Question 90
A convex quadrilateral $ABCD$ with $BC = CD$, $\angle CBA + \angle DAB \gt 180^\circ$ is given. Let the points $W$ and $Q$ (different from the quadrilateral's vertices) belong to the sides $BC$ and $DC$ respectively, and let $AD = QD$, $WQ \parallel AD$. It is known that the intersection point $M$ of segments $AQ$ and $BD$ is equidistant from the lines $AD$ and $BC$. Prove that $\angle BWD = \angle ADW$.