for any
particular state visit http://pglaf.org

While we cannot and do not solicit contributions from states where we
have not met the solicitation requirements, we know of no prohibition
against accepting unsolicited donations from donors in such states who
approach us with offers to donate.

International donations are gratefully accepted, but we cannot make
any statements concerning tax treatment of donations received from
outside the United States.  U.S. laws alone swamp our small staff.

Please check the Project Gutenberg Web pages for current donation
methods and addresses.  Donations are accepted in a number of other
ways including checks, online payments and credit card donations.
To donate, please visit: http://pglaf.org/donate


Section 5.  General Information About Project Gutenberg-tm electronic
works.

Professor Michael S. Hart is the originator of the Project Gutenberg-tm
concept of a library of electronic works that could be freely shared
with anyone.  For thirty years, he produced and distributed Project
Gutenberg-tm eBooks with only a loose network of volunteer support.


Project Gutenberg-tm eBooks are often created from several printed
editions, all of which are confirmed as Public Domain in the U.S.
unless a copyright notice is included.  Thus, we do not necessarily
keep eBooks in compliance with any particular paper edition.


Most people start at our Web site which has the main PG search facility:

     http://www.gutenberg.org

This Web site includes information about Project Gutenberg-tm,
including how to make donations to the Project Gutenberg Literary
Archive Foundation, how to help produce our new eBooks, and how to
subscribe to our email newsletter to hear about new eBooks.
\end{PGtext}

% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
%                                                                         %
% End of Project Gutenberg's Plane Geometry, by George Albert Wentworth   %
%                                                                         %
% *** END OF THIS PROJECT GUTENBERG EBOOK PLANE GEOMETRY ***              %
%                                                                         %
% ***** This file should be named 33063-t.tex or 33063-t.zip *****        %
% This and all associated files of various formats will be found in:      %
%         http://www.gutenberg.org/3/3/0/6/33063/                         %
%                                                                         %
% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %

\end{document}
###
@ControlwordReplace = (
  );

@MathEnvironments = (
  ['\\begin{supertabular}','\\end{supertabular}','']
  );

@ControlwordArguments = (
  ['\\part', 1, 1, '', ''],
  ['\\chapter', 0, 0, '', '', 1, 1, '', ''],
  ['\\section', 0, 0, '', '', 1, 1, '', ''],
  ['\\subsection', 1, 1, '', ''],
  ['\\scanpage', 1, 0, '', ''],
  ['\\figc', 1, 0, '<FIGURE>', '', 1, 0, '', ''],
  ['\\figcc', 1, 0, '<FIGURE>', '', 1, 0, '', ''],
  ['\\figccc', 1, 0, '<FIGURE>', '', 1, 0, '', '', 1, 0, '', ''],
  ['\\figcccc', 1, 0, '<FIGURE>', '', 1, 0, '', '', 1, 0, '', '', 1, 0, '', ''],
  ['\\phantom', 1, 0, '', ''],
  ['\\ex', 1, 1, 'Ex. ', ''],
  ['\\exheader', 1, 1, '', ''],
  ['\\defn', 1, 1, 'Def. ', ''],
  ['\\ax', 1, 1, 'Axiom. ', ''],
  ['\\thm', 1, 1, 'Theorem. ', ''],
  ['\\cor', 0, 0, '', '', 1, 1, 'Cor. ', ''],
  ['\\note', 0, 0, '', '', 1, 1, 'Note. ', ''],
  ['\\pp', 1, 1, '', ''],
  ['\\step', 0, 1, '', ' ', 1, 1, '', ' ', 1, 1, '', ''],
  ['\\eq', 0, 1, '', ' ', 1, 1, '', ' ', 1, 1, '', ' ', 1, 1, '', ''],
  ['\\pnote', 1, 1, '', ''],
  ['\\tablehead', 1, 0, '', ''],
  ['\\multicolumn', 1, 0, '', '', 1, 0, '', '', 1, 1, '', '']
  );
###
This is pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6) (format=pdflatex 2010.5.6)  3 JUL 2010 09:28
entering extended mode
 %&-line parsing enabled.
**33063-t.tex
(./33063-t.tex
LaTeX2e <2005/12/01>
Babel <v3.8h> and hyphenation patterns for english, usenglishmax, dumylang, noh
yphenation, arabic, farsi, croatian, ukrainian, russian, bulgarian, czech, slov
ak, danish, dutch, finnish, basque, french, german, ngerman, ibycus, greek, mon
ogreek, ancientgreek, hungarian, italian, latin, mongolian, norsk, icelandic, i
nterlingua, turkish, coptic, romanian, welsh, serbian, slovenian, estonian, esp
eranto, uppersorbian, indonesian, polish, portuguese, spanish, catalan, galicia
n, swedish, ukenglish, pinyin, loaded.
(/usr/share/texmf-texlive/tex/latex/base/book.cls
Document Class: book 2005/09/16 v1.4f Standard LaTeX document class
(/usr/share/texmf-texlive/tex/latex/base/bk12.clo
File: bk12.clo 2005/09/16 v1.4f Standard LaTeX file (size option)
)
\c@part=\count79
\c@chapter=\count80
\c@section=\count81
\c@subsection=\count82
\c@subsubsection=\count83
\c@paragraph=\count84
\c@subparagraph=\count85
\c@figure=\count86
\c@table=\count87
\abovecaptionskip=\skip41
\belowcaptionskip=\skip42
\bibindent=\dimen102
) (/usr/share/texmf-texlive/tex/latex/base/inputenc.sty
Package: inputenc 2006/05/05 v1.1b Input encoding file
\inpenc@prehook=\toks14
\inpenc@posthook=\toks15
(/usr/share/texmf-texlive/tex/latex/base/latin1.def
File: latin1.def 2006/05/05 v1.1b Input encoding file
)) (/usr/share/texmf-texlive/tex/latex/base/ifthen.sty
Package: ifthen 2001/05/26 v1.1c Standard LaTeX ifthen package (DPC)
) (/usr/share/texmf-texlive/tex/latex/amsmath/amsmath.sty
Package: amsmath 2000/07/18 v2.13 AMS math features
\@mathmargin=\skip43
For additional information on amsmath, use the `?' option.
(/usr/share/texmf-texlive/tex/latex/amsmath/amstext.sty
Package: amstext 2000/06/29 v2.01
(/usr/share/texmf-texlive/tex/latex/amsmath/amsgen.sty
File: amsgen.sty 1999/11/30 v2.0
\@emptytoks=\toks16
\ex@=\dimen103
)) (/usr/share/texmf-texlive/tex/latex/amsmath/amsbsy.sty
Package: amsbsy 1999/11/29 v1.2d
\pmbraise@=\dimen104
) (/usr/share/texmf-texlive/tex/latex/amsmath/amsopn.sty
Package: amsopn 1999/12/14 v2.01 operator names
)
\inf@bad=\count88
LaTeX Info: Redefining \frac on input line 211.
\uproot@=\count89
\leftroot@=\count90
LaTeX Info: Redefining \overline on input line 307.
\classnum@=\count91
\DOTSCASE@=\count92
LaTeX Info: Redefining \ldots on input line 379.
LaTeX Info: Redefining \dots on input line 382.
LaTeX Info: Redefining \cdots on input line 467.
\Mathstrutbox@=\box26
\strutbox@=\box27
\big@size=\dimen105
LaTeX Font Info:    Redeclaring font encoding OML on input line 567.
LaTeX Font Info:    Redeclaring font encoding OMS on input line 568.
\macc@depth=\count93
\c@MaxMatrixCols=\count94
\dotsspace@=\muskip10
\c@parentequation=\count95
\dspbrk@lvl=\count96
\tag@help=\toks17
\row@=\count97
\column@=\count98
\maxfields@=\count99
\andhelp@=\toks18
\eqnshift@=\dimen106
\alignsep@=\dimen107
\tagshift@=\dimen108
\tagwidth@=\dimen109
\totwidth@=\dimen110
\lineht@=\dimen111
\@envbody=\toks19
\multlinegap=\skip44
\multlinetaggap=\skip45
\mathdisplay@stack=\toks20
LaTeX Info: Redefining \[ on input line 2666.
LaTeX Info: Redefining \] on input line 2667.
) (/usr/share/texmf-texlive/tex/latex/amsfonts/amssymb.sty
Package: amssymb 2002/01/22 v2.2d
(/usr/share/texmf-texlive/tex/latex/amsfonts/amsfonts.sty
Package: amsfonts 2001/10/25 v2.2f
\symAMSa=\mathgroup4
\symAMSb=\mathgroup5
LaTeX Font Info:    Overwriting math alphabet `\mathfrak' in version `bold'
(Font)                  U/euf/m/n --> U/euf/b/n on input line 132.
)) (/usr/share/texmf-texlive/tex/latex/base/alltt.sty
Package: alltt 1997/06/16 v2.0g defines alltt environment
) (/usr/share/texmf-texlive/tex/latex/footmisc/footmisc.sty
Package: footmisc 2005/03/17 v5.3d a miscellany of footnote facilities
\FN@temptoken=\toks21
\footnotemargin=\dimen112
\c@pp@next@reset=\count100
Package footmisc Info: Declaring symbol style bringhurst on input line 817.
Package footmisc Info: Declaring symbol style chicago on input line 818.
Package footmisc Info: Declaring symbol style wiley on input line 819.
Package footmisc Info: Declaring symbol style lamport-robust on input line 823.

Package footmisc Info: Declaring symbol style lamport* on input line 831.
Package footmisc Info: Declaring symbol style lamport*-robust on input line 840
.
) (/usr/share/texmf-texlive/tex/latex/tools/indentfirst.sty
Package: indentfirst 1995/11/23 v1.03 Indent first paragraph (DPC)
) (/usr/share/texmf-texlive/tex/latex/fancyhdr/fancyhdr.sty
\fancy@headwidth=\skip46
\f@ncyO@elh=\skip47
\f@ncyO@erh=\skip48
\f@ncyO@olh=\skip49
\f@ncyO@orh=\skip50
\f@ncyO@elf=\skip51
\f@ncyO@erf=\skip52
\f@ncyO@olf=\skip53
\f@ncyO@orf=\skip54
) (/usr/share/texmf-texlive/tex/latex/graphics/graphicx.sty
Package: graphicx 1999/02/16 v1.0f Enhanced LaTeX Graphics (DPC,SPQR)
(/usr/share/texmf-texlive/tex/latex/graphics/keyval.sty
Package: keyval 1999/03/16 v1.13 key=value parser (DPC)
\KV@toks@=\toks22
) (/usr/share/texmf-texlive/tex/latex/graphics/graphics.sty
Package: graphics 2006/02/20 v1.0o Standard LaTeX Graphics (DPC,SPQR)
(/usr/share/texmf-texlive/tex/latex/graphics/trig.sty
Package: trig 1999/03/16 v1.09 sin cos tan (DPC)
) (/etc/texmf/tex/latex/config/graphics.cfg
File: graphics.cfg 2007/01/18 v1.5 graphics configuration of teTeX/TeXLive
)
Package graphics Info: Driver file: pdftex.def on input line 90.
(/usr/share/texmf-texlive/tex/latex/pdftex-def/pdftex.def
File: pdftex.def 2007/01/08 v0.04d Graphics/color for pdfTeX
\Gread@gobject=\count101
))
\Gin@req@height=\dimen113
\Gin@req@width=\dimen114
) (/usr/share/texmf-texlive/tex/latex/tools/enumerate.sty
Package: enumerate 1999/03/05 v3.00 enumerate extensions (DPC)
\@enLab=\toks23
) (/usr/share/texmf-texlive/tex/latex/supertabular/supertabular.sty
Package: supertabular 2004/02/20 v4.1e the supertabular environment
\c@tracingst=\count102
\ST@wd=\dimen115
\ST@rightskip=\skip55
\ST@leftskip=\skip56
\ST@parfillskip=\skip57
\ST@pageleft=\dimen116
\ST@headht=\dimen117
\ST@tailht=\dimen118
\ST@pagesofar=\dimen119
\ST@pboxht=\dimen120
\ST@lineht=\dimen121
\ST@stretchht=\dimen122
\ST@prevht=\dimen123
\ST@toadd=\dimen124
\ST@dimen=\dimen125
\ST@pbox=\box28
) (/usr/share/texmf-texlive/tex/latex/geometry/geometry.sty
Package: geometry 2002/07/08 v3.2 Page Geometry
\Gm@cnth=\count103
\Gm@cntv=\count104
\c@Gm@tempcnt=\count105
\Gm@bindingoffset=\dimen126
\Gm@wd@mp=\dimen127
\Gm@odd@mp=\dimen128
\Gm@even@mp=\dimen129
\Gm@dimlist=\toks24
(/usr/share/texmf-texlive/tex/xelatex/xetexconfig/geometry.cfg)) (/usr/share/te
xmf-texlive/tex/latex/hyperref/hyperref.sty
Package: hyperref 2007/02/07 v6.75r Hypertext links for LaTeX
\@linkdim=\dimen130
\Hy@linkcounter=\count106
\Hy@pagecounter=\count107
(/usr/share/texmf-texlive/tex/latex/hyperref/pd1enc.def
File: pd1enc.def 2007/02/07 v6.75r Hyperref: PDFDocEncoding definition (HO)
) (/etc/texmf/tex/latex/config/hyperref.cfg
File: hyperref.cfg 2002/06/06 v1.2 hyperref configuration of TeXLive
) (/usr/share/texmf-texlive/tex/latex/oberdiek/kvoptions.sty
Package: kvoptions 2006/08/22 v2.4 Connects package keyval with LaTeX options (
HO)
)
Package hyperref Info: Option `hyperfootnotes' set `false' on input line 2238.
Package hyperref Info: Option `bookmarks' set `true' on input line 2238.
Package hyperref Info: Option `linktocpage' set `false' on input line 2238.
Package hyperref Info: Option `pdfdisplaydoctitle' set `true' on input line 223
8.
Package hyperref Info: Option `pdfpagelabels' set `true' on input line 2238.
Package hyperref Info: Option `bookmarksopen' set `true' on input line 2238.
Package hyperref Info: Option `colorlinks' set `true' on input line 2238.
Package hyperref Info: Hyper figures OFF on input line 2288.
Package hyperref Info: Link nesting OFF on input line 2293.
Package hyperref Info: Hyper index ON on input line 2296.
Package hyperref Info: Plain pages OFF on input line 2303.
Package hyperref Info: Backreferencing OFF on input line 2308.
Implicit mode ON; LaTeX internals redefined
Package hyperref Info: Bookmarks ON on input line 2444.
(/usr/share/texmf-texlive/tex/latex/ltxmisc/url.sty
\Urlmuskip=\muskip11
Package: url 2005/06/27  ver 3.2  Verb mode for urls, etc.
)
LaTeX Info: Redefining \url on input line 2599.
\Fld@menulength=\count108
\Field@Width=\dimen131
\Fld@charsize=\dimen132
\Choice@toks=\toks25
\Field@toks=\toks26
Package hyperref Info: Hyper figures OFF on input line 3102.
Package hyperref Info: Link nesting OFF on input line 3107.
Package hyperref Info: Hyper index ON on input line 3110.
Package hyperref Info: backreferencing OFF on input line 3117.
Package hyperref Info: Link coloring ON on input line 3120.
\Hy@abspage=\count109
\c@Item=\count110
)
*hyperref using driver hpdftex*
(/usr/share/texmf-texlive/tex/latex/hyperref/hpdftex.def
File: hpdftex.def 2007/02/07 v6.75r Hyperref driver for pdfTeX
\Fld@listcount=\count111
)
\TmpLen=\skip58
\c@proposition=\count112
\c@point=\count113
\c@exnum=\count114
\dentwidth=\skip59
\eqalign=\skip60
\c@Lcount=\count115
(./33063-t.aux)
\openout1 = `33063-t.aux'.

