cong.tex

\chapter{Congruent triangles}\label{chap:cong}

\section{Side-angle-side}

Our next goal is to give conditions that guarantee the congruence of two triangles.

One such condition is given in Axiom~\ref{def:birkhoff-axioms:3}; it states that if two pairs of sides of two triangles are equal, and the included angles are equal up to sign, then the triangles are congruent.
This axiom is also called the {}\emph{side-angle-side congruence condition}, or briefly, \index{SAS congruence condition}\emph{SAS}.

\section{Angle-side-angle}

\begin{thm}[\abs]{ASA condition}\label{thm:ASA}\index{ASA congruence condition}
Assume that 
\begin{align*}
AB&=A'B',
&
\measuredangle A B C &= \pm\measuredangle A' B' C', 
&
\measuredangle C A B&=\pm\measuredangle C' A' B'
\end{align*}
 and $\triangle A' B' C'$ is nondegenerate.
Then 
$$\triangle A B C\cong\triangle A' B' C'.$$

\end{thm}

For degenerate triangles this statement does not hold.
For example, consider one triangle with sides $1$, $4$, $5$ 
and the other with sides $2$, $3$,~$5$.

\parit{Proof.} 
According to Theorem~\ref{thm:signs-of-triug},
either
\[
\measuredangle A B C = \measuredangle A' B' C'
\quad\text{and}\quad
\measuredangle C A B=\measuredangle C' A' B'
\eqlbl{eq:+angles}\]
or
\[
\measuredangle A B C = -\measuredangle A' B' C'
\quad\text{and}\quad
\measuredangle C A B=-\measuredangle C' A' B'.
\eqlbl{eq:-angles}\]
Let us assume that \ref{eq:+angles} holds; 
the case \ref{eq:-angles} is analogous.

\begin{wrapfigure}{o}{26mm}
\vskip-2mm
\centering
\includegraphics{mppics/pic-36}
\end{wrapfigure}

Let $C''$ be the point on the half-line $[A' C')$ such that $A' C''\z=A C$. 

By Axiom~\ref{def:birkhoff-axioms:3}, 
$\triangle A' B' C''\cong \triangle A B C$. 
Applying Axiom~\ref{def:birkhoff-axioms:3} again,
we get that
$$\measuredangle A' B' C'' = \measuredangle A B C=\measuredangle A' B' C'.$$
By Axiom~\ref{def:birkhoff-axioms:2a}, $[B'C')=[B C'')$. 
Hence
$C''$ lies on $(B' C')$ as well as on~$(A' C')$.

Since $\triangle A' B' C'$ is not degenerate, $(A' C')$ is distinct from~$(B' C')$.
Applying Axiom~\ref{def:birkhoff-axioms:1}, we get that $C''=C'$. 

Therefore, 
$\triangle A' B' C'=\triangle A' B' C''\cong\triangle A B C$.
\qeds

\section{Isosceles triangles}

A triangle with two equal sides is called \index{isosceles triangle}\emph{isosceles};
the remaining side is called the \index{base}\emph{base}.


\begin{thm}[\abs]{Theorem}\label{thm:isos}
Assume $\triangle A B C$ is an isosceles triangle with the base $[A B]$. 
Then 
$$\measuredangle A B C\equiv -\measuredangle B A C.$$
Moreover, the converse holds if $\triangle A B C$ is nondegenerate.
\end{thm}

The following proof is due to Pappus of Alexandria.

\begin{wrapfigure}[10]{o}{25mm}
\vskip-4mm
\centering
\includegraphics{mppics/pic-38}
\end{wrapfigure}

\parit{Proof.}
Note that
$$C A = C B,
\quad 
C B=C A,
\quad
\measuredangle A C B \equiv -\measuredangle B C A.$$
By Axiom~\ref{def:birkhoff-axioms:3},
$$\triangle C A B\cong\triangle C B A.$$
Applying the theorem on the signs of angles of triangles (\ref{thm:signs-of-triug}) and Axiom~\ref{def:birkhoff-axioms:3} again,
we get that 
$$\measuredangle B A C
\equiv -\measuredangle A B C.$$

To prove the converse, we assume that
$\measuredangle C A B \z\equiv - \measuredangle C B A$.
By the ASA condition (\ref{thm:ASA}), $\triangle C A B\z\cong\triangle CBA$.
Therefore,~$C A\z=C B$.
\qeds

A triangle with three equal sides is called \index{equilateral triangle}\emph{equilateral}. 

\begin{thm}{Exercise}\label{ex:equilateral}
Let $\triangle ABC$ be an equilateral triangle.
Show that 
\[\measuredangle ABC=\measuredangle BCA=\measuredangle CAB.\]

\end{thm}


\section{Side-side-side}

\begin{thm}[\abs]{SSS condition}\label{thm:SSS}\index{SSS congruence condition}
$\triangle A B C\cong\triangle A' B' C'$ if 
$$A' B'=A B,
\quad 
B' C'=B C,
\quad 
\text{and}
\quad 
C' A'=C A.$$

\end{thm}

Note that this condition is valid for degenerate triangles as well.

