OMCC 2012 Problem 6

Let $ABC$ be a triangle with $AB < BC$ , and let $E$ and $F$ be points in $AC$ and $AB$ such that $BF = BC = CE$ , both on the same halfplane as $A$ with respect to $BC$ .

Let $G$ be the intersection of $BE$ and $CF$ . Let $H$ be a point in the parallel through $G$ to $AC$ such that $HG = AF$ (with $H$ and $C$ in opposite halfplanes with respect to $BG$ ). Show that $\angle EHG = \frac{\angle BAC}{2}$ .