66th Czech and Slovak Mathematical Olympiad

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 $|\angle H_{3}ED|=|\angle HDE|=|\angle DHC|=\frac{1}{2}|\angle BHC|=90^{\circ }-\frac{1}{2}\alpha$. Let $G$ be the midpoint of arc $BAC$. Since the arc $CG$ of $k$ is one half of the arc $CAB$, the directed angle $\widehat{C{H}_{3}G}$ subtending arc $CG$ has the same measure $90^{\circ }-\frac{1}{2}\alpha$, that is $\widehat{C{H}_{3}G}=\widehat{DE{H}_3}$.

From $C{H}_3\parallel DE$ we infer that points $G$, $H_3$, and $E$ are collinear. Similarly we find that $G$, $H_2$, and $F$ are also collinear, therefore $\overrightarrow{EGF}=\overrightarrow{H_{3}G{H}_2}=\overrightarrow{H_{3}A{H}_2}=\overrightarrow{H_{3}AH}+\overrightarrow{HA{H}_2}=2\alpha =\overrightarrow{EAF}$, as we needed to show.