Slovenia — National Math Olympiad

The triangle $CAD$ is isosceles since $|AC| = |AD|$. The line $AA_1$ is the altitude in this isosceles triangle, so $\angle HDA = \angle ACH$. In the quadrilateral $HA_1CB_1$ we have $\angle CA_1H = \frac{\pi}{2} = \angle CB_1H$, so this quadrilateral is cyclic and $\angle B_1A_1H = \angle B_1CH$.

We have shown that $\angle FA_1A = \angle B_1A_1H = \angle B_1CH = \angle ACH = \angle HDA = \angle FDA$, so $A, D, A_1$ and $F$ are concyclic. This implies that $\angle AFD = \angle AA_1D = \frac{\pi}{2}$. The segments $AB_1$ and $HF$ are the altitudes in the triangle $AHG$ and they meet at $E$. So $E$ is the orthocentre of this triangle and $EG$ is perpendicular to $AH$. Now, $AH$ is perpendicular to $BC$, so $EG$ and $BC$ are parallel.

Let $T$ denote the intersection of the lines $EG$ and $AD$ and let $C_1$ be the foot of the altitude to the side $AB$ in the triangle $ABC$. Since $ET$ is parallel to $CD$, the quadrilateral $TDCE$ is in fact a trapezoid. Since $\angle TDC = \angle ECD$ this trapezoid is isosceles. We see that $\angle TCE = \angle TDE = \angle ADH = \angle ABH = \angle C_1BH = \angle HCE$, so $C, H$ and $T$ are collinear which was all that remained to be shown.