10.277. Find the bisector of the right angle of a triangle with legs equal to $a$ and $b$.
## Solution.

In $\triangle A B C$ (Fig. 10.72) $\angle C=90^{\circ}, C B=a, C A=b, C K-$ is the bisector of $\angle B C A$. Let $C K=x . S_{\triangle A B C}=S_{\triangle C B K}+S_{\triangle A C K} ; \frac{1}{2} A C \cdot C B=$ $=\frac{1}{2} A C \cdot C K \sin \angle A C K+\frac{1}{2} B C \cdot C K \sin \angle B C K$. Then $a b=b x \sin 45^{\circ}+a x \sin 45^{\circ}$, $x=\frac{a b \sqrt{2}}{a+b}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0763.jpg?height=389&width=719&top_left_y=64&top_left_x=321)

Fig. 10.72

Answer: $x=\frac{a b \sqrt{2}}{a+b}$.