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.$