10.079. Find the bisectors of the acute angles of a right triangle with legs of 24 and $18 \mathrm{~cm}$.
Solution.

$$
\left.\frac{DC}{AC}=\frac{BD}{AB} \text { (fig. } 10.77\right) ; AC=\sqrt{24^{2}+18^{2}}=30(\mathrm{~cm}), \frac{x}{30}=\frac{24-x}{18} \cdot \text { Hence }
$$

here $x=15, CD=15($ cm $), BD=24-15=9$ (cm).

The bisector of $\angle BAC: AD=\sqrt{AC \cdot AB-CD \cdot BD}=\sqrt{30 \cdot 18-15 \cdot 9}=9 \sqrt{5}$ (cm). $\frac{AM}{AC}=\frac{MB}{BC}, \frac{y}{30}=\frac{18-y}{24}$. Hence $y=10(\mathrm{~cm}), AM=10(\mathrm{~cm}), MB=8(\mathrm{~cm})$. The bisector of $\angle BCA$ :

$CM=\sqrt{AC \cdot BC-AM \cdot MB}=\sqrt{30 \cdot 24-8 \cdot 10}=8 \sqrt{10}$ (cm).

Answer: $9 \sqrt{5}$ and $8 \sqrt{10}$ cm.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-594.jpg?height=542&width=370&top_left_y=93&top_left_x=159)

Fig. 10.78

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-594.jpg?height=386&width=550&top_left_y=189&top_left_x=605)

Fig. 10.79