10. Triangle $A B C$ is inscribed in a circle $\omega$. Let the bisector of angle $A$ meet $\omega$ at $D$ and $B C$ at $E$. Let the reflections of $A$ across $D$ and $C$ be $D^{\prime}$ and $C^{\prime}$, respectively. Suppose that $\angle A=60^{\circ}, A B=3$, and $A E=4$. If the tangent to $\omega$ at $A$ meets line $B C$ at $P$, and the circumcircle of $A P D^{\prime}$ meets line $B C$ at $F$ (other than $P$ ), compute $F C^{\prime}$.
$\angle A E P$, so by the cyclic quadrilateral $A P D^{\prime} F, \angle E F D^{\prime}=\angle P A E=\angle P E A=\angle D^{\prime} E F$. Thus, $E D^{\prime} F$ is isosceles.

Define $B^{\prime}$ to be the reflection of $A$ about $B$, and observe that $B^{\prime} C^{\prime} \| E F$ and $B^{\prime} D^{\prime} C^{\prime}$ is isosceles. It follows that $B^{\prime} E F C^{\prime}$ is an isosceles trapezoid, so $F C^{\prime}=B^{\prime} E$, which by the law of cosines, is equal to $\sqrt{A B^{\prime 2}+A E^{2}-2 A B \cdot A E \cos 30}=2 \sqrt{13-6 \sqrt{3}}$