On the height $C D$, dropped from the vertex $C$ of the right triangle $A B C$ to the hypotenuse $A B$, a circle is constructed with $C D$ as its diameter, which intersects the leg $A C$ at point $E$ and the leg $B C$ at point $F$. Find the area of the quadrilateral $C F D E$, if the leg $A C$ is equal to $b$, and the leg $B C$ is equal to $a$.

#
To find the sides of the rectangle $C F D E$, use the theorem about the height of a right triangle drawn from the vertex of the right angle.

## Solution

Since $2 S_{\triangle \mathrm{ABC}}=A B \cdot C D=A C \cdot B C$, then

$$
C D=B C \cdot \frac{A C}{A B}=\frac{a b}{\sqrt{a^{2}+b^{2}}}
$$

Since $D F$ is the height of the right triangle $C D B$, drawn from the vertex of the right angle, then

$$
C F=\frac{C D^{2}}{B C}=\frac{a b^{2}}{\left(a^{2}+b^{2}\right)}
$$

Similarly, we find that $C E=\frac{a^{2} b}{\left(a^{2}+b^{2}\right)}$.

Since $C F D E$ is a rectangle, then

$$
S_{\mathrm{CFDE}}=C F \cdot C E=\frac{a^{3} b^{3}}{\left(a^{2}+b^{2}\right)^{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-27.jpg?height=345&width=614&top_left_y=-1&top_left_x=722)

## Answer

$\frac{a^{3} b^{3}}{\left(a^{2}+b^{2}\right)^{2}}$