LaTeX Font Info:    Checking defaults for OML/cmm/m/it on input line 485.
LaTeX Font Info:    ... okay on input line 485.
LaTeX Font Info:    Checking defaults for T1/cmr/m/n on input line 485.
LaTeX Font Info:    ... okay on input line 485.
LaTeX Font Info:    Checking defaults for OT1/cmr/m/n on input line 485.
LaTeX Font Info:    ... okay on input line 485.
LaTeX Font Info:    Checking defaults for OMS/cmsy/m/n on input line 485.
LaTeX Font Info:    ... okay on input line 485.
LaTeX Font Info:    Checking defaults for OMX/cmex/m/n on input line 485.
LaTeX Font Info:    ... okay on input line 485.
LaTeX Font Info:    Checking defaults for U/cmr/m/n on input line 485.
LaTeX Font Info:    ... okay on input line 485.
LaTeX Font Info:    Checking defaults for PD1/pdf/m/n on input line 485.
LaTeX Font Info:    ... okay on input line 485.
(/usr/share/texmf/tex/context/base/supp-pdf.tex
[Loading MPS to PDF converter (version 2006.09.02).]
\scratchcounter=\count116
\scratchdimen=\dimen133
\scratchbox=\box29
\nofMPsegments=\count117
\nofMParguments=\count118
\everyMPshowfont=\toks27
\MPscratchCnt=\count119
\MPscratchDim=\dimen134
\MPnumerator=\count120
\everyMPtoPDFconversion=\toks28
)
-------------------- Geometry parameters
paper: class default
landscape: --
twocolumn: --
twoside: true
asymmetric: --
h-parts: 18.0675pt, 397.48499pt, 18.0675pt
v-parts: 9.07278pt, 591.61302pt, 13.60919pt
hmarginratio: 1:1
vmarginratio: 2:3
lines: --
heightrounded: --
bindingoffset: 0.0pt
truedimen: --
includehead: true
includefoot: true
includemp: --
driver: pdftex
-------------------- Page layout dimensions and switches
\paperwidth  433.62pt
\paperheight 614.295pt
\textwidth  397.48499pt
\textheight 529.73923pt
\oddsidemargin  -54.20248pt
\evensidemargin -54.20248pt
\topmargin  -63.1972pt
\headheight 14.5pt
\headsep    19.8738pt
\footskip   30.0pt
\marginparwidth 98.0pt
\marginparsep   7.0pt
\columnsep  10.0pt
\skip\footins  10.8pt plus 4.0pt minus 2.0pt
\hoffset 0.0pt
\voffset 0.0pt
\mag 1000
\@twosidetrue \@mparswitchtrue 
(1in=72.27pt, 1cm=28.45pt)
-----------------------
(/usr/share/texmf-texlive/tex/latex/graphics/color.sty
Package: color 2005/11/14 v1.0j Standard LaTeX Color (DPC)
(/etc/texmf/tex/latex/config/color.cfg
File: color.cfg 2007/01/18 v1.5 color configuration of teTeX/TeXLive
)
Package color Info: Driver file: pdftex.def on input line 130.
)
Package hyperref Info: Link coloring ON on input line 485.
(/usr/share/texmf-texlive/tex/latex/hyperref/nameref.sty
Package: nameref 2006/12/27 v2.28 Cross-referencing by name of section
(/usr/share/texmf-texlive/tex/latex/oberdiek/refcount.sty
Package: refcount 2006/02/20 v3.0 Data extraction from references (HO)
)
\c@section@level=\count121
)
LaTeX Info: Redefining \ref on input line 485.
LaTeX Info: Redefining \pageref on input line 485.
(./33063-t.out) (./33063-t.out)
\@outlinefile=\write3
\openout3 = `33063-t.out'.

LaTeX Font Info:    Try loading font information for U+msa on input line 521.
(/usr/share/texmf-texlive/tex/latex/amsfonts/umsa.fd
File: umsa.fd 2002/01/19 v2.2g AMS font definitions
)
LaTeX Font Info:    Try loading font information for U+msb on input line 521.
(/usr/share/texmf-texlive/tex/latex/amsfonts/umsb.fd
File: umsb.fd 2002/01/19 v2.2g AMS font definitions
) [1

{/var/lib/texmf/fonts/map/pdftex/updmap/pdftex.map}] [2

]
LaTeX Font Info:    Try loading font information for OMS+cmr on input line 575.

(/usr/share/texmf-texlive/tex/latex/base/omscmr.fd
File: omscmr.fd 1999/05/25 v2.5h Standard LaTeX font definitions
)
LaTeX Font Info:    Font shape `OMS/cmr/m/n' in size <12> not available
(Font)              Font shape `OMS/cmsy/m/n' tried instead on input line 575.
[1


]
Underfull \hbox (badness 1199) in paragraph at lines 610--610
[]\OT1/cmr/m/n/12 GINN & COM-PANY \OMS/cmsy/m/n/12 ^^A \OT1/cmr/m/n/12 PRO-
 []

[2] [3

] [4] [5

] [6] (./33063-t.toc [7

])
\tf@toc=\write4
\openout4 = `33063-t.toc'.

[8]

Package hyperref Warning: Difference (2) between bookmark levels is greater 
(hyperref)                than one, level fixed on input line 776.

<./images/woodcutsmall.pdf, id=325, 171.64125pt x 144.54pt>
File: ./images/woodcutsmall.pdf Graphic file (type pdf)
<use ./images/woodcutsmall.pdf> [1


 <./images/woodcutsmall.pdf>] <./images/012aaZ10.pdf, id=334, 112.42pt x 66.247
5pt>
File: ./images/012aaZ10.pdf Graphic file (type pdf)
<use ./images/012aaZ10.pdf> [2 <./images/012aaZ10.pdf>] <./images/012bbZ14.pdf,
 id=350, 135.50626pt x 94.3525pt>
File: ./images/012bbZ14.pdf Graphic file (type pdf)
<use ./images/012bbZ14.pdf> [3 <./images/012bbZ14.pdf>] [4] [5] [6]
 .

Package hyperref Warning: Token not allowed in a PDFDocEncoded string,
(hyperref)                removing `\spacefactor' on input line 1099.


Package hyperref Warning: Token not allowed in a PDFDocEncoded string,
(hyperref)                removing `\@m' on input line 1099.

<./images/016aaZ37.pdf, id=379, 113.42375pt x 83.31125pt>
File: ./images/016aaZ37.pdf Graphic file (type pdf)
<use ./images/016aaZ37.pdf> [7

 <./images/016aaZ37.pdf>] <./images/018aaZ55.pdf, id=392, 169.63374pt x 36.135p
t>
File: ./images/018aaZ55.pdf Graphic file (type pdf)
<use ./images/018aaZ55.pdf> [8] <./images/018bbZ56.pdf, id=399, 174.6525pt x 29
.10875pt>
File: ./images/018bbZ56.pdf Graphic file (type pdf)
<use ./images/018bbZ56.pdf> <./images/018ccZ57.pdf, id=400, 86.3225pt x 80.3pt>
File: ./images/018ccZ57.pdf Graphic file (type pdf)
<use ./images/018ccZ57.pdf> [9 <./images/018aaZ55.pdf> <./images/018bbZ56.pdf>]
<./images/019aaZ58.pdf, id=420, 67.25125pt x 66.2475pt>
File: ./images/019aaZ58.pdf Graphic file (type pdf)
<use ./images/019aaZ58.pdf> <./images/019bbZ58.pdf, id=421, 91.34125pt x 61.228
74pt>
File: ./images/019bbZ58.pdf Graphic file (type pdf)
<use ./images/019bbZ58.pdf> [10 <./images/018ccZ57.pdf> <./images/019aaZ58.pdf>
 <./images/019bbZ58.pdf>] <./images/019ccZ62.pdf, id=451, 92.345pt x 69.25874pt
>
File: ./images/019ccZ62.pdf Graphic file (type pdf)
<use ./images/019ccZ62.pdf> <./images/020aaZ63.pdf, id=452, 82.3075pt x 77.2887
6pt>
File: ./images/020aaZ63.pdf Graphic file (type pdf)
<use ./images/020aaZ63.pdf> <./images/020bbZ66.pdf, id=453, 169.63374pt x 36.13
5pt>
File: ./images/020bbZ66.pdf Graphic file (type pdf)
<use ./images/020bbZ66.pdf> <./images/020ccZ68.pdf, id=454, 78.2925pt x 54.2025
pt>
File: ./images/020ccZ68.pdf Graphic file (type pdf)
<use ./images/020ccZ68.pdf> <./images/020ddZ69.pdf, id=455, 108.405pt x 56.21pt
>
File: ./images/020ddZ69.pdf Graphic file (type pdf)
<use ./images/020ddZ69.pdf> [11 <./images/019ccZ62.pdf> <./images/020aaZ63.pdf>
 <./images/020bbZ66.pdf>] <./images/021aaZ72.pdf, id=481, 105.39375pt x 114.427
5pt>
File: ./images/021aaZ72.pdf Graphic file (type pdf)
<use ./images/021aaZ72.pdf> [12 <./images/020ccZ68.pdf> <./images/020ddZ69.pdf>
] <./images/022aaZ75.pdf, id=500, 62.2325pt x 56.21pt>
File: ./images/022aaZ75.pdf Graphic file (type pdf)
<use ./images/022aaZ75.pdf> <./images/022bbZ76.pdf, id=501, 53.19875pt x 65.243
74pt>
File: ./images/022bbZ76.pdf Graphic file (type pdf)
<use ./images/022bbZ76.pdf> <./images/022ccZ77.pdf, id=502, 92.345pt x 69.25874
pt>
File: ./images/022ccZ77.pdf Graphic file (type pdf)
<use ./images/022ccZ77.pdf> [13 <./images/021aaZ72.pdf>] [14 <./images/022aaZ75
.pdf> <./images/022bbZ76.pdf> <./images/022ccZ77.pdf>] <./images/023aaZ79.pdf, 
id=546, 187.70125pt x 81.30376pt>
File: ./images/023aaZ79.pdf Graphic file (type pdf)
<use ./images/023aaZ79.pdf> <./images/023bcZ80.pdf, id=547, 378.41376pt x 94.35
25pt>
File: ./images/023bcZ80.pdf Graphic file (type pdf)
<use ./images/023bcZ80.pdf> [15 <./images/023aaZ79.pdf>] [16 <./images/023bcZ80
.pdf>] <./images/024aaZ81.pdf, id=571, 169.63374pt x 42.1575pt>
File: ./images/024aaZ81.pdf Graphic file (type pdf)
<use ./images/024aaZ81.pdf>
LaTeX Font Info:    Font shape `OMS/cmr/m/n' in size <10.95> not available
(Font)              Font shape `OMS/cmsy/m/n' tried instead on input line 1512.

