In triangle $ABC$, the lateral sides $AB$ and $BC$ are equal. A line parallel to the base $AC$ intersects side $AB$ at point $D$ and side $BC$ at point $E$, and each of the segments $AD$, $EC$, and $DE$ is equal to 2. Point $F$ is the midpoint of segment $AC$, and point $G$ is the midpoint of segment $EC$, connected by a straight line segment. It is known that the measure of angle $GFC$ is $\beta$. Find the area of triangle $ABC$.
Prove that the angle at the base of the given triangle is $2 \beta$.

## Solution

Since $F G$ is the midline of triangle $A E C$, $F G$ is parallel to $A E$. Therefore, $\angle E A C = \angle G F C = \beta$. In the isosceles triangle $A D E$, it is known that

$$
\angle D A E = \angle D E A = \angle E A C = \angle G F C = \beta
$$

Therefore, $\angle B C F = \angle B A C = 2 \beta$.

Let $K$ be the projection of point $D$ onto the base $A C$. Then

$$
\begin{gathered}
F K = 1, A K = A D \cos \angle B A C = 2 \cos 2 \beta \\
A F = K F + A K = 1 + 2 \cos 2 \beta .
\end{gathered}
$$

From the right triangle $B F C$, we find that

$$
B F = F C \tan \angle B C F = (1 + 2 \cos 2 \beta) \tan 2 \beta
$$

Therefore,

$$
S_{\triangle \mathrm{ABC}} = F C \cdot B F = (1 + 2 \cos 2 \beta)^{2} \tan 2 \beta .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-13.jpg?height=440&width=538&top_left_y=0&top_left_x=763)

## Answer

$(1 + 2 \cos 2 \beta)^{2} \tan 2 \beta$.