10.242. The base of an isosceles triangle is 8 cm, and the lateral side is 12 cm. Find the length of the segment connecting the points of intersection of the angle bisectors at the base with the lateral sides of the triangle.
Solution.

In $\triangle ABC$ (Fig. 10.49): $AB=BC=12, AC=8, AN$ and $CM$ are angle bisectors. Let $MN=x$. As in problem 10.240, $MN \| AC$. Since

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0742.jpg?height=382&width=660&top_left_y=7&top_left_x=340)

Fig. 10.50

$\angle ACM = \angle NCM$ and $\angle ACM = \angle NMC$, then $\triangle MNC$ is isosceles and $NC = MN = x$. Therefore, $BN = 12 - x$. By the angle bisector theorem: $\frac{BN}{NC} = \frac{AB}{AC}$, $\frac{12 - x}{x} = \frac{12}{8}$, $x = \frac{24}{5} = 4.8 \text{ cm}$.

Answer: 4.8 cm.