## Task A-4.4.

Let $\overline{B D}$ and $\overline{C E}$ be the altitudes of the acute triangle $A B C$. The circle with diameter $\overline{A C}$ intersects the segment $\overline{B D}$ at point $F$. The circle with diameter $\overline{A B}$ intersects the line $C E$ at points $G$ and $H$, with $G$ between $C$ and $E$. If $\varangle C H F=12^{\circ}$, determine $\varangle A G F$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_684c59f28a15b0aa883eg-19.jpg?height=820&width=1014&top_left_y=281&top_left_x=521)

The chord $\overline{G H}$ is perpendicular to the diameter $\overline{A B}$, so $A B$ is the perpendicular bisector of the segment $\overline{G H}$. Therefore, $|A G|=|A H|$. Triangle $A F C$ is a right triangle, so by Euclid's theorem, we have $|A F|^{2}=|A D| \cdot|A C|$.

Similarly, triangle $A B G$ is a right triangle, so $|A G|^{2}=|A E| \cdot|A B|$.

Angles $\varangle B D C$ and $\varangle B E C$ are right angles, so quadrilateral $B C D E$ is cyclic. By the power of a point theorem applied to point $A$ with respect to the circumcircle of quadrilateral $B C D E$, we conclude that $|A D| \cdot|A C|=|A E| \cdot|A B|$.

Therefore, $|A F|=|A G|=|A H|$, which means point $A$ is the circumcenter of triangle $G F H$.

Finally,

$$
\varangle A G F=\varangle G F A=\frac{1}{2}\left(180^{\circ}-\varangle F A G\right)=\frac{1}{2}\left(180^{\circ}-2 \varangle F H G\right)=\frac{1}{2}\left(180^{\circ}-24^{\circ}\right)=78^{\circ} .
$$