Question 176
The extension of median $C M$ of the triangle $A B C$ intersects its circumcircle $\omega$ at $N$. Let $P$ and $Q$ be the points on the rays $C A$ and $C B$ respectively such that $P M \parallel B N$ and $Q M \parallel A N$. Let $X$ and $Y$ be the points on the segments $P M$ and $Q M$ respectively such that $P Y$ and $Q X$ are tangent to $\omega$. The segments $P Y$ and $Q X$ intersect at $Z$. Prove that the quadrilateral $M X Z Y$ is circumscribed.