Czech-Polish-Slovak Match

Let $p$ be the line parallel to $BC$ that passes through $A$ and let $P = p \cap DG$. We denote the midpoint of $BC$ by $T$.
We project the harmonic ratio $(BTC_\infty) = -1$ from $A$ onto the line $DG$ and learn that $(EGFP) = -1$. Therefore by a well-known Apollonian property of harmonic ratios it suffices to prove $\angle PHA = 90^\circ$.

Now let $Q$ be the orthogonal projection of $D$ onto $AH$. The homothety centered at $G$ with factor $-\frac{1}{2}$ maps the line $p$ onto the line $BC$ and hence it maps $P$ to $D$. Moreover, it leaves the line $AH$ intact, so we just need to prove that it maps $H$ to $Q$.
As $H$ lies on the circumcircle of $ABC$, which is mapped to the nine-point circle $\omega$ of $ABC$ by the considered homothety, we just need to verify that $Q$ lies on $\omega$ and that $Q$ is not the “wrong” intersection point of $\omega$ and $AH$. But this other point is the midpoint of $BC$, which does not coincide with $Q$ when $AB \neq AC$.
So our current claim is just that $Q$ lies on $\omega$. To verify it, we denote the orthogonal projection of $A$ to $BC$ by $R$ and the midpoint of $AB$ by $S$. It is known that $R, S$, and $T$ lie on $\omega$. Further, the points $Q$ and $R$ lie on the circle with diameter $AD$. Hence
$$\angle RQT = \angle BDA = \beta - \angle BAD = \beta - \gamma = \alpha - (180^\circ - 2\beta) = \angle BST - \angle BSR = \angle RST$$
and we may conclude. $\square$