In the isosceles triangle $ABC$, with $AB=AC$, $D$ is a point on the extension of $CA$ such that $DB$ is perpendicular to $BC$, $E$ is a point on the extension of $BC$ such that $CE=2BC$, and $F$ is a point on $ED$ such that $FC$ is parallel to $AB$. Prove that $FA$ is parallel to $BC$.