ISL 2004 G8

Given a cyclic quadrilateral $ABCD$ , let $M$ be the midpoint of the side $CD$ , and let $N$ be a point on the circumcircle of triangle $ABM$ . Assume that the point $N$ is different from the point $M$ and satisfies $\frac{AN}{BN}=\frac{AM}{BM}$ . Prove that the points $E$ , $F$ , $N$ are collinear, where $E=AC\cap BD$ and $F=BC\cap DA$ .

Proposed by Dusan Dukic, Serbia and Montenegro