[ Area of a triangle (using two sides and the angle between them).] [ Auxiliary area. The area helps to solve the task. ]

In a right triangle $ABC$, the bisector of the right angle $CL$ is drawn. From vertex $A$ ($\angle A>45^{\circ}$) a perpendicular $AD$ is dropped to $CL$. Find the area of triangle $ABC$ if $AD=a$ and $CL=b$.
Using the formula for the area of a triangle, express the bisector $C L$ in terms of the legs of the triangle $A B C$.

## Solution

Note that $A C=a \sqrt{2}$. Let $B C=x$. Since $S_{\Delta \mathrm{ABC}}=S_{\Delta \mathrm{ACL}}+S_{\Delta \mathrm{BCL}}$, then

$$
\frac{a x \sqrt{2}}{2}=\frac{a b \sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}+\frac{b x \sqrt{2}}{4}
$$

From this, we find that $x=\frac{a b \sqrt{2}}{2(a-b)}$. Therefore,

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

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-47.jpg?height=380&width=774&top_left_y=-1&top_left_x=642)

## Answer

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