Question 42
Let $ABCD$ be a parallelogram and let $M$ be the intersection of its diagonals. The circumcircle of $\triangle ABM$ intersects the line segment $AD$ in $E \ne A$ and the circumcircle of $\triangle EMD$ intersects the line segment $BE$ in the point $F \ne E$.
Prove that $\angle ACB = \angle DCF$.
Prove that $\angle ACB = \angle DCF$.