\chapter{Triangles}   
   
\section{Congruent triangles}   
\begin{Def}   
Triangles $ABC$ and $DEF$ are said to be \nuvo{congruent} \pron{£hk£ongr£Rnt} iff%correction 2000-09-27   
$$AB= DE\text{, } BC= EF\text{, and } CA = FD.$$ 
   
The sides $[AB]$ and $[DE]$ are said to be \nuvo{corresponding} \pron{£bk£or£is£hp£ond£i£n} as are corresponding the 
vertices $A$ and $D$ and the angles $\angle BAC$ and $\angle EDF$. 
\end{Def}   
This definition will be referred to as the SSS rule (SSS for Side-Side-Side).\index{SSS rule} 
   
On the following figure, triangles $ABC$, $DEF$, and $GHI$ are congruent. 
   
\begin{fgre}
  \IG{trian01}
\end{fgre}
   
\begin{prop}   
  If $ABC$ and $DEF$ are congruent then their corresponding angles are equal: $\angle BAC=\angle EDF$, $\angle 
  ABC=\angle DEF$, and $\angle ACB=\angle DFE$. 
\end{prop}   
This property is known as the AAA rule (AAA for Angle-Angle-Angle). We can then sum up the preceding property whith: 
\framebox{SSS $\alors$ AAA}.\index{AAA rule} 
   
\begin{prop}[SAS rule]   
  If two sides and the angle between them (the included angle) are the same in both triangles then these triangles are 
  congruent. 
\end{prop}\index{SAS rule}   
The converse of that property is obviously true. So \framebox{SSS \Iff SAS}. 
   
\begin{prop}[ASA rule]   
  If two angles and the side between them are the same in both triangles then these triangles are congruent. 
\end{prop}\index{ASA rule}   
The converse of that property is obviously true. So \framebox{SSS \Iff ASA}. 
   
\begin{prop}   
  A triangle $ABC$ and its image $A'B'C'$ under an isometry \pron{£p£hs£om£itr£i} (translation, reflection, rotation)
  are congruent.
\end{prop}   
   
\section{Similar triangles}   
   
\begin{Def}   
  Triangles $ABC$ and $DEF$ are said to be \nuvo{similar} \pron{£hs£im£il£e£Z} iff 
  $$\angle BAC=\angle EDF\text{, } \angle ABC=\angle DEF\text{, and } \angle ACB=\angle DFE.$$ 
\end{Def}   
   
\begin{prop}   
  If $ABC$ and $DEF$ are similar, then their sides are in the same ratio \idest\  
  $$\dfrac{AB}{DE}= \dfrac{BC}{EF} = \dfrac{AC}{DF}.$$ 
\end{prop}   
The sides $[AB]$, $[BC]$, and $[CA]$ are said to be \nuvo{homologous} \pron{h£o£hm£ol£eg£es} to the sides $[DE]$, 
$[EF]$, and $[FD]$ respectively. 
   
\begin{fgre}
  \IG{trian02}
\end{fgre}

\begin{prop}   
  Let $ABC$ be any triangle. Let $M$ and $N$ be points on $(AB)$ and $(AC)$ respectively such that $(MN)$ is parallel to 
  $(AB)$. Then $ABC$ and $AMN$ are similar. % correction 2002-2003
\end{prop}   
   
\begin{prop}   
  Let $ABC$ and $DEF$ be triangles such that % correction 2002-2003
  $$\dfrac{AB}{DE}= \dfrac{BC}{EF} = \dfrac{AC}{DF}.$$ 
  Then $ABC$ and $DEF$ are similar. 
\end{prop}   
   
\begin{Def}   
  Let $ABC$ and $A'B'C'$ be similar triangles such that $$\dfrac{A'B'}{AB}= \dfrac{B'C'}{BC} = \dfrac{A'C'}{AC} = k.$$ 
% correction 2002-2003
\begin{itemize}   
\item If $k =1$ then the triangles are congruent. 
\item If $k \neq 1$ then $A'B'C'$ is said to be an \nuvo{enlargement} of $ABC$ by \nuvo{scale factor} $k$. 
\end{itemize}\end{Def}   
\begin{prop}   
  If $A'B'C'$ is an enlargement of $ABC$ by scale factor $k$ then the area of $A'B'C'$ is $k^{2}$ times the area of 
  $ABC$.
\end{prop} 

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$$ \includegraphics[keepaspectratio=true, totalheight=8cm]{figures/Vignette02}$$

