Let $M$ be the point of intersection of the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$. Let$ K$ be the intersection point of the extension of side $AB$ (from $A$) with the bisector of the $\angle ACD$. If $MA \cdot MC + MA \cdot CD = MB\cdot MD$ , prove that $\angle BKC = \angle CDB$.