Question 32
Let $ABCD$ be a parallelogram, and let $O$ be a point inside $ABCD$. Suppose the circumcircles of triangles $OAB$ and $OCD$ intersect at $P \neq O$, and the circumcircles of triangles $OBC$ and $OAD$ intersect at $Q \neq O$. Prove $\angle POQ$ equals one of the angles of quadrilateral $ABCD$.
