MathNet 2005 — Part 03 Question 8

Let $M$ be the intersection point of the diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$. The bisector of $\angle ACD$ meets $BA$ extended at $K$. If $MA \cdot MC + MA \cdot CD = MB \cdot MD$, show that $\angle BKC = \angle CDB$.