66th Czech and Slovak Mathematical Olympiad

Let $K$ be the intersection of segments $FG$ and $AE$ and $L$ the intersection of $EH$ and $AF$ (Fig. 2). Inscribed angles give

$$\angle AGF=\angle AEF=90^{\circ }-\angle DAE=90^{\circ }-\angle GAE,$$ Fig. 2

that is $\angle AGF+\angle GAE=90^{\circ }$, hence $$\angle AKG=180^{\circ }-(\angle AGF+\angle GAE)=90^{\circ }.$$ Line FK is therefore an altitude of the triangle AEF. Similarly we prove that EL is its altitude too, thus the intersection of FK and EL is the orthocenter of triangle AEF and it lies on its third altitude AD too.

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Alternative solution.

Let M be the second intersection of ray $AD$ and the circumcircle $k$ of triangle $AEF$ (Fig. 2). Since $AE$ is the bisector of angle $GAM$, the angles $GAE$ and $EAM$ inscribed in $k$ are equal and therefore the chords $GE$ and $EM$ are also equal. Furthermore, the corresponding inscribed angles $GFE$ and $EFM$ are equal too. The intersection of $FG$ and $AM$ is thus the reflection of $M$ about $EF$. The same argument applies to the intersection of $EH$ and $AM$, therefore the original intersections are identical.