In the cyclic quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at $P$. Let $O$ be the center of the circumcircle $ABCD$, and $E$ a point of the extension of $OC$ beyond $C$. A parallel line to $CD$ is drawn through $E$ that cuts the extension of $OD$ beyonf $D$ at $F$. Let $Q$ be a point interior to $ABCD$, such that $\angle AFQ = \angle BEQ$ and $\angle FAQ = \angle EBQ$. Prove that $PQ \perp CD$.