UA-UMO 2005 Problem 86

A convex quadrilateral $ABCD$ with $BC=CD$ and $\angle CBA+\angle DAB>180^{\circ}$ is given. Suppose that $W$ and $Q$ are points on the sides $BC$ and $DC$ respectively (distinct from the vertices) such that $AD=QD$ and $WQ \parallel AD.$ Also suppose that $AQ$ and $BD$ intersect at a point $M$ that is equidistant from the lines $AD$ and $BC$ . Prove that angle $\angle BWD=\angle ADW.$