We show first that the triangles $FDG$ and $FBE$ are similar. Since $ABCD$ is cyclic, the triangles $EAB$ and $EDC$ are similar, as well as $FAB$ and $FCD$. The parallelogram $ECGD$ yields $GD = EC$ and $\angle CDG = \angle DCE$; also $\angle DCE = \angle DCA = \angle DBA$ by inscribed angles. Therefore
$$\begin{gathered} \angle FDG = \angle FDC + \angle CDG = \angle FBA + \angle ABD = \angle FBE, \\ \frac{GD}{EB} = \frac{CE}{EB} = \frac{CD}{AB} = \frac{FD}{FB} . \end{gathered}$$
It follows that $FDG$ and $FBE$ are similar, and so $\angle FGD = \angle FEB$.

Since $H$ is the reflection of $E$ with respect to $FD$, we conclude that
$$\angle FHD = \angle FED = 180^{\circ} - \angle FEB = 180^{\circ} - \angle FGD .$$

This proves that $D$, $H$, $F$, $G$ are concyclic.