10.400 Calculate the area of a triangle given two sides $\boldsymbol{a}$ and $\boldsymbol{b}$ and the bisector $l$ of the angle between them.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-243.jpg?height=239&width=308&top_left_y=321&top_left_x=146)

Let $B D$ be the bisector of the given $\triangle A B C, a=A B, b=B C, l=B D, \angle B=2 \alpha$. By the cosine rule for $\triangle A B D$ and $\triangle D B C$, we have:

$$
\left\{\begin{array}{l}
A D^{2}=a^{2}+l^{2}-2 a \cdot l \cos \alpha \\
D C^{2}=b^{2}+l^{2}-2 b \cdot l \cos \alpha
\end{array} \Rightarrow\right.
$$

$\Rightarrow\left(\frac{A D}{D C}\right)^{2}=\frac{a^{2}+l^{2}-2 a \cdot l \cos \alpha}{b^{2}+l^{2}-2 b \cdot l \cos \alpha}=\frac{a^{2}}{b^{2}}$ (Theorem 4) $\Rightarrow \cos \alpha=\frac{l(a+b)}{2 a b}$,

$\sin \alpha=\frac{\sqrt{4 a^{2} b^{2}-l^{2}(a+b)^{2}}}{2 a b} \Rightarrow S_{A B C}=\frac{a \cdot l \sin \alpha}{2}+\frac{b \cdot l \sin \alpha}{2}=$

$=\frac{l(a+b)}{4 a b} \sqrt{4 a^{2} b^{2}-l^{2}(a+b)^{2}}$.

Answer: $\frac{l(a+b)}{4 a b} \sqrt{4 a^{2} b^{2}-l^{2}(a+b)^{2}}$.