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\title{Isosceles Triangles}
\maketitle


\textbf{Sophie} I suppose that almost every high school student
knows and can prove that any
isosceles triangle has equal base angles and any triangle with
two equal angles is an isosceles
triangle. Furthermore it is easy to prove that two bisectors of
the base angles of any isosceles
triangle are equal.  A challenging problem is to prove a converse
proposition, which seems to be true.

\vspace{0.3cm}
{\begin{figure}
{\centering \includegraphics{isoscel1.gif} \par}
\caption{\label{Iso1}}
\end{figure}\par}
\vspace{0.3cm}


\textbf{Problem 1}  Prove that if bisectors of two angles of a
triangle are equal, then the
triangle is isosceles.   


\textbf{Sophie}  In Figure 1, $BD$ and
$CE$ are respectively bisectors of angles
$B$ and $C$, and $|CD| = |BE|$. 
Our objective is to prove $|AB| = |AC|$. 
Considering all the line segments involved
in the conditions and goal, we find two
triangles $\triangle ABD$ and $\triangle
ACE$.  Moreover it is natural for us to try
to show $\triangle ABD$ and $\triangle
ACE$ are congruent.

\textbf{Gena}  Two triangles have the same
vertex angle and the bases of the same
length. But these are not enough to say that
they are congruent. 

\textbf{Sophie}  Yes.  And the difficult point in this problem is
that we cannot immediately
show that conditions given in the text books for two triangles to
be congruent are satisfied. 

\textbf{Sophie}  However, two things should be noted here.  The
first we have not used the
condition that $BD$ and $CE$ are bisectors of $\angle B$ and
$\angle C$.  Can you use it? 
Secondly can you find point out another equality between part of
$\triangle ABD$ and $\triangle ACE$?

\textbf{Luis}  The second question seems easier.  Let us name the
intersection of $BE$ and
$CF$ be $F$, then $AF$ is part of the two triangles.

\textbf{Sophie}  Good. Probably we can use it.  But what is $AF$? 


\textbf{Gena}  A theorem in text books of geometry says that the
three angle bisectors of a
triangle pass through the same point.  Therefore $AF$ is the
bisector of $\angle A$.

\textbf{Sophie}  Good, and you used the remaining condition.

\textbf{Sophie}  Now we have get to a substitute problem:

\textbf{Problem 2}  Prove that if two triangles have the vertex
angles of the same size, the
bases of the same length and the bisectors of the same length of
the vertex angles, the two
triangles are congruent.

\textbf{Sophie}  Problem 2 still looks hard, but we have three
equalities  for two triangles. 
We also know that conditions for two triangles to be congruent
consist of three equalities, as
we learned in a geometry course.  For example, the lengths of all
corresponding edges are
equal.  

\textbf{Luis}  It's really hard.

\textbf{Sophie}  We need to switch our
mind a little.  Since it is hard to prove the
congruence in their original positions or in
completely separate positions, let's move
the bases of two triangles in the same
position $BD$. Then the vertices are
separated to $A$ and $A'$, but $\angle A$
and $\angle A'$ are equal.  Have you not
seen such a figure before.  

\vspace{0.3cm}
{\begin{figure}
{\centering \includegraphics{isoscel2.gif} \par}
\caption{\label{Iso2}}
\end{figure}\par}
\vspace{0.3cm}



\textbf{Luis}  Yes, It reminds me of a
figure in the chapter of circles in a
geometry text book.  The four points $A,
A', B, D$ are on the same circle, since
$\angle A$ and $\angle A'$ are equal.

\vspace{0.3cm}
{\begin{figure}
{\centering \includegraphics{isoscel3.gif} \par}
\caption{\label{Iso2}}
\end{figure}\par}
\vspace{0.3cm}

\textbf{Sophie}  Right.  But what happened
to the bisector $AF$?  

\textbf{Luis}  Now the bisector $AF$ of $\triangle ACE$ moved to
$A'F'$ in the new
location.  Oh, if we extend $AF$ and $A'F'$ they bisects the arc
$BD$ at its middle point
say $G$, since "inscribed angles of the same size have their
intercepted arcs of the same
length."                

\textbf{Gena}  A great discovery!  Now, if the lengths of the
cords $AG$ and $A'G$ are
equal then the lengths of the arcs $AG$ and $A'G$ are equal. 
Therfore, the lengths of the
arcs $AD$ and $A'B$ are equal, so that $|AD|$ and $|A'B|$ are
equal, and also $|AB|$ and $|A'D|$ are equal.  Therefore, the two
triangles $ADB$ and $A'BD$ are congruent. 
However, the given condition is $|AF| = |A'F'|$ instead of $|AG|
= |A'G|$.

\textbf{Sophie}  Here we need another switching of our mind, for
a static approach to a
dynamic approach, by which I mean a relation between $|AG|$ and
$|AF|$ when $A$ moves along the arc $DB$.  Since $|AF| = |AG| -
|FG|$, $|AF|$ is
a function of $|AG|$ and $|FG|$.  $|AG|$ increases as $A$ moves
away from $D$, and $|FG|$ decreases as $A$ moves away from $D$
until $A$ reaches the antipodal point $H$ of $G$, where $HG$ is a
diameter of the circle.  Therefore $|AF|$ increases from $0$ as
$A$ moves away from $D$ until $A$ reaches $H$. Therefore, given a
value $a$, there is a unique point A such that $|AF| = a$. 
Therefore if $|AF| = |A'F'|$, then $A'$ is the mirror image of
$A$ through $HG$, so that $|AG| = |A'G|$. 