<./images/024bbZ83.pdf, id=572, 79.29625pt x 70.2625pt>
File: ./images/024bbZ83.pdf Graphic file (type pdf)
<use ./images/024bbZ83.pdf> [17 <./images/024aaZ81.pdf> <./images/024bbZ83.pdf>
] <./images/025aaZ86.pdf, id=586, 188.705pt x 81.30376pt>
File: ./images/025aaZ86.pdf Graphic file (type pdf)
<use ./images/025aaZ86.pdf> [18 <./images/025aaZ86.pdf>] <./images/026aaZ90.pdf
, id=596, 202.7575pt x 83.31125pt>
File: ./images/026aaZ90.pdf Graphic file (type pdf)
<use ./images/026aaZ90.pdf> <./images/027aaZ93.pdf, id=597, 169.63374pt x 85.31
876pt>
File: ./images/027aaZ93.pdf Graphic file (type pdf)
<use ./images/027aaZ93.pdf> [19 <./images/026aaZ90.pdf>] <./images/028aaZ95.pdf
, id=607, 188.705pt x 131.49126pt>
File: ./images/028aaZ95.pdf Graphic file (type pdf)
<use ./images/028aaZ95.pdf> [20 <./images/027aaZ93.pdf>] <./images/029aaZ96.pdf
, id=617, 94.3525pt x 130.4875pt>
File: ./images/029aaZ96.pdf Graphic file (type pdf)
<use ./images/029aaZ96.pdf> [21 <./images/028aaZ95.pdf>] [22 <./images/029aaZ96
.pdf>] <./images/030aaZ97.pdf, id=640, 94.3525pt x 130.4875pt>
File: ./images/030aaZ97.pdf Graphic file (type pdf)
<use ./images/030aaZ97.pdf> [23 <./images/030aaZ97.pdf>] <./images/031aa100.pdf
, id=653, 197.73875pt x 111.41624pt>
File: ./images/031aa100.pdf Graphic file (type pdf)
<use ./images/031aa100.pdf> [24 <./images/031aa100.pdf>] <./images/032aa101.pdf
, id=663, 149.55875pt x 131.49126pt>
File: ./images/032aa101.pdf Graphic file (type pdf)
<use ./images/032aa101.pdf> [25 <./images/032aa101.pdf>] <./images/033aa104.pdf
, id=673, 179.67125pt x 106.3975pt>
File: ./images/033aa104.pdf Graphic file (type pdf)
<use ./images/033aa104.pdf> <./images/034aa107.pdf, id=674, 207.77625pt x 112.4
2pt>
File: ./images/034aa107.pdf Graphic file (type pdf)
<use ./images/034aa107.pdf> [26 <./images/033aa104.pdf>] <./images/034bb109.pdf
, id=684, 170.6375pt x 132.495pt>
File: ./images/034bb109.pdf Graphic file (type pdf)
<use ./images/034bb109.pdf> [27 <./images/034aa107.pdf>] <./images/035aa110.pdf
, id=695, 264.99pt x 95.35625pt>
File: ./images/035aa110.pdf Graphic file (type pdf)
<use ./images/035aa110.pdf> [28 <./images/034bb109.pdf>] [29 <./images/035aa110
.pdf>] <./images/036aa111.pdf, id=716, 228.855pt x 130.4875pt>
File: ./images/036aa111.pdf Graphic file (type pdf)
<use ./images/036aa111.pdf> [30 <./images/036aa111.pdf>] <./images/037aa112.pdf
, id=726, 235.88126pt x 160.6pt>
File: ./images/037aa112.pdf Graphic file (type pdf)
<use ./images/037aa112.pdf> [31 <./images/037aa112.pdf>] <./images/038aa115.pdf
, id=736, 235.88126pt x 160.6pt>
File: ./images/038aa115.pdf Graphic file (type pdf)
<use ./images/038aa115.pdf> [32 <./images/038aa115.pdf>] <./images/039aa117.pdf
, id=746, 95.35625pt x 85.31876pt>
File: ./images/039aa117.pdf Graphic file (type pdf)
<use ./images/039aa117.pdf> <./images/039bb118.pdf, id=747, 132.495pt x 85.3187
6pt>
File: ./images/039bb118.pdf Graphic file (type pdf)
<use ./images/039bb118.pdf> <./images/039ce119.pdf, id=748, 285.065pt x 91.3412
5pt>
File: ./images/039ce119.pdf Graphic file (type pdf)
<use ./images/039ce119.pdf> <./images/039fi120.pdf, id=749, 385.44pt x 97.36375
pt>
File: ./images/039fi120.pdf Graphic file (type pdf)
<use ./images/039fi120.pdf> [33 <./images/039aa117.pdf> <./images/039bb118.pdf>
] [34 <./images/039ce119.pdf> <./images/039fi120.pdf>] <./images/041aa129.pdf, 
id=785, 208.78pt x 104.39pt>
File: ./images/041aa129.pdf Graphic file (type pdf)
<use ./images/041aa129.pdf> [35] [36 <./images/041aa129.pdf>] <./images/042aa13
8.pdf, id=799, 164.615pt x 66.2475pt>
File: ./images/042aa138.pdf Graphic file (type pdf)
<use ./images/042aa138.pdf> <./images/043ab139.pdf, id=800, 330.23375pt x 90.33
75pt>
File: ./images/043ab139.pdf Graphic file (type pdf)
<use ./images/043ab139.pdf> [37 <./images/042aa138.pdf>] <./images/044ab143.pdf
, id=810, 330.23375pt x 90.3375pt>
File: ./images/044ab143.pdf Graphic file (type pdf)
<use ./images/044ab143.pdf> [38 <./images/043ab139.pdf>] <./images/045aa145.pdf
, id=820, 158.5925pt x 113.42375pt>
File: ./images/045aa145.pdf Graphic file (type pdf)
<use ./images/045aa145.pdf> [39 <./images/044ab143.pdf>] <./images/046aa147.pdf
, id=831, 158.5925pt x 113.42375pt>
File: ./images/046aa147.pdf Graphic file (type pdf)
<use ./images/046aa147.pdf> [40 <./images/045aa145.pdf>] <./images/047ab150.pdf
, id=841, 323.2075pt x 150.5625pt>
File: ./images/047ab150.pdf Graphic file (type pdf)
<use ./images/047ab150.pdf> [41 <./images/046aa147.pdf>] [42 <./images/047ab150
.pdf>] <./images/048ac151.pdf, id=864, 338.26375pt x 115.43124pt>
File: ./images/048ac151.pdf Graphic file (type pdf)
<use ./images/048ac151.pdf> <./images/049aa152.pdf, id=865, 206.7725pt x 85.318
76pt>
File: ./images/049aa152.pdf Graphic file (type pdf)
<use ./images/049aa152.pdf> [43 <./images/048ac151.pdf>] <./images/050aa153.pdf
, id=878, 160.6pt x 95.35625pt>
File: ./images/050aa153.pdf Graphic file (type pdf)
<use ./images/050aa153.pdf> [44 <./images/049aa152.pdf>] <./images/051ac154.pdf
, id=888, 378.41376pt x 131.49126pt>
File: ./images/051ac154.pdf Graphic file (type pdf)
<use ./images/051ac154.pdf> [45 <./images/050aa153.pdf>] [46 <./images/051ac154
.pdf>] <./images/052ab155.pdf, id=908, 322.20375pt x 112.42pt>
File: ./images/052ab155.pdf Graphic file (type pdf)
<use ./images/052ab155.pdf> [47 <./images/052ab155.pdf>] <./images/054aa160.pdf
, id=918, 133.49875pt x 112.42pt>
File: ./images/054aa160.pdf Graphic file (type pdf)
<use ./images/054aa160.pdf> [48] [49 <./images/054aa160.pdf>] <./images/055aa16
2.pdf, id=932, 121.45375pt x 101.37875pt>
File: ./images/055aa162.pdf Graphic file (type pdf)
<use ./images/055aa162.pdf> [50 <./images/055aa162.pdf>] <./images/056ac166.pdf
, id=942, 355.3275pt x 75.28125pt>
File: ./images/056ac166.pdf Graphic file (type pdf)
<use ./images/056ac166.pdf> <./images/056dg170.pdf, id=943, 383.4325pt x 66.247
5pt>
File: ./images/056dg170.pdf Graphic file (type pdf)
<use ./images/056dg170.pdf> <./images/057aa174.pdf, id=944, 132.495pt x 68.255p
t>
File: ./images/057aa174.pdf Graphic file (type pdf)
<use ./images/057aa174.pdf> [51 <./images/056ac166.pdf> <./images/056dg170.pdf>
] <./images/057bb176.pdf, id=960, 190.7125pt x 87.32625pt>
File: ./images/057bb176.pdf Graphic file (type pdf)
<use ./images/057bb176.pdf> [52 <./images/057aa174.pdf>] <./images/058aa178.pdf
, id=970, 222.8325pt x 103.38625pt>
File: ./images/058aa178.pdf Graphic file (type pdf)
<use ./images/058aa178.pdf> [53 <./images/057bb176.pdf>] <./images/058bb181.pdf
, id=989, 192.72pt x 67.25125pt>
File: ./images/058bb181.pdf Graphic file (type pdf)
<use ./images/058bb181.pdf> [54 <./images/058aa178.pdf> <./images/058bb181.pdf>
] <./images/059aa182.pdf, id=1004, 222.8325pt x 103.38625pt>
File: ./images/059aa182.pdf Graphic file (type pdf)
<use ./images/059aa182.pdf> [55 <./images/059aa182.pdf>] <./images/060aa183.pdf
, id=1014, 222.8325pt x 103.38625pt>
File: ./images/060aa183.pdf Graphic file (type pdf)
<use ./images/060aa183.pdf> <./images/061aa184.pdf, id=1015, 222.8325pt x 103.3
8625pt>
File: ./images/061aa184.pdf Graphic file (type pdf)
<use ./images/061aa184.pdf> [56 <./images/060aa183.pdf>] [57 <./images/061aa184
.pdf>] <./images/062ab185.pdf, id=1035, 336.25626pt x 75.28125pt>
File: ./images/062ab185.pdf Graphic file (type pdf)
<use ./images/062ab185.pdf> <./images/063aa187.pdf, id=1036, 234.8775pt x 169.6
3374pt>
File: ./images/063aa187.pdf Graphic file (type pdf)
<use ./images/063aa187.pdf> [58 <./images/062ab185.pdf>] <./images/064aa188.pdf
, id=1049, 123.46124pt x 95.35625pt>
File: ./images/064aa188.pdf Graphic file (type pdf)
<use ./images/064aa188.pdf> [59 <./images/063aa187.pdf>] <./images/064bb190.pdf
, id=1059, 162.6075pt x 86.3225pt>
File: ./images/064bb190.pdf Graphic file (type pdf)
<use ./images/064bb190.pdf> [60 <./images/064aa188.pdf> <./images/064bb190.pdf>
] <./images/065ac192.pdf, id=1074, 388.45125pt x 112.42pt>
File: ./images/065ac192.pdf Graphic file (type pdf)
<use ./images/065ac192.pdf> [61 <./images/065ac192.pdf>] <./images/066ad203.pdf
, id=1086, 355.3275pt x 78.2925pt>
File: ./images/066ad203.pdf Graphic file (type pdf)
<use ./images/066ad203.pdf> <./images/067aa205.pdf, id=1087, 188.705pt x 160.6p
t>
File: ./images/067aa205.pdf Graphic file (type pdf)
<use ./images/067aa205.pdf> [62 <./images/066ad203.pdf>] <./images/068aa207.pdf
, id=1097, 157.58875pt x 172.645pt>
File: ./images/068aa207.pdf Graphic file (type pdf)
<use ./images/068aa207.pdf> [63 <./images/067aa205.pdf>] [64 <./images/068aa207
.pdf>] <./images/069ac208.pdf, id=1119, 379.4175pt x 123.46124pt>
File: ./images/069ac208.pdf Graphic file (type pdf)
<use ./images/069ac208.pdf> <./images/069dd211.pdf, id=1120, 124.465pt x 113.42
375pt>
File: ./images/069dd211.pdf Graphic file (type pdf)
<use ./images/069dd211.pdf> [65 <./images/069ac208.pdf>] <./images/070aa212.pdf
, id=1133, 218.8175pt x 151.56625pt>
File: ./images/070aa212.pdf Graphic file (type pdf)
<use ./images/070aa212.pdf> [66 <./images/069dd211.pdf>] <./images/071aa213.pdf
, id=1146, 226.8475pt x 151.56625pt>
File: ./images/071aa213.pdf Graphic file (type pdf)
<use ./images/071aa213.pdf> [67 <./images/070aa212.pdf>] [68 <./images/071aa213
.pdf>] [69] [70] <./images/074aaZ19.pdf, id=1208, 89.33376pt x 96.36pt>
File: ./images/074aaZ19.pdf Graphic file (type pdf)
<use ./images/074aaZ19.pdf> [71] [72 <./images/074aaZ19.pdf>] <./images/074bbZ2
0.pdf, id=1224, 77.28876pt x 76.285pt>
File: ./images/074bbZ20.pdf Graphic file (type pdf)
<use ./images/074bbZ20.pdf> <./images/075aaZ21.pdf, id=1225, 112.42pt x 76.285p
t>
File: ./images/075aaZ21.pdf Graphic file (type pdf)
<use ./images/075aaZ21.pdf> <./images/075bbZ22.pdf, id=1226, 112.42pt x 76.285p
t>
File: ./images/075bbZ22.pdf Graphic file (type pdf)
<use ./images/075bbZ22.pdf> [73 <./images/074bbZ20.pdf>] <./images/075ceZ23.pdf
, id=1236, 270.00874pt x 57.21375pt>
File: ./images/075ceZ23.pdf Graphic file (type pdf)
<use ./images/075ceZ23.pdf> [74 <./images/075aaZ21.pdf> <./images/075bbZ22.pdf>
] <./images/076aaZ24.pdf, id=1257, 95.35625pt x 70.2625pt>
File: ./images/076aaZ24.pdf Graphic file (type pdf)
<use ./images/076aaZ24.pdf> <./images/076bbZ25.pdf, id=1258, 86.3225pt x 67.251
25pt>
File: ./images/076bbZ25.pdf Graphic file (type pdf)
<use ./images/076bbZ25.pdf> [75 <./images/075ceZ23.pdf>] <./images/076ccZ26.pdf
, id=1272, 126.4725pt x 86.3225pt>
File: ./images/076ccZ26.pdf Graphic file (type pdf)
<use ./images/076ccZ26.pdf> <./images/076ddZ27.pdf, id=1273, 93.34875pt x 83.31
125pt>
File: ./images/076ddZ27.pdf Graphic file (type pdf)
<use ./images/076ddZ27.pdf> [76 <./images/076aaZ24.pdf> <./images/076bbZ25.pdf>
] <./images/077adZ28.pdf, id=1291, 337.26pt x 65.24374pt>
File: ./images/077adZ28.pdf Graphic file (type pdf)
<use ./images/077adZ28.pdf> [77 <./images/076ccZ26.pdf> <./images/076ddZ27.pdf>
 <./images/077adZ28.pdf>] <./images/077eeZ34.pdf, id=1314, 54.2025pt x 56.21pt>
File: ./images/077eeZ34.pdf Graphic file (type pdf)
<use ./images/077eeZ34.pdf> <./images/077ffZ35.pdf, id=1315, 72.27pt x 85.31876
pt>
File: ./images/077ffZ35.pdf Graphic file (type pdf)
<use ./images/077ffZ35.pdf> <./images/078aaZ41.pdf, id=1316, 88.33pt x 80.3pt>
File: ./images/078aaZ41.pdf Graphic file (type pdf)
<use ./images/078aaZ41.pdf> [78 <./images/077eeZ34.pdf> <./images/077ffZ35.pdf>
] <./images/078bbZ43.pdf, id=1331, 58.2175pt x 75.28125pt>
File: ./images/078bbZ43.pdf Graphic file (type pdf)
<use ./images/078bbZ43.pdf> <./images/078ccZ46.pdf, id=1332, 95.35625pt x 67.25
125pt>
File: ./images/078ccZ46.pdf Graphic file (type pdf)
<use ./images/078ccZ46.pdf> [79 <./images/078aaZ41.pdf> <./images/078bbZ43.pdf>
] <./images/078ddZ49.pdf, id=1348, 62.2325pt x 65.24374pt>
File: ./images/078ddZ49.pdf Graphic file (type pdf)
<use ./images/078ddZ49.pdf> <./images/079adZ51.pdf, id=1349, 400.49625pt x 95.3
5625pt>
File: ./images/079adZ51.pdf Graphic file (type pdf)
<use ./images/079adZ51.pdf> [80 <./images/078ccZ46.pdf> <./images/078ddZ49.pdf>
] <./images/079eeZ57.pdf, id=1364, 53.19875pt x 67.25125pt>
File: ./images/079eeZ57.pdf Graphic file (type pdf)
<use ./images/079eeZ57.pdf> [81 <./images/079adZ51.pdf>] <./images/080aaZ59.pdf
, id=1375, 111.41624pt x 87.32625pt>
File: ./images/080aaZ59.pdf Graphic file (type pdf)
<use ./images/080aaZ59.pdf> <./images/080bbZ60.pdf, id=1376, 92.345pt x 93.3487
5pt>
File: ./images/080bbZ60.pdf Graphic file (type pdf)
<use ./images/080bbZ60.pdf> [82 <./images/079eeZ57.pdf> <./images/080aaZ59.pdf>
 <./images/080bbZ60.pdf>] <./images/080ccZ61.pdf, id=1396, 87.32625pt x 84.315p
t>
File: ./images/080ccZ61.pdf Graphic file (type pdf)
<use ./images/080ccZ61.pdf> <./images/080ddZ62.pdf, id=1397, 66.2475pt x 66.247
5pt>
File: ./images/080ddZ62.pdf Graphic file (type pdf)
<use ./images/080ddZ62.pdf> <./images/080eeZ63.pdf, id=1398, 83.31125pt x 75.28
125pt>
File: ./images/080eeZ63.pdf Graphic file (type pdf)
<use ./images/080eeZ63.pdf> <./images/080ffZ64.pdf, id=1399, 53.19875pt x 67.25
125pt>
File: ./images/080ffZ64.pdf Graphic file (type pdf)
<use ./images/080ffZ64.pdf> <./images/081aaZ65.pdf, id=1400, 79.29625pt x 91.34
125pt>
File: ./images/081aaZ65.pdf Graphic file (type pdf)
<use ./images/081aaZ65.pdf> <./images/081bbZ66.pdf, id=1401, 72.27pt x 67.25125
pt>
File: ./images/081bbZ66.pdf Graphic file (type pdf)
<use ./images/081bbZ66.pdf> [83 <./images/080ccZ61.pdf> <./images/080ddZ62.pdf>
 <./images/080eeZ63.pdf> <./images/080ffZ64.pdf>] <./images/081ccZ67.pdf, id=14
26, 81.30376pt x 67.25125pt>
File: ./images/081ccZ67.pdf Graphic file (type pdf)
<use ./images/081ccZ67.pdf> <./images/081ddZ68.pdf, id=1427, 88.33pt x 58.2175p
t>
File: ./images/081ddZ68.pdf Graphic file (type pdf)
<use ./images/081ddZ68.pdf> <./images/081eeZ69.pdf, id=1428, 95.35625pt x 49.18
375pt>
File: ./images/081eeZ69.pdf Graphic file (type pdf)
<use ./images/081eeZ69.pdf> [84 <./images/081aaZ65.pdf> <./images/081bbZ66.pdf>
 <./images/081ccZ67.pdf> <./images/081ddZ68.pdf>] <./images/081ffZ73.pdf, id=14
53, 92.345pt x 57.21375pt>
File: ./images/081ffZ73.pdf Graphic file (type pdf)
<use ./images/081ffZ73.pdf> <./images/082adZ74.pdf, id=1454, 380.42125pt x 94.3
525pt>
File: ./images/082adZ74.pdf Graphic file (type pdf)
<use ./images/082adZ74.pdf> [85 <./images/081eeZ69.pdf> <./images/081ffZ73.pdf>
] <./images/082ehZ78.pdf, id=1469, 382.42876pt x 94.3525pt>
File: ./images/082ehZ78.pdf Graphic file (type pdf)
<use ./images/082ehZ78.pdf> <./images/082iiZ82.pdf, id=1470, 91.34125pt x 84.31
5pt>
File: ./images/082iiZ82.pdf Graphic file (type pdf)
<use ./images/082iiZ82.pdf> [86 <./images/082adZ74.pdf> <./images/082ehZ78.pdf>
] <./images/083adZ83.pdf, id=1485, 417.56pt x 84.315pt>
File: ./images/083adZ83.pdf Graphic file (type pdf)
<use ./images/083adZ83.pdf> [87 <./images/082iiZ82.pdf> <./images/083adZ83.pdf>
] <./images/083eeZ88.pdf, id=1501, 109.40875pt x 94.3525pt>
File: ./images/083eeZ88.pdf Graphic file (type pdf)
<use ./images/083eeZ88.pdf> <./images/083ffZ89.pdf, id=1502, 63.23625pt x 67.25
125pt>
File: ./images/083ffZ89.pdf Graphic file (type pdf)
<use ./images/083ffZ89.pdf> [88 <./images/083eeZ88.pdf> <./images/083ffZ89.pdf>
]
 .