\parit{Proof.} 
Choose $C''$ so that $A' C''= A' C'$ and $\measuredangle B' A' C''= \measuredangle B A C$.
According to Axiom~\ref{def:birkhoff-axioms:3},
$$\triangle A' B' C''\cong\triangle A B C.$$

It will suffice to
prove that 
$$\triangle A' B' C'\cong\triangle A' B' C''.\eqlbl{eq:A'B'C'simA'B'C''}$$
The condition \ref{eq:A'B'C'simA'B'C''} holds trivially if $C''\z=C'$.
Thus, it remains to consider the case $C''\z\ne C'$.

\begin{wrapfigure}{o}{30mm}
\centering
\includegraphics{mppics/pic-40}
\end{wrapfigure}

Clearly, the corresponding sides of $\triangle A' B' C'$ and $\triangle A' B' C''$ are equal.
Hence the triangles
$\triangle C' A' C''$ and $\triangle C' B' C''$ are isosceles.
By Theorem~\ref{thm:isos}, we have 
\begin{align*}
 \measuredangle A' C'' C'&\equiv -\measuredangle A' C' C'',
\\
\measuredangle C' C'' B'&\equiv -\measuredangle C'' C' B'.
\end{align*}
Adding them, we get that
$$\measuredangle A' C'' B'
\equiv -\measuredangle A' C' B'.$$
Applying Axiom~\ref{def:birkhoff-axioms:3} again,
we get \ref{eq:A'B'C'simA'B'C''}.
\qeds

\begin{thm}[\abs]{Corollary}\label{cor:degenerate-trig}
If $AB+BC=AC$, then $B\in [AC]$.
\end{thm}

\parit{Proof.}
Since $AB+BC=AC$, we can choose $B'\in [AC]$ such that $AB=AB'$;
observe that $BC\z=B'C$.

We may assume that $AB>0$ and $BC>0$;
otherwise, $A=B$ or $B=C$, and the statement follows.
In this case, $\measuredangle AB'C=\pi$.

By SSS, 
\[\triangle ABC\cong \triangle AB'C.\]
Therefore $\measuredangle ABC=\pi$.
By Theorem~\ref{thm:straight-angle}, $B$ lies between $A$ and $C$.
\qeds



\begin{thm}{Advanced exercise}\label{ex:SMS}
Consider two triangles $A B C$ and $A' B' C'$.
Let $M$ be the midpoint of $[A B]$, and
$M'$ be the midpoint of $[A' B']$.
Assume $C' A'=C A$, $C' B'= C B$, and $C' M'\z= C M$.
Prove that
\[\triangle A' B' C'\cong\triangle A B C.\]

\end{thm}

{

\begin{wrapfigure}[6]{r}{26mm}
\vskip-0mm
\centering
\includegraphics{mppics/pic-42}
\end{wrapfigure}

\begin{thm}{Exercise}\label{ex:isos-sides}
Let $\triangle A B C$ be an isosceles triangle with the base $[A B]$.
Suppose that point $A'$ lies between $B$ and $C$,
point $B'$ lies between $A$ and $C$,
and $C A'\z=C B'$.
Show that
\begin{enumerate}[(a)]
\item $\triangle A A' C\cong \triangle B B' C$;
\item $\triangle A B B'\cong \triangle B A A'$.
\end{enumerate}

\end{thm}

\begin{thm}{Exercise}\label{ex:ABC-motion}
Let $\triangle ABC$ be a nondegenerate triangle, and
let $f$ be a motion of the plane 
such that 
$$f(A)=A,
\quad 
f(B)=B,
\quad 
\text{and}
\quad
f(C)=C.$$

Show that $f$ is the identity map;
that is, $f(X)=X$ for any point $X$ on the plane.
\end{thm}

}

%{

%\begin{wrapfigure}{r}{28mm}\vskip-11mm\centering\includegraphics{mppics/pic-43}\end{wrapfigure}

%\begin{thm}{Exercise}\label{ex:3-isos}
%Suppose that $\triangle ABC'$, $\triangle BCA'$, and $\triangle CAB'$ are isosceles triangles that lie outside of $\triangle ABC$;
%that is,
%$A$ and $A'$ lie on the opposite sides of $(BC)$,
%$B$ and $B'$ lie on the opposite sides of $(CA)$,
%$C$ and $C'$ lie on the opposite sides of $(AB)$.

%Show that $AA'=BB'=CC'$.
%\end{thm}

%}

\section{On side-side-angle and side-angle-angle}

In each of the conditions SAS, ASA, and SSS we specify three corresponding parts of the triangles.
Now, let us discuss other triples of corresponding parts.

