TR-MO 1994 Problem 2

Let $ABCD$ be a cyclic quadrilateral $\angle{BAD}< 90^\circ$ and $\angle BCA = \angle DCA$ . Point $E$ is taken on segment $DA$ such that $BD=2DE$ . The line through $E$ parallel to $CD$ intersects the diagonal $AC$ at $F$ . Prove that $$\frac{AC\cdot BD}{AB\cdot FC}=2.$$