## [

The Law of Sines

A circle inscribed in an isosceles triangle \(ABC\) touches the base \(AC\) at point \(D\) and the lateral side \(AB\) at point \(E\). Point \(F\) is the midpoint of side \(AB\), and point \(G\) is the intersection of the circle and segment \(FD\), different from \(D\). The tangent to the circle passing through point \(G\) intersects side \(AB\) at point \(H\). Find the angle \(BCA\) if it is known that \(FH:HE = 2:3\).

#
Express the angles of triangle $F G H$ through the desired angle and apply the Law of Sines. Consider both cases of the possible position of point $H$ - on segment $B E$ and on segment $A E$.

## Solution

Since $F$ is the midpoint of $B C$ and $D$ is the midpoint of $A C$, $F D$ is the midline of triangle $A B C$. Therefore, $D F \| A B$. Thus, triangle $A F D$ is also isosceles.

Let point $H$ lie on segment $B E$. Denote $\angle B A C=\angle A C B=\alpha, F H=2 x, H E=3 x$. If $O$ is the center of the circle, then

\[
\begin{gathered}
\angle A F D=180^{\circ}-2 \angle D A F=180^{\circ}-2 \alpha, \angle D O E=180^{\circ}-\angle C A B=180^{\circ}-\alpha, \\
\angle D G E=\frac{1}{2} \angle D O E=90^{\circ}-\frac{\alpha}{2}, \\
\angle F E G=\angle D G E-\angle E F G=90^{\circ}-\frac{\alpha}{2}-180^{\circ}+2 \alpha=\frac{3 \alpha}{2}-90^{\circ} .
\end{gathered}
\]

Triangle $E G H$ is isosceles, so $G H=E H=3 x$ and

\[
\angle G E H=\angle E G H=\frac{3 \alpha}{2}-90^{\circ}
\]

Angle $G H F$ is the exterior angle of triangle $E G H$, so

\[
\angle G H F=2 \angle G E H=3 \alpha-180^{\circ} .
\]

Then

\[
\angle F G H=180^{\circ}-\angle G H F-\angle G F H=180^{\circ}-3 \alpha+180^{\circ}-180^{\circ}+2 \alpha=180^{\circ}-\alpha
\]

By the Law of Sines for triangle $F G H$:

\[
\frac{G H}{\sin \angle G F H}=\frac{F H}{\sin \angle F G H} \text {, or } \frac{3 x}{\sin 2 \alpha}=\frac{2 x}{\sin \alpha} \text {. }
\]

From this equation, we find that $\cos \alpha=\frac{3}{4}$.

If point $H$ lies on segment $A E$, then reasoning similarly, we will arrive at the same result.

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-49.jpg?height=540&width=380&top_left_y=159&top_left_x=839)

## Answer

$\arccos \frac{3}{4}$.