Package hyperref Warning: Token not allowed in a PDFDocEncoded string,
(hyperref)                removing `\spacefactor' on input line 4330.


Package hyperref Warning: Token not allowed in a PDFDocEncoded string,
(hyperref)                removing `\@m' on input line 4330.

<./images/084aa220.pdf, id=1517, 123.46124pt x 115.43124pt>
File: ./images/084aa220.pdf Graphic file (type pdf)
<use ./images/084aa220.pdf> [89

 <./images/084aa220.pdf>] <./images/085ac224.pdf, id=1529, 373.395pt x 124.465p
t>
File: ./images/085ac224.pdf Graphic file (type pdf)
<use ./images/085ac224.pdf> [90 <./images/085ac224.pdf>] <./images/086aa235.pdf
, id=1545, 108.405pt x 124.465pt>
File: ./images/086aa235.pdf Graphic file (type pdf)
<use ./images/086aa235.pdf> [91 <./images/086aa235.pdf>] <./images/087ab236.pdf
, id=1555, 263.98625pt x 104.39pt>
File: ./images/087ab236.pdf Graphic file (type pdf)
<use ./images/087ab236.pdf> [92 <./images/087ab236.pdf>] [93] <./images/089ab24
1.pdf, id=1573, 250.9375pt x 98.3675pt>
File: ./images/089ab241.pdf Graphic file (type pdf)
<use ./images/089ab241.pdf> [94] [95 <./images/089ab241.pdf>] <./images/090ab24
3.pdf, id=1590, 250.9375pt x 98.3675pt>
File: ./images/090ab243.pdf Graphic file (type pdf)
<use ./images/090ab243.pdf> [96] [97 <./images/090ab243.pdf>] <./images/091aa24
5.pdf, id=1607, 108.405pt x 123.46124pt>
File: ./images/091aa245.pdf Graphic file (type pdf)
<use ./images/091aa245.pdf> [98 <./images/091aa245.pdf>] <./images/092aa249.pdf
, id=1617, 114.4275pt x 112.42pt>
File: ./images/092aa249.pdf Graphic file (type pdf)
<use ./images/092aa249.pdf> [99 <./images/092aa249.pdf>] <./images/093aa250.pdf
, id=1628, 101.37875pt x 98.3675pt>
File: ./images/093aa250.pdf Graphic file (type pdf)
<use ./images/093aa250.pdf> [100 <./images/093aa250.pdf>] <./images/094aa251.pd
f, id=1638, 101.37875pt x 98.3675pt>
File: ./images/094aa251.pdf Graphic file (type pdf)
<use ./images/094aa251.pdf> [101 <./images/094aa251.pdf>] <./images/095aa253.pd
f, id=1648, 169.63374pt x 115.43124pt>
File: ./images/095aa253.pdf Graphic file (type pdf)
<use ./images/095aa253.pdf> [102 <./images/095aa253.pdf>] <./images/096ac257.pd
f, id=1658, 412.54124pt x 104.39pt>
File: ./images/096ac257.pdf Graphic file (type pdf)
<use ./images/096ac257.pdf> [103] [104 <./images/096ac257.pdf>] <./images/097aa
258.pdf, id=1677, 121.45375pt x 117.43875pt>
File: ./images/097aa258.pdf Graphic file (type pdf)
<use ./images/097aa258.pdf> [105 <./images/097aa258.pdf>] <./images/098aa261.pd
f, id=1688, 188.705pt x 111.41624pt>
File: ./images/098aa261.pdf Graphic file (type pdf)
<use ./images/098aa261.pdf> [106 <./images/098aa261.pdf>] <./images/099aa264.pd
f, id=1698, 147.55125pt x 93.34875pt>
File: ./images/099aa264.pdf Graphic file (type pdf)
<use ./images/099aa264.pdf> [107 <./images/099aa264.pdf>] <./images/100aa265.pd
f, id=1716, 159.59625pt x 108.405pt>
File: ./images/100aa265.pdf Graphic file (type pdf)
<use ./images/100aa265.pdf> [108 <./images/100aa265.pdf>] [109] [110] <./images
/103aa272.pdf, id=1742, 269.005pt x 24.09pt>
File: ./images/103aa272.pdf Graphic file (type pdf)
<use ./images/103aa272.pdf> <./images/103bb273.pdf, id=1743, 145.54375pt x 145.
54375pt>
File: ./images/103bb273.pdf Graphic file (type pdf)
<use ./images/103bb273.pdf> [111 <./images/103aa272.pdf>] [112 <./images/103bb2
73.pdf>] [113] [114] [115] [116] <./images/108aa286.pdf, id=1786, 255.95625pt x
 63.23625pt>
File: ./images/108aa286.pdf Graphic file (type pdf)
<use ./images/108aa286.pdf> [117 <./images/108aa286.pdf>] [118] <./images/109ac
287.pdf, id=1801, 353.32pt x 108.405pt>
File: ./images/109ac287.pdf Graphic file (type pdf)
<use ./images/109ac287.pdf> [119

