Brazilian Mathematical Olympiad

Let $O$ be the center of the incircle. We show first that $AE \cdot CF = AO^2$. Let $\angle AOE = \theta$. Then if $OX$ is the perpendicular from $O$ to $AB$, we have $\angle AOX = B/2$ and hence $\angle XOE = \theta - B/2$. If $EF$ touches the circle at $Y$, $\angle EOY = \angle XOE$, so $\angle BEF = 2(\theta - B/2) = 2\theta - B$. Hence $\angle BFE = 180^\circ - 2\theta$. Hence $\angle CFE = 2\theta$. So $\angle CFO = \theta$. So we have established that $\angle AOE = \angle CFO$. But $\angle EAO = \angle OCF$ (since $ABCD$ is a rhombus), so $AOE$ and $CFO$ are similar. Hence $AE/AO = CO/CF \Leftrightarrow AE \cdot CF = AO^2$.

Similarly, $AH \cdot CG = AO^2$. Hence $AH/AE = CF/CG$. So $AHE$ and $CFG$ are similar. So $\angle AEH = \angle CGF$, so $EH$ and $FG$ are parallel.