In a right triangle $A B C$, the height dropped to the hypotenuse $A B$ is $a$, and the bisector of the right angle is $b$. Find the area of triangle $A B C$.
Using the formula for the area of a triangle, express the indicated bisector in terms of the legs of the triangle.

## Solution

Let $C L$ and $C H$ be the bisector and the height of triangle $A B C$, respectively. Denote $B C=x, A C=y$.

Since $A B \cdot C H = B C \cdot A C$, then

$$
a \sqrt{x^{2}+y^{2}}=x y, \text{ or } x^{2} y^{2}=a\left(x^{2}+y^{2}\right)=a\left((x+y)^{2}-2 x y\right).
$$

Since $S_{\triangle \mathrm{ABC}}=S_{\triangle \mathrm{BCL}}+S_{\Delta \mathrm{ACL}}$, then $x+y=\frac{x y \sqrt{2}}{b}$.

Substituting the found expression for $x+y$ into the first equation, we get that

$$
x y=\frac{2 a^{2} b^{2}}{2 a^{2}-b^{2}}
$$

Therefore,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} x y=\frac{a^{2} b^{2}}{2 a^{2}-b^{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_0e468145bb05e34a17aag-43.jpg?height=366&width=740&top_left_y=732&top_left_x=662)

## Answer

$\frac{a^{2} b^{2}}{2 a^{2}-b^{2}}$.