Question 135
A quadrilateral $ABCD$ is inscribed into a circle $\Gamma$ centered at $O$. Its diagonals $AC$ and $BD$ are perpendicular to each other; let $P$ be their meeting point (the point $O$ lies inside the triangle $BPC$). A point $H$ is chosen on the segment $BO$ so that $\angle BHP = 90^\circ$. The circumcircle $\omega$ of the triangle $PHD$ meets again the segment $PC$ at $Q$. Prove that $AP = CQ$.