Let $M$ be the point of intersection of diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$. Let $K$ be the point of intersection of the extension of side $AB$ (beyond$A$) with the bisector of the angle $ACD$. Let $L$ be the intersection of $KC$ and $BD$. If $MA \cdot CD = MB \cdot LD$, prove that the angle $BKC$ is equal to the angle $CDB$.