ELMO SL 2013 G8

Let $ABC$ be a triangle, and let $D$ , $A$ , $B$ , $E$ be points on line $AB$ , in that order, such that $AC=AD$ and $BE=BC$ . Let $\omega_1, \omega_2$ be the circumcircles of $\triangle ABC$ and $\triangle CDE$ , respectively, which meet at a point $F \neq C$ . If the tangent to $\omega_2$ at $F$ cuts $\omega_1$ again at $G$ , and the foot of the altitude from $G$ to $FC$ is $H$ , prove that $\angle AGH=\angle BGH$ .

Proposed by David Stoner