 <./images/109ac287.pdf>] [120] <./images/111ac289.pdf, id=1822, 366.36874pt x 
120.45pt>
File: ./images/111ac289.pdf Graphic file (type pdf)
<use ./images/111ac289.pdf> [121] [122 <./images/111ac289.pdf>] <./images/112ac
290.pdf, id=1838, 370.38374pt x 107.40125pt>
File: ./images/112ac290.pdf Graphic file (type pdf)
<use ./images/112ac290.pdf> [123 <./images/112ac290.pdf>] <./images/113aa294.pd
f, id=1851, 114.4275pt x 98.3675pt>
File: ./images/113aa294.pdf Graphic file (type pdf)
<use ./images/113aa294.pdf> [124 <./images/113aa294.pdf>] <./images/114aa295.pd
f, id=1861, 151.56625pt x 106.3975pt>
File: ./images/114aa295.pdf Graphic file (type pdf)
<use ./images/114aa295.pdf> <./images/115ac296.pdf, id=1862, 340.27126pt x 114.
4275pt>
File: ./images/115ac296.pdf Graphic file (type pdf)
<use ./images/115ac296.pdf> [125 <./images/114aa295.pdf>] <./images/115dd297.pd
f, id=1872, 93.34875pt x 95.35625pt>
File: ./images/115dd297.pdf Graphic file (type pdf)
<use ./images/115dd297.pdf> <./images/115ee297.pdf, id=1873, 105.39375pt x 118.
4425pt>
File: ./images/115ee297.pdf Graphic file (type pdf)
<use ./images/115ee297.pdf> [126 <./images/115ac296.pdf>] <./images/116ag299.pd
f, id=1883, 423.5825pt x 78.2925pt>
File: ./images/116ag299.pdf Graphic file (type pdf)
<use ./images/116ag299.pdf> [127 <./images/115dd297.pdf> <./images/115ee297.pdf
>] [128 <./images/116ag299.pdf>] [129] <./images/117aa104.pdf, id=1942, 68.255p
t x 66.2475pt>
File: ./images/117aa104.pdf Graphic file (type pdf)
<use ./images/117aa104.pdf> <./images/117bb105.pdf, id=1943, 70.2625pt x 56.21p
t>
File: ./images/117bb105.pdf Graphic file (type pdf)
<use ./images/117bb105.pdf> <./images/118aa107.pdf, id=1944, 105.39375pt x 65.2
4374pt>
File: ./images/118aa107.pdf Graphic file (type pdf)
<use ./images/118aa107.pdf> <./images/118be108.pdf, id=1945, 410.53375pt x 93.3
4875pt>
File: ./images/118be108.pdf Graphic file (type pdf)
<use ./images/118be108.pdf> [130 <./images/117aa104.pdf> <./images/117bb105.pdf
> <./images/118aa107.pdf>] <./images/119aa120.pdf, id=1966, 108.405pt x 74.2775
pt>
File: ./images/119aa120.pdf Graphic file (type pdf)
<use ./images/119aa120.pdf> <./images/119bb121.pdf, id=1967, 69.25874pt x 50.18
75pt>
File: ./images/119bb121.pdf Graphic file (type pdf)
<use ./images/119bb121.pdf> [131 <./images/118be108.pdf>] <./images/119cc122.pd
f, id=1977, 75.28125pt x 52.195pt>
File: ./images/119cc122.pdf Graphic file (type pdf)
<use ./images/119cc122.pdf> <./images/119dd123.pdf, id=1978, 105.39375pt x 78.2
925pt>
File: ./images/119dd123.pdf Graphic file (type pdf)
<use ./images/119dd123.pdf> <./images/119ee125.pdf, id=1979, 74.2775pt x 74.277
5pt>
File: ./images/119ee125.pdf Graphic file (type pdf)
<use ./images/119ee125.pdf> <./images/119ff126.pdf, id=1980, 92.345pt x 64.24pt
>
File: ./images/119ff126.pdf Graphic file (type pdf)
<use ./images/119ff126.pdf> [132 <./images/119aa120.pdf> <./images/119bb121.pdf
> <./images/119cc122.pdf> <./images/119dd123.pdf>] <./images/120aa130.pdf, id=2
008, 80.3pt x 80.3pt>
File: ./images/120aa130.pdf Graphic file (type pdf)
<use ./images/120aa130.pdf> <./images/120bb131.pdf, id=2009, 64.24pt x 73.27374
pt>
File: ./images/120bb131.pdf Graphic file (type pdf)
<use ./images/120bb131.pdf> [133 <./images/119ee125.pdf> <./images/119ff126.pdf
>] <./images/120cd132.pdf, id=2024, 307.1475pt x 130.4875pt>
File: ./images/120cd132.pdf Graphic file (type pdf)
<use ./images/120cd132.pdf> [134 <./images/120aa130.pdf> <./images/120bb131.pdf
>] <./images/121aa300.pdf, id=2042, 206.7725pt x 123.46124pt>
File: ./images/121aa300.pdf Graphic file (type pdf)
<use ./images/121aa300.pdf> [135 <./images/120cd132.pdf>] <./images/122ab301.pd
f, id=2053, 340.27126pt x 128.48pt>
File: ./images/122ab301.pdf Graphic file (type pdf)
<use ./images/122ab301.pdf> [136 <./images/121aa300.pdf>] <./images/123aa302.pd
f, id=2065, 160.6pt x 142.5325pt>
File: ./images/123aa302.pdf Graphic file (type pdf)
<use ./images/123aa302.pdf> [137 <./images/122ab301.pdf>] <./images/123bb303.pd
f, id=2079, 150.5625pt x 131.49126pt>
File: ./images/123bb303.pdf Graphic file (type pdf)
<use ./images/123bb303.pdf> [138 <./images/123aa302.pdf>] <./images/124aa304.pd
f, id=2089, 152.57pt x 178.6675pt>
File: ./images/124aa304.pdf Graphic file (type pdf)
<use ./images/124aa304.pdf> [139 <./images/123bb303.pdf>] <./images/125ab305.pd
f, id=2099, 292.09125pt x 90.3375pt>
File: ./images/125ab305.pdf Graphic file (type pdf)
<use ./images/125ab305.pdf> [140 <./images/124aa304.pdf>] <./images/125cc306.pd
f, id=2111, 206.7725pt x 127.47626pt>
File: ./images/125cc306.pdf Graphic file (type pdf)
<use ./images/125cc306.pdf> [141 <./images/125ab305.pdf>] <./images/126aa307.pd
f, id=2122, 273.02pt x 113.42375pt>
File: ./images/126aa307.pdf Graphic file (type pdf)
<use ./images/126aa307.pdf> [142 <./images/125cc306.pdf>] <./images/127ab308.pd
f, id=2132, 246.9225pt x 123.46124pt>
File: ./images/127ab308.pdf Graphic file (type pdf)
<use ./images/127ab308.pdf> [143 <./images/126aa307.pdf>] <./images/127cd309.pd
f, id=2142, 355.3275pt x 131.49126pt>
File: ./images/127cd309.pdf Graphic file (type pdf)
<use ./images/127cd309.pdf> [144 <./images/127ab308.pdf>] <./images/128ab310.pd
f, id=2154, 377.41pt x 98.3675pt>
File: ./images/128ab310.pdf Graphic file (type pdf)
<use ./images/128ab310.pdf> [145 <./images/127cd309.pdf>] <./images/129ab311.pd
f, id=2166, 305.14pt x 111.41624pt>
File: ./images/129ab311.pdf Graphic file (type pdf)
<use ./images/129ab311.pdf> [146 <./images/128ab310.pdf>] <./images/129cc311.pd
f, id=2178, 148.555pt x 86.3225pt>
File: ./images/129cc311.pdf Graphic file (type pdf)
<use ./images/129cc311.pdf> <./images/129dd311.pdf, id=2179, 155.58125pt x 90.3
375pt>
File: ./images/129dd311.pdf Graphic file (type pdf)
<use ./images/129dd311.pdf> <./images/130aa311.pdf, id=2180, 101.37875pt x 54.2
025pt>
File: ./images/130aa311.pdf Graphic file (type pdf)
<use ./images/130aa311.pdf> [147 <./images/129ab311.pdf> <./images/129cc311.pdf
> <./images/129dd311.pdf>] <./images/130bb311.pdf, id=2208, 113.42375pt x 54.20
25pt>
File: ./images/130bb311.pdf Graphic file (type pdf)
<use ./images/130bb311.pdf> <./images/130cc311.pdf, id=2209, 88.33pt x 53.19875
pt>
File: ./images/130cc311.pdf Graphic file (type pdf)
<use ./images/130cc311.pdf> <./images/130dd311.pdf, id=2210, 112.42pt x 51.1912
5pt>
File: ./images/130dd311.pdf Graphic file (type pdf)
<use ./images/130dd311.pdf> <./images/131aa312.pdf, id=2211, 274.02374pt x 93.3
4875pt>
File: ./images/131aa312.pdf Graphic file (type pdf)
<use ./images/131aa312.pdf> [148 <./images/130aa311.pdf> <./images/130bb311.pdf
> <./images/130cc311.pdf> <./images/130dd311.pdf>] <./images/131bb313.pdf, id=2
245, 352.31625pt x 98.3675pt>
File: ./images/131bb313.pdf Graphic file (type pdf)
<use ./images/131bb313.pdf> [149 <./images/131aa312.pdf>] <./images/132aa314.pd
f, id=2255, 117.43875pt x 120.45pt>
File: ./images/132aa314.pdf Graphic file (type pdf)
<use ./images/132aa314.pdf> [150 <./images/131bb313.pdf>] <./images/133aa315.pd
f, id=2265, 285.065pt x 86.3225pt>
File: ./images/133aa315.pdf Graphic file (type pdf)
<use ./images/133aa315.pdf> [151 <./images/132aa314.pdf>] <./images/133bb316.pd
f, id=2275, 157.58875pt x 147.55125pt>
File: ./images/133bb316.pdf Graphic file (type pdf)
<use ./images/133bb316.pdf> [152 <./images/133aa315.pdf>] <./images/134aa317.pd
f, id=2285, 258.9675pt x 93.34875pt>
File: ./images/134aa317.pdf Graphic file (type pdf)
<use ./images/134aa317.pdf> [153 <./images/133bb316.pdf>] <./images/135aa318.pd
f, id=2296, 137.51375pt x 100.375pt>
File: ./images/135aa318.pdf Graphic file (type pdf)
<use ./images/135aa318.pdf> [154 <./images/134aa317.pdf>] [155 <./images/135aa3
18.pdf>] <./images/137aa147.pdf, id=2315, 119.44624pt x 91.34125pt>
File: ./images/137aa147.pdf Graphic file (type pdf)
<use ./images/137aa147.pdf> [156] <./images/137bb148.pdf, id=2321, 210.7875pt x
 78.2925pt>
File: ./images/137bb148.pdf Graphic file (type pdf)
<use ./images/137bb148.pdf> [157 <./images/137aa147.pdf>] [158 <./images/137bb1
48.pdf>] <./images/138aa158.pdf, id=2340, 50.1875pt x 52.195pt>
File: ./images/138aa158.pdf Graphic file (type pdf)
<use ./images/138aa158.pdf> <./images/138bb165.pdf, id=2341, 71.26625pt x 55.20
625pt>
File: ./images/138bb165.pdf Graphic file (type pdf)
<use ./images/138bb165.pdf> [159 <./images/138aa158.pdf>] <./images/139aa167.pd
f, id=2355, 80.3pt x 44.165pt>
File: ./images/139aa167.pdf Graphic file (type pdf)
<use ./images/139aa167.pdf> <./images/139bb168.pdf, id=2356, 60.225pt x 49.1837
5pt>
File: ./images/139bb168.pdf Graphic file (type pdf)
<use ./images/139bb168.pdf> <./images/139cc169.pdf, id=2357, 54.2025pt x 42.157
5pt>
File: ./images/139cc169.pdf Graphic file (type pdf)
<use ./images/139cc169.pdf> [160 <./images/138bb165.pdf> <./images/139aa167.pdf
> <./images/139bb168.pdf> <./images/139cc169.pdf>] <./images/139dd174.pdf, id=2
382, 138.5175pt x 62.2325pt>
File: ./images/139dd174.pdf Graphic file (type pdf)
<use ./images/139dd174.pdf> <./images/139ee175.pdf, id=2383, 103.38625pt x 58.2
175pt>
File: ./images/139ee175.pdf Graphic file (type pdf)
<use ./images/139ee175.pdf> <./images/139gg176.pdf, id=2384, 68.255pt x 52.195p
t>
File: ./images/139gg176.pdf Graphic file (type pdf)
<use ./images/139gg176.pdf> <./images/139ff177.pdf, id=2385, 97.36375pt x 91.34
125pt>
File: ./images/139ff177.pdf Graphic file (type pdf)
<use ./images/139ff177.pdf> <./images/140aa178.pdf, id=2386, 64.24pt x 57.21375
pt>
File: ./images/140aa178.pdf Graphic file (type pdf)
<use ./images/140aa178.pdf> <./images/140bb182.pdf, id=2387, 100.375pt x 54.202
5pt>
File: ./images/140bb182.pdf Graphic file (type pdf)
<use ./images/140bb182.pdf> [161 <./images/139dd174.pdf> <./images/139ee175.pdf
> <./images/139gg176.pdf> <./images/139ff177.pdf> <./images/140aa178.pdf>] <./i
mages/140cc191.pdf, id=2417, 82.3075pt x 87.32625pt>
File: ./images/140cc191.pdf Graphic file (type pdf)
<use ./images/140cc191.pdf> [162 <./images/140bb182.pdf> <./images/140cc191.pdf
>] <./images/141aa206.pdf, id=2432, 72.27pt x 51.19125pt>
File: ./images/141aa206.pdf Graphic file (type pdf)
<use ./images/141aa206.pdf> <./images/141bb207.pdf, id=2433, 102.3825pt x 121.4
5375pt>
File: ./images/141bb207.pdf Graphic file (type pdf)
<use ./images/141bb207.pdf> [163] [164 <./images/141aa206.pdf> <./images/141bb2
07.pdf>] <./images/142aa223.pdf, id=2452, 59.22125pt x 53.19875pt>
File: ./images/142aa223.pdf Graphic file (type pdf)
<use ./images/142aa223.pdf> [165 <./images/142aa223.pdf>] <./images/143aa241.pd
f, id=2463, 60.225pt x 49.18375pt>
File: ./images/143aa241.pdf Graphic file (type pdf)
<use ./images/143aa241.pdf> <./images/143bb242.pdf, id=2464, 92.345pt x 66.2475
pt>
File: ./images/143bb242.pdf Graphic file (type pdf)
<use ./images/143bb242.pdf> <./images/143cc243.pdf, id=2465, 91.34125pt x 76.28
5pt>
File: ./images/143cc243.pdf Graphic file (type pdf)
<use ./images/143cc243.pdf> <./images/143dd244.pdf, id=2466, 96.36pt x 62.2325p
t>
File: ./images/143dd244.pdf Graphic file (type pdf)
<use ./images/143dd244.pdf> [166 <./images/143aa241.pdf> <./images/143bb242.pdf
> <./images/143cc243.pdf> <./images/143dd244.pdf>] <./images/143ee245.pdf, id=2
491, 238.8925pt x 99.37125pt>
File: ./images/143ee245.pdf Graphic file (type pdf)
<use ./images/143ee245.pdf> [167 <./images/143ee245.pdf>]
 .

Package hyperref Warning: Token not allowed in a PDFDocEncoded string,
(hyperref)                removing `\spacefactor' on input line 7214.


Package hyperref Warning: Token not allowed in a PDFDocEncoded string,
(hyperref)                removing `\@m' on input line 7214.


Package hyperref Warning: Token not allowed in a PDFDocEncoded string,
(hyperref)                removing `\spacefactor' on input line 7214.


Package hyperref Warning: Token not allowed in a PDFDocEncoded string,
(hyperref)                removing `\@m' on input line 7214.

