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 $(BT{C}_{\infty })=-1$ 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 $AB\ne AC$. 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. $\square$