3.18. In an isosceles triangle, the angle at the base is equal to $\alpha$, the radius of the inscribed circle is $r$. A line is drawn through the vertex of the angle at the base and the center of the inscribed circle. Find the segment of this line that is contained within the triangle.
3.18. By the condition, $A B=A C, O$ is the center of the inscribed circle in $\triangle A B C$, $\angle A B C=\alpha, O D \perp B C, O D=r$, $B O \cap A C=B_{1}$ (Fig. 3.20). Since $B B_{1}$ is the bisector of $\angle A B C$, then $\angle B_{1} B C=\frac{\alpha}{2}, \angle B B_{1} C=\pi-\frac{3 \alpha}{2}$. From $\triangle O D B$ we find $B D=r \operatorname{ctg} \frac{\alpha}{2}$. Considering that $D$ is the point of tangency of the base $B C$ of the isosceles triangle $A B C$ with the circle, we get $B C=2 B D=2 r \operatorname{ctg} \frac{\alpha}{2}$. Finally, in $\triangle B B_{1} C$ by the Law of Sines we have

$$
\frac{B B_{1}}{\sin \alpha}=\frac{2 r \operatorname{ctg} \frac{\alpha}{2}}{\sin \left(\pi-\frac{3 \alpha}{2}\right)}, \text { i.e. } B B_{1}=\frac{2 r \operatorname{ctg} \frac{\alpha}{2} \sin \alpha}{\sin \frac{3 \alpha}{2}}=\frac{4 r \cos ^{2} \frac{\alpha}{2}}{\sin \frac{3 \alpha}{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-511.jpg?height=288&width=515&top_left_y=400&top_left_x=56)

Fig. 3.21