IZhO 2009 Problem 5

Given a quadrilateral $ABCD$ with $\angle B=\angle D=90^{\circ}$ . Point $M$ is chosen on segment $AB$ so taht $AD=AM$ . Rays $DM$ and $CB$ intersect at point $N$ . Points $H$ and $K$ are feet of perpendiculars from points $D$ and $C$ to lines $AC$ and $AN$ , respectively. Prove that $\angle MHN=\angle MCK$ .