10.343. Determine the area of a triangle if two of its sides are 35 and 14 cm, and the bisector of the angle between them is 12 cm.
Solution.

Let $B D$ be the bisector of $\triangle A B C$ (Fig. 10.126), $A B=14$ cm, $B C=35$ cm, $B D=12$ cm, $\angle A B D=\angle C B D=\alpha$. $S_{\triangle A B C}=S_{\triangle A B D}+S_{\triangle C B D}$.

Then $\frac{1}{2} A B \cdot B C \sin \angle A B C=\frac{1}{2} A B \cdot B D \sin \angle A B D+\frac{1}{2} B C \cdot B D \sin \angle C B D$;

$14 \cdot 35 \cdot 2 \sin \alpha \cos \alpha=14 \cdot 12 \sin \alpha+35 \cdot 12 \sin \alpha ; \cos \alpha=\frac{3}{5} . \sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\frac{4}{5}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0805.jpg?height=537&width=1165&top_left_y=75&top_left_x=96)

Fig. 10.127

Fig. 10.128

$$
\sin 2 \alpha=2 \sin \alpha \cos \alpha=\frac{24}{25} \text{. Then } S_{\triangle A B C}=\frac{1}{2} 35 \cdot 14 \cdot \frac{24}{25}=235.2 \text{ cm}^{2} \text{. }
$$

Answer: $235.2 \text{ cm}^{2}$.