5. As shown in Figure 1, let $A$ be a point on the circle with diameter $BC$, $AD \perp BC$ at point $D$, point $E$ is on segment $DC$, and point $F$ is on the extension of $CB$, such that $\angle BAF = \angle CAE$. Given $BC = 15$, $BF = 6$, $BD = 3$, then $AE =$ ( ).
(A) $4 \sqrt{3}$
(B) $2 \sqrt{13}$
(C) $2 \sqrt{14}$
(D) $2 \sqrt{15}$
5. B.
Since $\angle B A F=\angle C A E$, we have $\angle B A F+\angle B A E=\angle C A E+\angle B A E$, which means $\angle F A E=\angle B A C=90^{\circ}$.
Also, $A D \perp B C$, so $A D^{2}=D E \cdot D F=D B \cdot D C$.
It is easy to see that $D F=B F+B D=9$, $D C=B C-B D=12$.
Thus, $A D^{2}=D E \cdot 9=3 \times 12$
$$
\Rightarrow A D=6, D E=4 \text {. }
$$

Therefore, $A E=\sqrt{A D^{2}+D E^{2}}=2 \sqrt{13}$.