[168

] [169] [170] [171] [172] [173] [174] [175] <./images/151aa342.pdf, id=2537, 35
3.32pt x 103.38625pt>
File: ./images/151aa342.pdf Graphic file (type pdf)
<use ./images/151aa342.pdf> [176] [177 <./images/151aa342.pdf>] <./images/152aa
344.pdf, id=2552, 90.3375pt x 75.28125pt>
File: ./images/152aa344.pdf Graphic file (type pdf)
<use ./images/152aa344.pdf> [178 <./images/152aa344.pdf>] <./images/153aa345.pd
f, id=2562, 197.73875pt x 98.3675pt>
File: ./images/153aa345.pdf Graphic file (type pdf)
<use ./images/153aa345.pdf> [179 <./images/153aa345.pdf>] <./images/154aa346.pd
f, id=2572, 236.885pt x 20.075pt>
File: ./images/154aa346.pdf Graphic file (type pdf)
<use ./images/154aa346.pdf> <./images/154bb346.pdf, id=2573, 221.82875pt x 18.0
675pt>
File: ./images/154bb346.pdf Graphic file (type pdf)
<use ./images/154bb346.pdf> <./images/155aa348.pdf, id=2574, 174.6525pt x 83.31
125pt>
File: ./images/155aa348.pdf Graphic file (type pdf)
<use ./images/155aa348.pdf> [180 <./images/154aa346.pdf> <./images/154bb346.pdf
>] [181 <./images/155aa348.pdf>] <./images/156aa349.pdf, id=2604, 170.6375pt x 
93.34875pt>
File: ./images/156aa349.pdf Graphic file (type pdf)
<use ./images/156aa349.pdf> <./images/157aa351.pdf, id=2607, 193.72375pt x 76.2
85pt>
File: ./images/157aa351.pdf Graphic file (type pdf)
<use ./images/157aa351.pdf> [182 <./images/156aa349.pdf>] <./images/157bb353.pd
f, id=2621, 313.17pt x 36.135pt>
File: ./images/157bb353.pdf Graphic file (type pdf)
<use ./images/157bb353.pdf> <./images/158aa354.pdf, id=2622, 218.8175pt x 88.33
pt>
File: ./images/158aa354.pdf Graphic file (type pdf)
<use ./images/158aa354.pdf> [183 <./images/157aa351.pdf> <./images/157bb353.pdf
>] <./images/159aa357.pdf, id=2646, 218.8175pt x 88.33pt>
File: ./images/159aa357.pdf Graphic file (type pdf)
<use ./images/159aa357.pdf> [184 <./images/158aa354.pdf>] <./images/160aa358.pd
f, id=2659, 218.8175pt x 88.33pt>
File: ./images/160aa358.pdf Graphic file (type pdf)
<use ./images/160aa358.pdf> [185 <./images/159aa357.pdf>] <./images/161aa359.pd
f, id=2672, 283.0575pt x 100.375pt>
File: ./images/161aa359.pdf Graphic file (type pdf)
<use ./images/161aa359.pdf> [186 <./images/160aa358.pdf>] [187 <./images/161aa3
59.pdf>] <./images/162aa361.pdf, id=2697, 282.05376pt x 80.3pt>
File: ./images/162aa361.pdf Graphic file (type pdf)
<use ./images/162aa361.pdf> <./images/163aa362.pdf, id=2698, 275.0275pt x 106.3
975pt>
File: ./images/163aa362.pdf Graphic file (type pdf)
<use ./images/163aa362.pdf> [188 <./images/162aa361.pdf>] <./images/164aa363.pd
f, id=2711, 101.37875pt x 112.42pt>
File: ./images/164aa363.pdf Graphic file (type pdf)
<use ./images/164aa363.pdf> [189 <./images/163aa362.pdf>] <./images/165aa364.pd
f, id=2725, 278.03876pt x 135.50626pt>
File: ./images/165aa364.pdf Graphic file (type pdf)
<use ./images/165aa364.pdf> [190 <./images/164aa363.pdf>] [191 <./images/165aa3
64.pdf>] <./images/166aa365.pdf, id=2744, 278.03876pt x 135.50626pt>
File: ./images/166aa365.pdf Graphic file (type pdf)
<use ./images/166aa365.pdf> [192] [193 <./images/166aa365.pdf>] <./images/167aa
366.pdf, id=2758, 278.03876pt x 135.50626pt>
File: ./images/167aa366.pdf Graphic file (type pdf)
<use ./images/167aa366.pdf> [194 <./images/167aa366.pdf>] <./images/168aa263.pd
f, id=2768, 63.23625pt x 68.255pt>
File: ./images/168aa263.pdf Graphic file (type pdf)
<use ./images/168aa263.pdf> [195] [196 <./images/168aa263.pdf>] <./images/169aa
367.pdf, id=2783, 153.57375pt x 83.31125pt>
File: ./images/169aa367.pdf Graphic file (type pdf)
<use ./images/169aa367.pdf> [197

 <./images/169aa367.pdf>] <./images/170aa370.pdf, id=2799, 109.40875pt x 64.24p
t>
File: ./images/170aa370.pdf Graphic file (type pdf)
<use ./images/170aa370.pdf> [198 <./images/170aa370.pdf>] <./images/171aa371.pd
f, id=2809, 152.57pt x 83.31125pt>
File: ./images/171aa371.pdf Graphic file (type pdf)
<use ./images/171aa371.pdf> <./images/171bb373.pdf, id=2810, 60.225pt x 70.2625
pt>
File: ./images/171bb373.pdf Graphic file (type pdf)
<use ./images/171bb373.pdf> <./images/171cc374.pdf, id=2811, 132.495pt x 69.258
74pt>
File: ./images/171cc374.pdf Graphic file (type pdf)
<use ./images/171cc374.pdf> [199 <./images/171aa371.pdf> <./images/171bb373.pdf
> <./images/171cc374.pdf>] <./images/172aa375.pdf, id=2831, 287.0725pt x 112.42
pt>
File: ./images/172aa375.pdf Graphic file (type pdf)
<use ./images/172aa375.pdf> <./images/173aa376.pdf, id=2832, 155.58125pt x 112.
42pt>
File: ./images/173aa376.pdf Graphic file (type pdf)
<use ./images/173aa376.pdf> [200 <./images/172aa375.pdf>] [201 <./images/173aa3
76.pdf>] <./images/174aa377.pdf, id=2852, 206.7725pt x 75.28125pt>
File: ./images/174aa377.pdf Graphic file (type pdf)
<use ./images/174aa377.pdf> <./images/175aa378.pdf, id=2853, 114.4275pt x 101.3
7875pt>
File: ./images/175aa378.pdf Graphic file (type pdf)
<use ./images/175aa378.pdf> [202 <./images/174aa377.pdf>] [203 <./images/175aa3
78.pdf>] <./images/176aa381.pdf, id=2875, 105.39375pt x 126.4725pt>
File: ./images/176aa381.pdf Graphic file (type pdf)
<use ./images/176aa381.pdf> [204 <./images/176aa381.pdf>] <./images/177aa383.pd
f, id=2888, 111.41624pt x 108.405pt>
File: ./images/177aa383.pdf Graphic file (type pdf)
<use ./images/177aa383.pdf> [205 <./images/177aa383.pdf>] <./images/178aa384.pd
f, id=2901, 111.41624pt x 99.37125pt>
File: ./images/178aa384.pdf Graphic file (type pdf)
<use ./images/178aa384.pdf> [206 <./images/178aa384.pdf>] <./images/179aa270.pd
f, id=2914, 95.35625pt x 64.24pt>
File: ./images/179aa270.pdf Graphic file (type pdf)
<use ./images/179aa270.pdf> [207 <./images/179aa270.pdf>] <./images/180aa280.pd
f, id=2925, 48.18pt x 51.19125pt>
File: ./images/180aa280.pdf Graphic file (type pdf)
<use ./images/180aa280.pdf> <./images/180bb281.pdf, id=2926, 69.25874pt x 50.18
75pt>
File: ./images/180bb281.pdf Graphic file (type pdf)
<use ./images/180bb281.pdf> [208 <./images/180aa280.pdf> <./images/180bb281.pdf
>] [209] <./images/181aa385.pdf, id=2945, 174.6525pt x 107.40125pt>
File: ./images/181aa385.pdf Graphic file (type pdf)
<use ./images/181aa385.pdf> <./images/182aa386.pdf, id=2946, 159.59625pt x 106.
3975pt>
File: ./images/182aa386.pdf Graphic file (type pdf)
<use ./images/182aa386.pdf> [210

 <./images/181aa385.pdf>] <./images/183aa387.pdf, id=2956, 192.72pt x 113.42375
pt>
File: ./images/183aa387.pdf Graphic file (type pdf)
<use ./images/183aa387.pdf> [211 <./images/182aa386.pdf>] <./images/184aa388.pd
f, id=2966, 324.21124pt x 101.37875pt>
File: ./images/184aa388.pdf Graphic file (type pdf)
<use ./images/184aa388.pdf> [212 <./images/183aa387.pdf>] <./images/185aa390.pd
f, id=2976, 336.25626pt x 115.43124pt>
File: ./images/185aa390.pdf Graphic file (type pdf)
<use ./images/185aa390.pdf> [213 <./images/184aa388.pdf>] <./images/186aa391.pd
f, id=2987, 268.00125pt x 125.46875pt>
File: ./images/186aa391.pdf Graphic file (type pdf)
<use ./images/186aa391.pdf> [214 <./images/185aa390.pdf>] [215 <./images/186aa3
91.pdf>] <./images/187aa291.pdf, id=3009, 118.4425pt x 72.27pt>
File: ./images/187aa291.pdf Graphic file (type pdf)
<use ./images/187aa291.pdf> <./images/187bb292.pdf, id=3010, 64.24pt x 56.21pt>
File: ./images/187bb292.pdf Graphic file (type pdf)
<use ./images/187bb292.pdf> <./images/187cc293.pdf, id=3011, 82.3075pt x 74.277
5pt>
File: ./images/187cc293.pdf Graphic file (type pdf)
<use ./images/187cc293.pdf> <./images/187dd294.pdf, id=3012, 76.285pt x 73.2737
4pt>
File: ./images/187dd294.pdf Graphic file (type pdf)
<use ./images/187dd294.pdf> <./images/187ee295.pdf, id=3013, 106.3975pt x 76.28
5pt>
File: ./images/187ee295.pdf Graphic file (type pdf)
<use ./images/187ee295.pdf> <./images/187ff296.pdf, id=3014, 89.33376pt x 79.29
625pt>
File: ./images/187ff296.pdf Graphic file (type pdf)
<use ./images/187ff296.pdf> <./images/187gg297.pdf, id=3015, 71.26625pt x 80.3p
t>
File: ./images/187gg297.pdf Graphic file (type pdf)
<use ./images/187gg297.pdf> <./images/187hh298.pdf, id=3016, 108.405pt x 71.266
25pt>
File: ./images/187hh298.pdf Graphic file (type pdf)
<use ./images/187hh298.pdf> [216 <./images/187aa291.pdf> <./images/187bb292.pdf
> <./images/187cc293.pdf> <./images/187dd294.pdf> <./images/187ee295.pdf> <./im
ages/187ff296.pdf> <./images/187gg297.pdf> <./images/187hh298.pdf>] <./images/1
88aa303.pdf, id=3064, 65.24374pt x 57.21375pt>
File: ./images/188aa303.pdf Graphic file (type pdf)
<use ./images/188aa303.pdf> [217 <./images/188aa303.pdf>] <./images/189aa312.pd
f, id=3074, 71.26625pt x 74.2775pt>
File: ./images/189aa312.pdf Graphic file (type pdf)
<use ./images/189aa312.pdf> [218] [219 <./images/189aa312.pdf>] <./images/189bb
313.pdf, id=3090, 71.26625pt x 61.22874pt>
File: ./images/189bb313.pdf Graphic file (type pdf)
<use ./images/189bb313.pdf> <./images/190aa314.pdf, id=3091, 104.39pt x 97.3637
5pt>
File: ./images/190aa314.pdf Graphic file (type pdf)
<use ./images/190aa314.pdf> [220 <./images/189bb313.pdf>] <./images/190bb315.pd
f, id=3101, 104.39pt x 90.3375pt>
File: ./images/190bb315.pdf Graphic file (type pdf)
<use ./images/190bb315.pdf> [221 <./images/190aa314.pdf>] [222 <./images/190bb3
15.pdf>] [223] [224] [225]
 .

Package hyperref Warning: Token not allowed in a PDFDocEncoded string,
(hyperref)                removing `\spacefactor' on input line 9509.


Package hyperref Warning: Token not allowed in a PDFDocEncoded string,
(hyperref)                removing `\@m' on input line 9509.

