13. $\triangle A B C$ is acute-angled with $A B=13$ and $B C=7 . D$ and $E$ are points on $A B$ and $A C$ respectively such that $B D=B C$ and $\angle D E B=\angle C E B$. Find the product of all possible values of the length of $A E$.
(2 marks)
$A B C$ 是銳角三角形, 其中 $A B=13$ 及 $B C=7$ 。設 $D$ 和 $E$ 分別為 $A B$ 和 $A C$ 上的點, 使得 $B D=B C$ 及 $\angle D E B=\angle C E B$ 。求 $A E$ 的長度的所有可能值之積。
13. $\frac{507}{10}$
13. Obviously one possible position of $E$ arises from the case when $B E \perp C D$ (or equivalently, $E D=E C$ ). It is denoted by $E_{1}$ in the figure, in which case $E_{1} D B C$ is a kite. In particular, $B E_{1}$ bisects $\angle C B A$ and hence $\frac{A E_{1}}{E_{1} C}=\frac{A B}{B C}=\frac{13}{7}$. With $b$ denoting the length of $A C$, we get $A E_{1}=\frac{13 b}{20}$.
If $E D \neq E C$, then the internal bisector of $\angle D E C$ intersects the perpendicular bisector of $C D$ on the circumcircle of $\triangle C D E$. But we know that this intersection is $B$ and so $E$ must be the second intersection of the circumcircle of $\triangle B C D$ with side $A C$. This is the point $E_{2}$ in the figure. The power chord theorem thus gives $A E_{2} \times A C=A D \times A B$ and so $A E_{2}=\frac{A D \times A B}{A C}=\frac{(13-7) 13}{b}=\frac{78}{b}$.
It follows that the answer is $\frac{13 b}{20} \times \frac{78}{b}=\frac{507}{10}$.

Remark. The two cases are the same, i.e. $E_{1}=E_{2}$, only if $\angle A C B=90^{\circ}$, which is excluded since the question assumed that $\triangle A B C$ is acute. On the other hand, the question intended to mean that $\triangle A B C$ is fixed; indeed the length of $A E$ itself has infinitely many possible values, but with each fixed $\triangle A B C$ there are only two and the product of them is the same regardless of the choice of $\triangle A B C$.