Question 169
Let $ABCD$ be a parallelogram and let $M$ be the intersection point of the diagonals. The circumcircle of $\triangle ABM$ intersects the line segment $AD$ at $E \neq A$, and the circumcircle of $\triangle EMD$ intersects the line segment $BE$ at the point $F \neq E$.
Prove that $\angle ACB = \angle DCF$.
Prove that $\angle ACB = \angle DCF$.