10.241. In an isosceles triangle with a lateral side equal to $b$, the angle bisectors of the angles at the base are drawn. The segment of the line between the points of intersection of the bisectors with the lateral sides is equal to $m$. Determine the base of the triangle.
Solution.

Since $\triangle A B C$ is isosceles, then $C E=A D$ and $D E \| A C$ (Fig. 10.48).

From the similarity of triangles $A B C$ and $D B E$, it follows that $\frac{m}{A C}=\frac{B D}{b}$. Considering that $C D$ is the bisector, we have $\frac{B D}{D A}=\frac{B C}{A C}$. Let $A C=x, B D=y$. Then we have $\left\{\begin{array}{l}\frac{m}{x}=\frac{y}{b}, \\ \frac{y}{b-y}=\frac{b}{x},\end{array}\right.$ from which $y=\frac{b m}{x}$ and $\frac{b m}{x\left(b-\frac{b m}{x}\right)}=\frac{b}{x}$, i.e., $x=\frac{b m}{b-m}$.

Answer: $\frac{b m}{b-m}$.