Middle European Mathematical Olympiad 2015 Problem T-6

Let $I$ be the incentre of triangle $ABC$ with $AB > AC$ and let the line $AI$ intersect the side $BC$ at $D$. Suppose that point $P$ lies on the segment $BC$ and satisfies $PI = PD$. Further, let $J$ be the point obtained by reflecting $I$ over the perpendicular bisector of $BC$, and let $Q$ be the other intersection of the circumcircles of the triangles $ABC$ and $APD$. Prove that $\angle BAQ = \angle CAJ$.