Let the bisector of the outside angle of $A$ of triangle $ABC$ and the circumcircle of triangle $ABC$ meet at point $P$. The circle passing through points $A$ and $P$ intersects segments $BP$ and $CP$ at points $E$ and $F$ respectively. Let $AD$ is the angle bisector of triangle $ABC$. Prove that $\angle PED = \angle PFD$.
Problem
Source: 2019 Saudi Arabia IMO Training Test 5.2
Tags: geometry, angle bisector, equal angles
