In a right triangle $A B C$, from the vertex of the right angle $C$, the bisector $C L$ and the median $C M$ are drawn.

Find the area of triangle $A B C$, if $L M=a, C M=b$.
Note that $A M=M B=b$. Let $B C=x, A C=y$. Suppose $x<y$. Then, by the property of the angle bisector of a triangle, $\frac{x}{y}=\frac{B L}{A L}=\frac{b-a}{b+a}$. Therefore,

$\frac{x-y}{x+y}=\frac{a}{b}$. Consequently, $\frac{2 x y}{x^{2}+y^{2}}=\frac{(y+x)^{2}-(y-x)^{2}}{(y+x)^{2}+(y-x)^{2}}=\frac{b^{2}-a^{2}}{b^{2}+a^{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-05.jpg?height=234&width=443&top_left_y=1964&top_left_x=819)

By the Pythagorean theorem, $x^{2}+y^{2}=4 b^{2}$, so $S_{A B C}=\frac{1}{2} x y=\frac{b^{2}\left(b^{2}-a^{2}\right)}{b^{2}+a^{2}}$.

## Answer

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