The altitudes $A D$ 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.
Let $\angle F A H = \angle H B F = \alpha$. Right triangles $A D C$ and $E C B$ share the common angle $C$, so $\angle E B C = \alpha$.

Thus, $B E$ is both the altitude 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, $C D = D G$. Therefore, $E D$ is the midline of triangle $F C G$. Hence, $F G = 2 D E$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-07.jpg?height=523&width=732&top_left_y=527&top_left_x=663)

Answer

10 cm.