Question 77
Convex quadrilateral $ABCD$ has right angles $\angle A$ and $\angle C$ and is such that $AB = BC$ and $AD = CD$. The diagonals $AC$ and $BD$ intersect at point $M$. Points $P$ and $Q$ lie on the circumcircle of triangle $AMB$ and segment $CD$, respectively, such that points $P$, $M$, and $Q$ are collinear. Suppose that $m \angle ABC = 160^\circ$ and $m \angle QMC = 40^\circ$. Find $MP \cdot MQ$, given that $MC = 6$.
