Question 172
In a convex quadrilateral $ABCD$, let $I$ and $J$ be the incenters of $\triangle ABC$ and $\triangle ADC$, respectively. Assume that $IJ$, $AC$, $BD$ meet at point $P$. The line through $P$ perpendicular to $BD$ meets the exterior angle bisector of $\angle BAD$ at $E$, and meets the exterior angle bisector of $\angle BCD$ at $F$. Prove that $PE = PF$.
