Clearly, the directed angle $EAF$ is $\widehat{EAF} = 2\alpha$. Let $H_2$, $H_3$ be the reflections of the orthocenter $H$ about the sides $AC$, $AB$, respectively. It is well-known that $H_2$ and $H_3$ lie on the circumcircle $k$ of triangle $ABC$.

Fig. 1

As $DH$ is the bisector of angle $BHC$ and $CH \parallel DE$, we have $|\angle H_3ED| = |\angle HDE| = |\angle DHC| = \frac{1}{2}|\angle BHC| = 90^\circ - \frac{1}{2}\alpha$. Let $G$ be the midpoint of arc $BAC$. Since the arc $CG$ of $k$ is one half of the arc $CAB$, the directed angle $\widehat{CH_3G}$ subtending arc $CG$ has the same measure $90^\circ - \frac{1}{2}\alpha$, that is $\widehat{CH_3G} = \widehat{DEH_3}$.

From $CH_3 \parallel DE$ we infer that points $G$, $H_3$, and $E$ are collinear. Similarly we find that $G$, $H_2$, and $F$ are also collinear, therefore $\overrightarrow{EGF} = \overrightarrow{H_3GH_2} = \overrightarrow{H_3AH_2} = \overrightarrow{H_3AH} + \overrightarrow{HAH_2} = 2\alpha = \overrightarrow{EAF}$, as we needed to show.