Hong Kong Preliminary Selection Contest

Obviously one possible position of $E$ arises from the case when $BE\perp CD$ (or equivalently, $ED=EC$). It is denoted by $E_1$ in the figure, in which case $E_{1}DBC$ is a kite. In particular, $B{E}_1$ bisects $\angle CBA$, and hence $\frac{A{E}_1}{E_{1}C}=\frac{AB}{BC}=\frac{13}{7}$. With $b$ denoting the length of $AC$, we get $A{E}_1=\frac{13b}{20}$.

If $ED\ne EC$, then the internal bisector of $\angle DEC$ intersects the perpendicular bisector of $CD$ on the circumcircle of $△CDE$. But we know that this intersection point is $B$ and so $E$ must be the second intersection point of the circumcircle of $△BCD$ and the side $AC$. This is the point $E_2$ in the figure. The chord theorem thus gives $A{E}_2\times AC=AD\times AB$ and so $A{E}_2=\frac{AD\times AB}{AC}=\frac{(13-7)13}{b}=\frac{78}{b}$.

It follows that the answer is $\frac{13b}{20}\times \frac{78}{b}=\frac{507}{10}$.