Let $ABCD$ be a parallelogram, such that the point $M$ is the midpoint of the side $CD$ and lies on the bisector of the angle $\angle BAD$. Prove that $\angle AMB = 90^o$.
Problem
Source: V.A. Yasinsky Geometry Olympiad 2018 VIII-IX p2 [Ukraine]
Tags: geometry, parallelogram, midpoint, angle bisector