9.5. The heights $A D$ and $B E$ of an acute-angled triangle $A B C$ intersect at point $H$. The circumcircle of triangle $A B H$ intersects sides $A C$ and $B C$ at points $F$ and $G$ respectively. Find $F G$, if $D E=$ $=5$ cm.
Answer: 10 cm.

Solution. Let $\angle H B F=\alpha$ (see Fig. 9.5). Then $\angle F A H=\angle H B F=\alpha$ (inscribed angles subtending the same arc). From the right triangle $A D C: \angle C=90^{\circ}-\alpha$, and from the right triangle $E C B: \angle E B C=90^{\circ}-\angle C=\alpha$.

Thus, $B E$ is the height and the angle bisector of triangle $F B C$, which means this triangle is isosceles and $B E$ is its median, i.e., $F E = E C$. Similarly, it can be shown that $C D = D G$. Therefore, $E D$ is the midline of triangle $F C G$. Hence, $F G = 2 D E = 10$ (cm).

## Grading Criteria:

+     - a complete and well-reasoned solution
- only the answer provided

![](https://cdn.mathpix.com/cropped/2024_05_06_f571041068196027ec27g-2.jpg?height=358&width=512&top_left_y=796&top_left_x=1525)

Fig. 9.5