XXV OBM

Let $O$ be the center of the incircle. We show first that $AE\cdot CF=A{O}^2$. 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 $AE/AO=CO/CF\iff AE\cdot CF=A{O}^2$.

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