Let $I$ and $J$ be points on $CF$ such that $AI \parallel EJ \parallel BC$. Since $\dfrac{IJ}{JC}=\dfrac{AE}{EB}=\dfrac{DG}{GC}$ and $\dfrac{IF}{FC}=\dfrac{AF}{FB}=\dfrac{DH}{HC}$, we get $ID \parallel JG \parallel FH$. Let $K$ be the foot of the perpendicular from $A$ to $ID$. Also, $\dfrac{AD}{BC}=\dfrac{AF}{FB}=\dfrac{AI}{BC} \implies AD=AI$. Hence, $K$ is the midpoint of $ID$. Note that $\angle AIK = \angle EJG$. Then $\dfrac{AI}{EJ}=\dfrac{FI}{FJ}=\dfrac{HD}{HG}$ and $\dfrac{IK}{JG}=\dfrac{ID}{2JG}=\dfrac{DC}{2GC}$. From $\dfrac{HD}{HC}=\dfrac{GD}{GC}$, we have $HC \cdot GD = HD \cdot GC$. Then $HG \cdot DC = (HD + DG)(DG + GC) = HD \cdot CG + DG(HD+DG+GC) = HD \cdot GC + DG \cdot HC = HD \cdot GC + HD \cdot GC = 2HD \cdot GC$. Therefore, $\dfrac{AI}{EJ}=\dfrac{IK}{JG}$, and together with $\angle AIK = \angle EJG$, we get $\triangle AIK \sim \triangle EJG \implies \angle EGJ = \angle AKI = 90^{\circ}$. Since $EG \perp GJ$ and $GJ \parallel FH$, we conclude that $FH \perp EG$.