The first triple is called {}\emph{side-side-angle}, or briefly SSA;
it specifies two sides and a non-included angle.
This condition is not sufficient for congruence.
In other words, there exist two nondegenerate triangles $ABC$ and $A'B'C'$ such that
\[AB=A'B',\quad BC=B'C',\quad \measuredangle BAC\equiv\pm \measuredangle B'A'C',\]
but $\triangle ABC\not\cong\triangle A'B'C'$ and $AC\ne A'C'$.

\begin{wrapfigure}{r}{35mm}
\vskip-6mm
\centering
\includegraphics{mppics/pic-44}
\end{wrapfigure}

We will not use this negative statement, so there is no need to prove it formally.
You can guess an example from the picture.

The second triple is {}\emph{side-angle-angle}, or briefly SAA;
it specifies one side and two angles one of which is opposite to the side.
This provides a congruence condition; 
that is, if one of the triangles $ABC$ or $A'B'C'$ is nondegenerate and
$AB=A'B'$, $\measuredangle ABC\equiv\pm \measuredangle A'B'C'$, $\measuredangle BCA\z\equiv\pm \measuredangle B'C'A'$,
then $\triangle ABC\cong\triangle A'B'C'$.

The SAA condition will not be used directly in the sequel.
One proof of this condition can be obtained from ASA and the theorem on the sum of angles of a triangle, which is proved below (see~\ref{thm:3sum}). 
For a more direct proof, see Exercise~\ref{ex:SAA}.

Another triple is called {}\emph{angle-angle-angle}, or briefly AAA.
By Axiom~\ref{def:birkhoff-axioms:4}, it is not a congruence condition in the Euclidean plane.
However, in the hyperbolic plane it is; see \ref{thm:AAA}.

\section{Constructions}

The construction problems provide a valuable source of exercises in geometry,
which we will use further in the book.
The whole Chapter~\ref{chap:car} will be devoted to the subject.

{

\begin{wrapfigure}{o}{33mm}
\vskip-4mm
\centering
\includegraphics{mppics/pic-445}
\bigskip
\includegraphics{mppics/pic-446}
\bigskip
\includegraphics{mppics/pic-447}
\bigskip
\includegraphics{mppics/pic-448}
\bigskip
\includegraphics{mppics/pic-449}
\end{wrapfigure}

A \index{ruler-and-compass construction}\emph{ruler-and-compass construction} is a construction with an idealized ruler and compass.
The idealized ruler can be used only to draw a line thru the given two points.
The idealized compass can be used only to draw a circle with a given center and radius.
That is, given three points $A$, $B$, and $O$
we can draw the circle of radius $AB$ centered at~$O$.
We may also mark new points in the plane,
as well as on the constructed lines, circles,
and their intersections (assuming that such points exist).

We may also look at the different sets of construction tools.
For example,
we may only use the ruler,
or we may invent a new tool,
say a tool that produces a midpoint for any given two points,
or the reflection of one point across the other.

As an example, let us consider the following problem:

\begin{thm}{Problem}
Given two lines $\ell$ and $m$ and a point $M$ not on these lines,
construct a line segment with ends on $\ell$ and $m$ that is bisected by~$M$.
\end{thm}

The stages of the following construction are shown in the picture.

\parit{Construction.}
\begin{enumerate}[1.]
\item Draw a circle centered at $M$ that crosses line $\ell$ at two points, say $P$ and $Q$.
\item Draw the line $(PM)$ and mark by $P'$ its other point of intersection with the circle.
Similarly, draw $(QM)$ and mark the other intersection point by $Q'$.
\item Draw the line $\ell'=(P'Q')$ and mark by $A$ its point of intersection with $m$ (if it exists).
\item Draw the line $(MA)$;
mark by $B$ its point of intersection with $\ell$.
Then $[AB]$ is the required segment.
\end{enumerate}

}

It is often required to do \emph{problem analysis};
proving that the result of the construction satisfies the required conditions and also determine when the construction is possible and how many solutions may exist.
Let us do it for the example above.


\parit{Problem analysis.}
Notice that $P'$ and $Q'$ are reflections of $P$ and $Q$ across $M$.
Therefore, the line $\ell'=(P'Q')$ is the reflection of $\ell=(PQ)$ across $M$.

Since $A$ lies on $\ell'$,
the point $B$ is the reflection of $A$ across $M$.
Indeed, $B$ is the (necessarily unique) point of intersection of $(MA)$ and $\ell$, and $\ell'$ is the reflection of $\ell$.
By the definition of reflection, $M$ is the midpoint of $[AB]$.

The line $\ell'$ may have no points of intersection with $m$.
In this case, the problem has no solution.
Also, it might happen that $\ell'=m$;
in this case, we have an infinite set of solutions --- any point on $m$ can be marked as~$A$.
In all the remaining cases, we have exactly one solution.
\qeds

\begin{thm}{Exercise}\label{ex:motion}
Suppose $\triangle ABC$ and $\triangle A'B'C'$ are congruent nondegenerate triangles;
in particular, there is a motion $f$ such that
\[f\colon A\mapsto A',\quad f\colon B\mapsto B',\quad\text{and}\quad f\colon C\mapsto C'.\]

Given a point $X$, construct $X'=f(X)$.
\end{thm}