<./images/193aa395.pdf, id=3133, 357.335pt x 71.26625pt>
File: ./images/193aa395.pdf Graphic file (type pdf)
<use ./images/193aa395.pdf> [226

 <./images/193aa395.pdf>] <./images/194aa395.pdf, id=3143, 290.08376pt x 67.251
25pt>
File: ./images/194aa395.pdf Graphic file (type pdf)
<use ./images/194aa395.pdf> [227 <./images/194aa395.pdf>] <./images/195aa397.pd
f, id=3153, 341.275pt x 77.28876pt>
File: ./images/195aa397.pdf Graphic file (type pdf)
<use ./images/195aa397.pdf> <./images/196aa398.pdf, id=3154, 180.675pt x 90.337
5pt>
File: ./images/196aa398.pdf Graphic file (type pdf)
<use ./images/196aa398.pdf> [228 <./images/195aa397.pdf>] <./images/196bb399.pd
f, id=3167, 136.51pt x 80.3pt>
File: ./images/196bb399.pdf Graphic file (type pdf)
<use ./images/196bb399.pdf> <./images/197aa400.pdf, id=3168, 365.365pt x 75.281
25pt>
File: ./images/197aa400.pdf Graphic file (type pdf)
<use ./images/197aa400.pdf> [229 <./images/196aa398.pdf> <./images/196bb399.pdf
>] <./images/198aa403.pdf, id=3183, 170.6375pt x 81.30376pt>
File: ./images/198aa403.pdf Graphic file (type pdf)
<use ./images/198aa403.pdf> [230 <./images/197aa400.pdf>] <./images/199aa407.pd
f, id=3193, 177.66376pt x 107.40125pt>
File: ./images/199aa407.pdf Graphic file (type pdf)
<use ./images/199aa407.pdf> [231 <./images/198aa403.pdf>] <./images/199bb409.pd
f, id=3204, 172.645pt x 105.39375pt>
File: ./images/199bb409.pdf Graphic file (type pdf)
<use ./images/199bb409.pdf> [232 <./images/199aa407.pdf>] <./images/200aa410.pd
f, id=3217, 129.48375pt x 132.495pt>
File: ./images/200aa410.pdf Graphic file (type pdf)
<use ./images/200aa410.pdf> [233 <./images/199bb409.pdf>] [234 <./images/200aa4
10.pdf>] <./images/201ab411.pdf, id=3236, 150.5625pt x 147.55125pt>
File: ./images/201ab411.pdf Graphic file (type pdf)
<use ./images/201ab411.pdf> [235

 <./images/201ab411.pdf>] <./images/202ab412.pdf, id=3249, 140.525pt x 240.9pt>
File: ./images/202ab412.pdf Graphic file (type pdf)
<use ./images/202ab412.pdf> [236 <./images/202ab412.pdf>] <./images/203aa415.pd
f, id=3262, 126.4725pt x 170.6375pt>
File: ./images/203aa415.pdf Graphic file (type pdf)
<use ./images/203aa415.pdf> [237] [238 <./images/203aa415.pdf>] <./images/204aa
356.pdf, id=3277, 81.30376pt x 79.29625pt>
File: ./images/204aa356.pdf Graphic file (type pdf)
<use ./images/204aa356.pdf> <./images/204bb357.pdf, id=3278, 81.30376pt x 106.3
975pt>
File: ./images/204bb357.pdf Graphic file (type pdf)
<use ./images/204bb357.pdf> <./images/204cc358.pdf, id=3279, 80.3pt x 95.35625p
t>
File: ./images/204cc358.pdf Graphic file (type pdf)
<use ./images/204cc358.pdf> [239 <./images/204aa356.pdf> <./images/204bb357.pdf
> <./images/204cc358.pdf>] [240] [241] <./images/206aa417.pdf, id=3307, 317.185
pt x 77.28876pt>
File: ./images/206aa417.pdf Graphic file (type pdf)
<use ./images/206aa417.pdf> <./images/207aa418.pdf, id=3308, 285.065pt x 74.277
5pt>
File: ./images/207aa418.pdf Graphic file (type pdf)
<use ./images/207aa418.pdf> [242

 <./images/206aa417.pdf>] [243 <./images/207aa418.pdf>] <./images/208aa419.pdf,
 id=3334, 380.42125pt x 110.4125pt>
File: ./images/208aa419.pdf Graphic file (type pdf)
<use ./images/208aa419.pdf> [244 <./images/208aa419.pdf>] <./images/209aa420.pd
f, id=3347, 188.705pt x 112.42pt>
File: ./images/209aa420.pdf Graphic file (type pdf)
<use ./images/209aa420.pdf> <./images/210aa421.pdf, id=3348, 385.44pt x 82.3075
pt>
File: ./images/210aa421.pdf Graphic file (type pdf)
<use ./images/210aa421.pdf> [245 <./images/209aa420.pdf>] <./images/211aa424.pd
f, id=3358, 217.81375pt x 73.27374pt>
File: ./images/211aa424.pdf Graphic file (type pdf)
<use ./images/211aa424.pdf> [246 <./images/210aa421.pdf>] <./images/212aa425.pd
f, id=3368, 355.3275pt x 116.435pt>
File: ./images/212aa425.pdf Graphic file (type pdf)
<use ./images/212aa425.pdf> [247 <./images/211aa424.pdf>] <./images/213aa426.pd
f, id=3378, 293.095pt x 87.32625pt>
File: ./images/213aa426.pdf Graphic file (type pdf)
<use ./images/213aa426.pdf> [248 <./images/212aa425.pdf>] <./images/214aa427.pd
f, id=3391, 301.125pt x 123.46124pt>
File: ./images/214aa427.pdf Graphic file (type pdf)
<use ./images/214aa427.pdf> [249 <./images/213aa426.pdf>] [250 <./images/214aa4
27.pdf>] <./images/215aa428.pdf, id=3414, 270.00874pt x 87.32625pt>
File: ./images/215aa428.pdf Graphic file (type pdf)
<use ./images/215aa428.pdf> [251 <./images/215aa428.pdf>] [252] <./images/217aa
404.pdf, id=3431, 80.3pt x 86.3225pt>
File: ./images/217aa404.pdf Graphic file (type pdf)
<use ./images/217aa404.pdf> <./images/217bb405.pdf, id=3432, 95.35625pt x 66.24
75pt>
File: ./images/217bb405.pdf Graphic file (type pdf)
<use ./images/217bb405.pdf> <./images/217cc406.pdf, id=3433, 113.42375pt x 99.3
7125pt>
File: ./images/217cc406.pdf Graphic file (type pdf)
<use ./images/217cc406.pdf> [253 <./images/217aa404.pdf> <./images/217bb405.pdf
> <./images/217cc406.pdf>] [254] [255] [256] [257]
 .

Package hyperref Warning: Token not allowed in a PDFDocEncoded string,
(hyperref)                removing `\spacefactor' on input line 10726.


Package hyperref Warning: Token not allowed in a PDFDocEncoded string,
(hyperref)                removing `\@m' on input line 10726.

<./images/220aa430.pdf, id=3473, 109.40875pt x 103.38625pt>
File: ./images/220aa430.pdf Graphic file (type pdf)
<use ./images/220aa430.pdf> <./images/221aa431.pdf, id=3474, 109.40875pt x 105.
39375pt>
File: ./images/221aa431.pdf Graphic file (type pdf)
<use ./images/221aa431.pdf> [258

 <./images/220aa430.pdf>] [259 <./images/221aa431.pdf>] <./images/223aa439.pdf,
 id=3493, 149.55875pt x 141.52875pt>
File: ./images/223aa439.pdf Graphic file (type pdf)
<use ./images/223aa439.pdf> [260] <./images/224aa441.pdf, id=3498, 112.42pt x 1
03.38625pt>
File: ./images/224aa441.pdf Graphic file (type pdf)
<use ./images/224aa441.pdf> <./images/224bb442.pdf, id=3499, 94.3525pt x 94.352
5pt>
File: ./images/224bb442.pdf Graphic file (type pdf)
<use ./images/224bb442.pdf> <./images/224cc443.pdf, id=3500, 96.36pt x 92.345pt
>
File: ./images/224cc443.pdf Graphic file (type pdf)
<use ./images/224cc443.pdf> [261 <./images/223aa439.pdf>] <./images/225aa445.pd
f, id=3511, 193.72375pt x 103.38625pt>
File: ./images/225aa445.pdf Graphic file (type pdf)
<use ./images/225aa445.pdf> [262 <./images/224aa441.pdf> <./images/224bb442.pdf
> <./images/224cc443.pdf>] <./images/226aa447.pdf, id=3538, 193.72375pt x 103.3
8625pt>
File: ./images/226aa447.pdf Graphic file (type pdf)
<use ./images/226aa447.pdf> [263 <./images/225aa445.pdf>] [264 <./images/226aa4
47.pdf>] <./images/227aa449.pdf, id=3563, 80.3pt x 84.315pt>
File: ./images/227aa449.pdf Graphic file (type pdf)
<use ./images/227aa449.pdf> [265 <./images/227aa449.pdf>] <./images/228aa451.pd
f, id=3573, 100.375pt x 66.2475pt>
File: ./images/228aa451.pdf Graphic file (type pdf)
<use ./images/228aa451.pdf> <./images/229aa454.pdf, id=3574, 117.43875pt x 125.
46875pt>
File: ./images/229aa454.pdf Graphic file (type pdf)
<use ./images/229aa454.pdf> [266 <./images/228aa451.pdf>] [267 <./images/229aa4
54.pdf>] <./images/231aa455.pdf, id=3597, 230.8625pt x 118.4425pt>
File: ./images/231aa455.pdf Graphic file (type pdf)
<use ./images/231aa455.pdf> [268] [269 <./images/231aa455.pdf>] <./images/232aa
459.pdf, id=3614, 131.49126pt x 117.43875pt>
File: ./images/232aa459.pdf Graphic file (type pdf)
<use ./images/232aa459.pdf> <./images/233aa461.pdf, id=3615, 107.40125pt x 103.
38625pt>
File: ./images/233aa461.pdf Graphic file (type pdf)
<use ./images/233aa461.pdf> [270 <./images/232aa459.pdf>] [271 <./images/233aa4
61.pdf>] <./images/234aa466.pdf, id=3634, 139.52126pt x 83.31125pt>
File: ./images/234aa466.pdf Graphic file (type pdf)
<use ./images/234aa466.pdf> [272] [273 <./images/234aa466.pdf>] <./images/235aa
467.pdf, id=3652, 99.37125pt x 99.37125pt>
File: ./images/235aa467.pdf Graphic file (type pdf)
<use ./images/235aa467.pdf> <./images/236aa469.pdf, id=3653, 113.42375pt x 99.3
7125pt>
File: ./images/236aa469.pdf Graphic file (type pdf)
<use ./images/236aa469.pdf> [274

 <./images/235aa467.pdf>] <./images/237aa472.pdf, id=3663, 104.39pt x 106.3975p
t>
File: ./images/237aa472.pdf Graphic file (type pdf)
<use ./images/237aa472.pdf> [275 <./images/236aa469.pdf>] [276 <./images/237aa4
72.pdf>] <./images/239aa475.pdf, id=3682, 104.39pt x 106.3975pt>
File: ./images/239aa475.pdf Graphic file (type pdf)
<use ./images/239aa475.pdf> [277] <./images/240aa477.pdf, id=3687, 222.8325pt x
 101.37875pt>
File: ./images/240aa477.pdf Graphic file (type pdf)
<use ./images/240aa477.pdf> [278 <./images/239aa475.pdf>] <./images/241aa478.pd
f, id=3697, 99.37125pt x 112.42pt>
File: ./images/241aa478.pdf Graphic file (type pdf)
<use ./images/241aa478.pdf> [279 <./images/240aa477.pdf>] [280 <./images/241aa4
78.pdf>] <./images/242aa480.pdf, id=3720, 99.37125pt x 99.37125pt>
File: ./images/242aa480.pdf Graphic file (type pdf)
<use ./images/242aa480.pdf> [281 <./images/242aa480.pdf>] <./images/243aa484.pd
f, id=3728, 158.5925pt x 84.315pt>
File: ./images/243aa484.pdf Graphic file (type pdf)
<use ./images/243aa484.pdf> [282 <./images/243aa484.pdf>] <./images/244aa485.pd
f, id=3738, 153.57375pt x 153.57375pt>
File: ./images/244aa485.pdf Graphic file (type pdf)
<use ./images/244aa485.pdf> [283 <./images/244aa485.pdf>] <./images/245aa486.pd
f, id=3750, 228.855pt x 103.38625pt>
File: ./images/245aa486.pdf Graphic file (type pdf)
<use ./images/245aa486.pdf> <./images/246aa487.pdf, id=3751, 229.85875pt x 113.
42375pt>
File: ./images/246aa487.pdf Graphic file (type pdf)
<use ./images/246aa487.pdf> [284 <./images/245aa486.pdf>] <./images/247aa488.pd
f, id=3761, 117.43875pt x 124.465pt>
File: ./images/247aa488.pdf Graphic file (type pdf)
<use ./images/247aa488.pdf> [285 <./images/246aa487.pdf>] <./images/248aa490.pd
f, id=3775, 192.72pt x 84.315pt>
File: ./images/248aa490.pdf Graphic file (type pdf)
<use ./images/248aa490.pdf> [286 <./images/247aa488.pdf>] <./images/249aa491.pd
f, id=3785, 300.12125pt x 86.3225pt>
File: ./images/249aa491.pdf Graphic file (type pdf)
<use ./images/249aa491.pdf> [287 <./images/248aa490.pdf>] [288 <./images/249aa4
91.pdf>] [289] [290] [291] [292] [293] [294] [295] [296] [297] [298] [299] [300
] [301]
 .
[302

] [303] [304]
 .
[305


] [306] [307] [308

] [309



] [310] [311] [312] [313] [314] [315] [316] (./33063-t.aux)

