Problem 101. The center of the inscribed circle divides the height of an isosceles triangle, drawn to the base, into segments of 5 cm and 3 cm, measured from the vertex. Find the sides of the triangle.

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-077.jpg?height=523&width=626&top_left_y=1623&top_left_x=361)
Solution. Let in triangle $A B C \quad b=c$ and $\frac{A I}{I L_{1}}=\frac{5}{3}$ (Fig. 102). Then $\frac{b+c}{a}=\frac{2 b}{a}=\frac{5}{3}$, or $a=\frac{6}{5} b$. Then $\quad C L_{1}=B L_{1}=\frac{a}{2}=\frac{3}{5} b . \quad$ By the Pythagorean theorem for triangle $A L_{1} C \quad A L_{1}=\frac{4}{5} b$. Thus:

76

$$
\frac{4}{5} b=8 \text { cm, from which } b=10 \text { cm. Therefore } a=\frac{6}{5} b=12 \text { cm. }
$$

We will demonstrate how the property $\frac{A I}{I L_{1}}=\frac{b+c}{a}$ can be applied in trigonometry.