MEMO 2015 Problem 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 $\angle BAQ=\angle CAJ$ .