15. In the figure below, $B C$ is the diameter of a semicircle centered at $O$, which intersects $A B$ and $A C$ at $D$ and $E$ respectively. Suppose that $A D=9, D B=4$, and $\angle A C D=\angle D O B$. Find the length of $A E$.
(a) $\frac{117}{16}$
(b) $\frac{39}{5}$
(c) $2 \sqrt{13}$
(d) $3 \sqrt{13}$
15. $\mathrm{B}$

Solution. Let $\angle D O B=\angle E O C=\alpha$. Note that $\angle D C B=\frac{\alpha}{2}$. Also, note that $\tan \frac{\alpha}{2}=\frac{4}{D C}$ and $\tan \alpha=\frac{9}{D C}=\frac{9}{4} \tan \frac{\alpha}{2}$. Let $x=\tan \frac{\alpha}{2}$. By the double-angle formula,
$$
\begin{aligned}
\frac{9}{4} x & =\frac{2 x}{1-x^{2}} \\
\frac{9}{4} x-\frac{9}{4} x^{3} & =2 x \\
\frac{1}{4} x\left(1-9 x^{2}\right) & =0
\end{aligned}
$$
and thus $x=0$ or $x= \pm \frac{1}{3}$. Clearly, only $x=\frac{1}{3}$ is possible here. Thus, $C D=12$. Note also that $\angle C D B=\angle A D C=90^{\circ}$, and so by the Pythagorean theorem, $A C=\sqrt{9^{2}+12^{2}}=15$.
Finally, by the power of a point theorem, we have $A E \cdot A C=A D \cdot A B$ and so $A E \cdot 15=9(9+4)$, which gives us $A E=\frac{117}{15}=\frac{39}{5}$.