OMM 2000 Problem 6

Let $ABC$ be a triangle with $\angle B > 90^o$ such that there is a point $H$ on side $AC$ with $AH = BH$ and BH perpendicular to $BC$ . Let $D$ and $E$ be the midpoints of $AB$ and $BC$ respectively. A line through $H$ parallel to $AB$ cuts $DE$ at $F$ . Prove that $\angle BCF = \angle ACD$ .