The sidelines $AB$ and $CD$ of a trapezoid meet at point $P$, and the diagonals of this trapezoid meet at point $Q$. Point $M$ on the smallest base $BC$ is such that $AM=MD$. Prove that $\angle PMB=\angle QMB$.
Problem
Source: Sharygin Geometry Olympiad, Final Round 2016, Problem 6 grade 9
Tags: geometry, trapezoid