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 \[ \frac{AC\cdot BD}{AB\cdot FC}=2.\]
Problem
Source: Turkish Mathematical Olympiad 2nd Round 1994
Tags: geometry, cyclic quadrilateral, geometry unsolved