Croatia_2018

The chord $\overline{GH}$ is perpendicular to $\overline{AB}$, so $AB$ is the bisector of the segment $\overline{GH}$. Hence $|AG|=|AH|$.

Since $AFC$ is a right-angled triangle, Euclid's theorem gives us $|AF{|}^2=|AD|\cdot |AC|$.

Analogously, since $ABG$ is a right-angled triangle, we have $|AG{|}^2=|AE|\cdot |AB|$.



Angles $\angle BDC$ and $\angle BEC$ are right angles, so the quadrilateral $BCDE$ is cyclic. By using the power-of-a-point theorem applied to $A$ with respect to the circle circumscribed to the quadrilateral $BCDE$, we conclude that $|AD|\cdot |AC|=|AE|\cdot |AB|$.

Hence $|AF|=|AG|=|AH|$, i.e. the point $A$ is the circumcentre of the triangle $GFH$.

Finally, $$\angle AGF=\angle GFA=\frac{1}{2}(180^{\circ }-\angle FAG)=\frac{1}{2}(180^{\circ }-2\angle FHG)=\frac{1}{2}(180^{\circ }-24^{\circ })=78^{\circ }.$$