The diagonals of a convex quadrilateral $ABCD$ intersect at $M$. The bisector of $\angle ACD$ intersects the ray $BA$ at $K$. Prove that if $MA\cdot MC + MA\cdot CD = MB \cdot MD $, then $\angle BKC = \angle BDC$
Source: Slovenia TST 2005 p1
Tags: equal angles, geometry, diagonals
The diagonals of a convex quadrilateral $ABCD$ intersect at $M$. The bisector of $\angle ACD$ intersects the ray $BA$ at $K$. Prove that if $MA\cdot MC + MA\cdot CD = MB \cdot MD $, then $\angle BKC = \angle BDC$