Question 105
The diagonals $AC$ and $BD$ of a cyclic quadrilateral $ABCD$ meet at $P$. The point $Q$ is chosen on the segment $BC$ so that $PQ \perp AC$. Prove that the line of centers of the circumcircles of the triangles $APD$ and $BQD$ is parallel to $AD$. (A. Kuznetsov)