 *File List*
    book.cls    2005/09/16 v1.4f Standard LaTeX document class
    bk12.clo    2005/09/16 v1.4f Standard LaTeX file (size option)
inputenc.sty    2006/05/05 v1.1b Input encoding file
  latin1.def    2006/05/05 v1.1b Input encoding file
  ifthen.sty    2001/05/26 v1.1c Standard LaTeX ifthen package (DPC)
 amsmath.sty    2000/07/18 v2.13 AMS math features
 amstext.sty    2000/06/29 v2.01
  amsgen.sty    1999/11/30 v2.0
  amsbsy.sty    1999/11/29 v1.2d
  amsopn.sty    1999/12/14 v2.01 operator names
 amssymb.sty    2002/01/22 v2.2d
amsfonts.sty    2001/10/25 v2.2f
   alltt.sty    1997/06/16 v2.0g defines alltt environment
footmisc.sty    2005/03/17 v5.3d a miscellany of footnote facilities
indentfirst.sty    1995/11/23 v1.03 Indent first paragraph (DPC)
fancyhdr.sty    
graphicx.sty    1999/02/16 v1.0f Enhanced LaTeX Graphics (DPC,SPQR)
  keyval.sty    1999/03/16 v1.13 key=value parser (DPC)
graphics.sty    2006/02/20 v1.0o Standard LaTeX Graphics (DPC,SPQR)
    trig.sty    1999/03/16 v1.09 sin cos tan (DPC)
graphics.cfg    2007/01/18 v1.5 graphics configuration of teTeX/TeXLive
  pdftex.def    2007/01/08 v0.04d Graphics/color for pdfTeX
enumerate.sty    1999/03/05 v3.00 enumerate extensions (DPC)
supertabular.sty    2004/02/20 v4.1e the supertabular environment
geometry.sty    2002/07/08 v3.2 Page Geometry
geometry.cfg
hyperref.sty    2007/02/07 v6.75r Hypertext links for LaTeX
  pd1enc.def    2007/02/07 v6.75r Hyperref: PDFDocEncoding definition (HO)
hyperref.cfg    2002/06/06 v1.2 hyperref configuration of TeXLive
kvoptions.sty    2006/08/22 v2.4 Connects package keyval with LaTeX options (HO
)
     url.sty    2005/06/27  ver 3.2  Verb mode for urls, etc.
 hpdftex.def    2007/02/07 v6.75r Hyperref driver for pdfTeX
supp-pdf.tex
   color.sty    2005/11/14 v1.0j Standard LaTeX Color (DPC)
   color.cfg    2007/01/18 v1.5 color configuration of teTeX/TeXLive
 nameref.sty    2006/12/27 v2.28 Cross-referencing by name of section
refcount.sty    2006/02/20 v3.0 Data extraction from references (HO)
 33063-t.out
 33063-t.out
    umsa.fd    2002/01/19 v2.2g AMS font definitions
    umsb.fd    2002/01/19 v2.2g AMS font definitions
  omscmr.fd    1999/05/25 v2.5h Standard LaTeX font definitions
./images/woodcutsmall.pdf
./images/012aaZ10.pdf
./images/012bbZ14.pdf
./images/016aaZ37.pdf
./images/018aaZ55.pdf
./images/018bbZ56.pdf
./images/018ccZ57.pdf
./images/019aaZ58.pdf
./images/019bbZ58.pdf
./images/019ccZ62.pdf
./images/020aaZ63.pdf
./images/020bbZ66.pdf
./images/020ccZ68.pdf
./images/020ddZ69.pdf
./images/021aaZ72.pdf
./images/022aaZ75.pdf
./images/022bbZ76.pdf
./images/022ccZ77.pdf
./images/023aaZ79.pdf
./images/023bcZ80.pdf
./images/024aaZ81.pdf
./images/024bbZ83.pdf
./images/025aaZ86.pdf
./images/026aaZ90.pdf
./images/027aaZ93.pdf
./images/028aaZ95.pdf
./images/029aaZ96.pdf
./images/030aaZ97.pdf
./images/031aa100.pdf
./images/032aa101.pdf
./images/033aa104.pdf
./images/034aa107.pdf
./images/034bb109.pdf
./images/035aa110.pdf
./images/036aa111.pdf
./images/037aa112.pdf
./images/038aa115.pdf
./images/039aa117.pdf
./images/039bb118.pdf
./images/039ce119.pdf
./images/039fi120.pdf
./images/041aa129.pdf
./images/042aa138.pdf
./images/043ab139.pdf
./images/044ab143.pdf
./images/045aa145.pdf
./images/046aa147.pdf
./images/047ab150.pdf
./images/048ac151.pdf
./images/049aa152.pdf
./images/050aa153.pdf
./images/051ac154.pdf
./images/052ab155.pdf
./images/054aa160.pdf
./images/055aa162.pdf
./images/056ac166.pdf
./images/056dg170.pdf
./images/057aa174.pdf
./images/057bb176.pdf
./images/058aa178.pdf
./images/058bb181.pdf
./images/059aa182.pdf
./images/060aa183.pdf
./images/061aa184.pdf
./images/062ab185.pdf
./images/063aa187.pdf
./images/064aa188.pdf
./images/064bb190.pdf
./images/065ac192.pdf
./images/066ad203.pdf
./images/067aa205.pdf
./images/068aa207.pdf
./images/069ac208.pdf
./images/069dd211.pdf
./images/070aa212.pdf
./images/071aa213.pdf
./images/074aaZ19.pdf
./images/074bbZ20.pdf
./images/075aaZ21.pdf
./images/075bbZ22.pdf
./images/075ceZ23.pdf
./images/076aaZ24.pdf
./images/076bbZ25.pdf
./images/076ccZ26.pdf
./images/076ddZ27.pdf
./images/077adZ28.pdf
./images/077eeZ34.pdf
./images/077ffZ35.pdf
./images/078aaZ41.pdf
./images/078bbZ43.pdf
./images/078ccZ46.pdf
./images/078ddZ49.pdf
./images/079adZ51.pdf
./images/079eeZ57.pdf
./images/080aaZ59.pdf
./images/080bbZ60.pdf
./images/080ccZ61.pdf
./images/080ddZ62.pdf
./images/080eeZ63.pdf
./images/080ffZ64.pdf
./images/081aaZ65.pdf
./images/081bbZ66.pdf
./images/081ccZ67.pdf
./images/081ddZ68.pdf
./images/081eeZ69.pdf
./images/081ffZ73.pdf
./images/082adZ74.pdf
./images/082ehZ78.pdf
./images/082iiZ82.pdf
./images/083adZ83.pdf
./images/083eeZ88.pdf
./images/083ffZ89.pdf
./images/084aa220.pdf
./images/085ac224.pdf
./images/086aa235.pdf
./images/087ab236.pdf
./images/089ab241.pdf
./images/090ab243.pdf
./images/091aa245.pdf
./images/092aa249.pdf
./images/093aa250.pdf
./images/094aa251.pdf
./images/095aa253.pdf
./images/096ac257.pdf
./images/097aa258.pdf
./images/098aa261.pdf
./images/099aa264.pdf
./images/100aa265.pdf
./images/103aa272.pdf
./images/103bb273.pdf
./images/108aa286.pdf
./images/109ac287.pdf
./images/111ac289.pdf
./images/112ac290.pdf
./images/113aa294.pdf
./images/114aa295.pdf
./images/115ac296.pdf
./images/115dd297.pdf
./images/115ee297.pdf
./images/116ag299.pdf
./images/117aa104.pdf
./images/117bb105.pdf
./images/118aa107.pdf
./images/118be108.pdf
./images/119aa120.pdf
./images/119bb121.pdf
./images/119cc122.pdf
./images/119dd123.pdf
./images/119ee125.pdf
./images/119ff126.pdf
./images/120aa130.pdf
./images/120bb131.pdf
./images/120cd132.pdf
./images/121aa300.pdf
./images/122ab301.pdf
./images/123aa302.pdf
./images/123bb303.pdf
./images/124aa304.pdf
./images/125ab305.pdf
./images/125cc306.pdf
./images/126aa307.pdf
./images/127ab308.pdf
./images/127cd309.pdf
./images/128ab310.pdf
./images/129ab311.pdf
./images/129cc311.pdf
./images/129dd311.pdf
./images/130aa311.pdf
./images/130bb311.pdf
./images/130cc311.pdf
./images/130dd311.pdf
./images/131aa312.pdf
./images/131bb313.pdf
./images/132aa314.pdf
./images/133aa315.pdf
./images/133bb316.pdf
./images/134aa317.pdf
./images/135aa318.pdf
./images/137aa147.pdf
./images/137bb148.pdf
./images/138aa158.pdf
./images/138bb165.pdf
./images/139aa167.pdf
./images/139bb168.pdf
./images/139cc169.pdf
./images/139dd174.pdf
./images/139ee175.pdf
./images/139gg176.pdf
./images/139ff177.pdf
./images/140aa178.pdf
./images/140bb182.pdf
./images/140cc191.pdf
./images/141aa206.pdf
./images/141bb207.pdf
./images/142aa223.pdf
./images/143aa241.pdf
./images/143bb242.pdf
./images/143cc243.pdf
./images/143dd244.pdf
./images/143ee245.pdf
./images/151aa342.pdf
./images/152aa344.pdf
./images/153aa345.pdf
./images/154aa346.pdf
./images/154bb346.pdf
./images/155aa348.pdf
./images/156aa349.pdf
./images/157aa351.pdf
./images/157bb353.pdf
./images/158aa354.pdf
./images/159aa357.pdf
./images/160aa358.pdf
./images/161aa359.pdf
./images/162aa361.pdf
./images/163aa362.pdf
./images/164aa363.pdf
./images/165aa364.pdf
./images/166aa365.pdf
./images/167aa366.pdf
./images/168aa263.pdf
./images/169aa367.pdf
./images/170aa370.pdf
./images/171aa371.pdf
./images/171bb373.pdf
./images/171cc374.pdf
./images/172aa375.pdf
./images/173aa376.pdf
./images/174aa377.pdf
./images/175aa378.pdf
./images/176aa381.pdf
./images/177aa383.pdf
./images/178aa384.pdf
./images/179aa270.pdf
./images/180aa280.pdf
./images/180bb281.pdf
./images/181aa385.pdf
./images/182aa386.pdf
./images/183aa387.pdf
./images/184aa388.pdf
./images/185aa390.pdf
./images/186aa391.pdf
./images/187aa291.pdf
./images/187bb292.pdf
./images/187cc293.pdf
./images/187dd294.pdf
./images/187ee295.pdf
./images/187ff296.pdf
./images/187gg297.pdf
./images/187hh298.pdf
./images/188aa303.pdf
./images/189aa312.pdf
./images/189bb313.pdf
./images/190aa314.pdf
./images/190bb315.pdf
./images/193aa395.pdf
./images/194aa395.pdf
./images/195aa397.pdf
./images/196aa398.pdf
./images/196bb399.pdf
./images/197aa400.pdf
./images/198aa403.pdf
./images/199aa407.pdf
./images/199bb409.pdf
./images/200aa410.pdf
./images/201ab411.pdf
./images/202ab412.pdf
./images/203aa415.pdf
./images/204aa356.pdf
./images/204bb357.pdf
./images/204cc358.pdf
./images/206aa417.pdf
./images/207aa418.pdf
./images/208aa419.pdf
./images/209aa420.pdf
./images/210aa421.pdf
./images/211aa424.pdf
./images/212aa425.pdf
./images/213aa426.pdf
./images/214aa427.pdf
./images/215aa428.pdf
./images/217aa404.pdf
./images/217bb405.pdf
./images/217cc406.pdf
./images/220aa430.pdf
./images/221aa431.pdf
./images/223aa439.pdf
./images/224aa441.pdf
./images/224bb442.pdf
./images/224cc443.pdf
./images/225aa445.pdf
./images/226aa447.pdf
./images/227aa449.pdf
./images/228aa451.pdf
./images/229aa454.pdf
./images/231aa455.pdf
./images/232aa459.pdf
./images/233aa461.pdf
./images/234aa466.pdf
./images/235aa467.pdf
./images/236aa469.pdf
./images/237aa472.pdf
./images/239aa475.pdf
./images/240aa477.pdf
./images/241aa478.pdf
./images/242aa480.pdf
./images/243aa484.pdf
./images/244aa485.pdf
./images/245aa486.pdf
./images/246aa487.pdf
./images/247aa488.pdf
./images/248aa490.pdf
./images/249aa491.pdf
 ***********

 ) 
Here is how much of TeX's memory you used:
 7259 strings out of 94074
 109629 string characters out of 1165154
 167599 words of memory out of 1500000
 9671 multiletter control sequences out of 10000+50000
 19537 words of font info for 73 fonts, out of 1200000 for 2000
 645 hyphenation exceptions out of 8191
 29i,14n,43p,264b,486s stack positions out of 5000i,500n,6000p,200000b,5000s
</usr/share/texmf-texlive/fonts/type1/bluesky/cm/cmbsy10.pfb></usr/share/texm
f-texlive/fonts/type1/bluesky/ams/cmbsy7.pfb></usr/share/texmf-texlive/fonts/ty
pe1/bluesky/cm/cmbx10.pfb></usr/share/texmf-texlive/fonts/type1/bluesky/cm/cmbx
12.pfb></usr/share/texmf-texlive/fonts/type1/bluesky/cm/cmcsc10.pfb></usr/share
/texmf-texlive/fonts/type1/bluesky/cm/cmex10.pfb></usr/share/texmf-texlive/font
s/type1/bluesky/cm/cmmi10.pfb></usr/share/texmf-texlive/fonts/type1/bluesky/cm/
cmmi12.pfb></usr/share/texmf-texlive/fonts/type1/bluesky/cm/cmmi6.pfb></usr/sha
re/texmf-texlive/fonts/type1/bluesky/cm/cmmi8.pfb></usr/share/texmf-texlive/fon
ts/type1/bluesky/cm/cmmib10.pfb></usr/share/texmf-texlive/fonts/type1/bluesky/a
ms/cmmib7.pfb></usr/share/texmf-texlive/fonts/type1/bluesky/cm/cmr10.pfb></usr/
share/texmf-texlive/fonts/type1/bluesky/cm/cmr12.pfb></usr/share/texmf-texlive/
fonts/type1/bluesky/cm/cmr17.pfb></usr/share/texmf-texlive/fonts/type1/bluesky/
cm/cmr6.pfb></usr/share/texmf-texlive/fonts/type1/bluesky/cm/cmr8.pfb></usr/sha
re/texmf-texlive/fonts/type1/bluesky/cm/cmsy10.pfb></usr/share/texmf-texlive/fo
nts/type1/bluesky/cm/cmsy6.pfb></usr/share/texmf-texlive/fonts/type1/bluesky/cm
/cmsy7.pfb></usr/share/texmf-texlive/fonts/type1/bluesky/cm/cmsy8.pfb></usr/sha
re/texmf-texlive/fonts/type1/bluesky/cm/cmti10.pfb></usr/share/texmf-texlive/fo
nts/type1/bluesky/cm/cmti12.pfb></usr/share/texmf-texlive/fonts/type1/bluesky/c
m/cmtt9.pfb></usr/share/texmf-texlive/fonts/type1/bluesky/latex-fonts/line10.pf
b></usr/share/texmf-texlive/fonts/type1/bluesky/ams/msam10.pfb>
Output written on 33063-t.pdf (326 pages, 3009904 bytes).
PDF statistics:
 4377 PDF objects out of 5155 (max. 8388607)
 474 named destinations out of 1000 (max. 131072)
 2001 words of extra memory for PDF output out of 10000 (max. 10000000)

