Školjka

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 $|\angle{BAQ}| = |\angle{CAJ}|$.