\vspace*{\stretch{1}}   

\endinput 
   
\section{Particular figures in triangles}   
   
A triangle is a three-sided plane figure. Three non-collinear points determine a triangle. If $A$, $B$ and $C$ are not 
collinear, they are the \emph{vertices} \pron{£hv£Et£is£Iz} (plural of `vertex' \pron{£hv£Eteks}) of the triangle $ABC$. 
Triangle $ABC$ can be denoted by $\triangle ABC$.\index{vertices|see{vertex}}\index{vertex}\index{vertex!of a figure} 
   
Triangles can be classified by the relative \pron{£hrel£et£iv} lengths of their sides. A \emph{scalene} \pron{£hsk£ql£In} 
triangle has no sides of equal length; an \emph{isosceles} \pron{£p£hs£os£il£Iz} triangle has at least two equal sides; 
an \emph{equilateral} \pron{£b£Ikw£i£hl£at(£e)r(£e)l} triangle has three equal sides. 
A \emph{right} or \emph{right-angled} triangle has one angle of 90\degre.%   
\index{scalene}\index{isosceles}\index{equilateral}\index{right}\index{right-angles}\index{triangle} 
   
Pythagoras'\,\footnote{Pythagoras (6th-c \textsc{bc}) Philosopher and mathematician, born in Samos \pron{£hs£qm£os}, 
  Greece.  He settled at Crotona, Magna Graecia \pron{£bm£agn£e £hgr£Is£Q} (c.  \oldstylenums{530} \textsc{bc}) where he 
  founded a moral and religious school. He eventually fled from ther because of persecution, settling at Megapontum in 
  Lucania \pron{l£U£hk£qn£Q}. The famous theorem attributed to him was probably developed later by members of the 
  Pythagorean school, which is best known for its studies of the relations between numbers.}  theorem \pron{p£p£h£f£ag£er£as 
£h£f£Qr£em} (also known as Pythagorean \pron{p£p£b£f£ag£e£hr£Qn} theorem) states that if $\triangle ABC$ (triangle 
$ABC$) is a right triangle with \emph{legs} 
of lengths $a$ and $b$, and \emph{hypotenuse} \pron{h£p£hp£ot£enj£Uz} of length $c$, then $\dps c^2= a^2+b^2$.%   
\index{hypotenuse}\index{leg}\index{Pythagoras}\index{pythagorean} 
   
An \emph{altitude} \pron{£h£alt£itj£Ud} is a segment between a vertex and the opposite \pron{£h£op£ez£it} side or 
\emph{base} \pron{b£qs}, that is perpendicular to the side. It is also the length of that segment.  The common point of 
the three altitudes is the \emph{orthocentre} \pron{£b£O£f£w£hsent£e£Z}. 
   
A \emph{median} \pron{£hm£Idj£en} is a line joining a vertex of a triangle to the midpoint of the opposite side.  All 
three such lines coincide in the \emph{centroid} \pron{£hsentr£rd}. Archimedes \pron{£b£Ak£i£hm£Id£Iz} (c. 
\detantatant{287}{212} \textsc{b.c.}) obtained it as the center of gravity \pron{£hgr£av£it£i} of a triangular 
\pron{tr£p£h£a£ngj£ul£e£Z} plate of uniform density \pron{£hdens£et£i}.  The centroid of the triangle is also the point 
of trisection \pron{tr£p£hsek£sn} of the three medians of the triangle. 
   
Line \Li is called the \nuvo{perpendicular bisector} \pron{£bp£Epen£hd£ikj£ul£e£Z b£p£hsekt£e£Z} or \nuvo{mid-perpendicular} 
\pron{m£id£bp£Epen£hd£ikj£ul£e£Z} of $[AB]$ if and only if \Li is perpendicular to the line $(AB)$ and intersects it in 
the midpoint of $[AB]$. 
   
The mid-perpendiculars of the sides of a triangle intersect at the \emph{circumcentre} \pron{£bs£Ek£em£hsent£e£Z} which is 
the \emph{centre} \pron{£hsent£e£Z} of the \emph{circumcircle} \pron{£hs£Ek£em£bs£Ekl} or \emph{circumscribed} 
\pron{£hs£Ek£em£bskr£pbd} \emph{circle}.%   
\index{altitude}\index{base}\index{median}\index{centroid}\index{orthocentre}\index{circumcentre}%   
\index{circumcircle}\index{circumscribed} 

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