It suffices to show that $\angle H E F=\angle H G E$ (see Figure 1), since in circle $E G H$ the angle over the chord $E H$ at $G$ equals the angle between the tangent at $E$ and $E H$. First, $\angle B A D=180^{\circ}-\angle D C B=\angle F C D$. Since triangles $F A B$ and $F C D$ have also a common interior angle at $F$, they are similar. ! Figure 1 Denote by $\mathcal{T}$ the transformation consisting of a reflection at the bisector of $\angle D F C$ followed by a dilation with center $F$ and factor of $\frac{F A}{F C}$. Then $\mathcal{T}$ maps $F$ to $F, C$ to $A, D$ to $B$, and $H$ to $G$. To see this, note that $\triangle F C A \sim \triangle F D B$, so $\frac{F A}{F C}=\frac{F B}{F D}$. Moreover, as $\angle A D B=\angle A C B$, the image of the line $D E$ under $\mathcal{T}$ is parallel to $A C$ (and passes through $B$) and similarly the image of $C E$ is parallel to $D B$ and passes through $A$. Hence $E$ is mapped to the point $X$ which is the fourth vertex of the parallelogram $B E A X$. Thus, in particular $\angle H E F=\angle F X G$. As $G$ is the midpoint of the diagonal $A B$ of the parallelogram $B E A X$, it is also the midpoint of $E X$. In particular, $E, G, X$ are collinear, and $E X=2 \cdot E G$. Denote by $Y$ the fourth vertex of the parallelogram $D E C Y$. By an analogous reasoning as before, it follows that $\mathcal{T}$ maps $Y$ to $E$, thus $E, H, Y$ are collinear with $E Y=2 \cdot E H$. Therefore, by the intercept theorem, $H G \| X Y$. From the construction of $\mathcal{T}$ it is clear that the lines $F X$ and $F E$ are symmetric with respect to the bisector of $\angle D F C$, as are $F Y$ and $F E$. Thus, $F, X, Y$ are collinear, which together with $H G \| X Y$ implies $\angle F X E=\angle H G E$. This completes the proof.