A line parallel to the hypotenuse $AB$ of a right triangle $ABC$ intersects the leg $AC$ at point $D$ and the leg $BC$ at point $E$, with $DE=2$ and $BE=1$. A point $F$ is taken on the hypotenuse such that $BF=1$. It is also known that $\angle FCB=\alpha$. Find the area of triangle $ABC$.
Let $H$ be the midpoint of $DE$. Then $HFBE$ is a rhombus.

![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-50.jpg?height=478&width=580&top_left_y=204&top_left_x=745)

Since $CH = \frac{1}{2} DE = 1$, triangle $CHF$ is isosceles. Therefore, $\angle HCF = \angle HFC = \angle FCB = \alpha$. Hence, $\angle B = \angle DEC = \angle HCB = 2\alpha$.

Thus, $CE = 2 \cos 2\alpha$, $BC = 2 \cos 2\alpha + 1$, and $AC = BC \tan \angle B = (2 \cos 2\alpha + 1) \tan 2\alpha$.

Therefore, $S_{ABC} = \frac{1}{2} BC \cdot AC = \frac{1}{2} (2 \cos 2\alpha + 1)^2 \tan 2\alpha$.

## Answer

$\frac{1}{2} (2 \cos 2\alpha + 1)^2 \tan 2\alpha$