In an isosceles triangle $A B C$, angle $B$ is a right angle, and $A B=B C=2$. A circle is tangent to both legs at their midpoints and intersects the hypotenuse in a chord $D E$. Find the area of triangle $B D E$.

#
Let point $E$ lie between points $D$ and $C$. Denote $D E=x$; $P$ is the point of tangency of the circle with the leg $B C$. Since $C D \cdot C E=C P^{2}$, we have

$$
\left(\sqrt{2}+\frac{x}{2}\right)\left(\sqrt{2}-\frac{x}{2}\right)=1
$$

From this, we find that $x=\frac{1}{2}$. Therefore,

$$
S_{\triangle \mathrm{BDE}}=\frac{D E}{A C} S_{\Delta \mathrm{ABC}}=\sqrt{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-09.jpg?height=412&width=560&top_left_y=746&top_left_x=749)

Answer

$\sqrt{2}$. Submit a comment

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