Question 42
In a triangle $ABC$, $D$ is the intersection of the internal bisector of $\angle BAC$ with side $BC$. Let $P$ be the second intersection of the external bisector of $\angle BAC$ with the circumcircle of $\triangle ABC$. A circle through $A$ and $P$ intersects segment $BP$ in its interior in $E$ and segment $CP$ in its interior in $F$. Prove that $\angle DEP = \angle DFP$.