3.3. The base of an isosceles triangle is equal to $a$, and the angle at the vertex is $\alpha$. Find the length of the bisector drawn to the lateral side.
3.3. By the condition, $A B=A C, B C=a, \angle B A C=\alpha, \angle A B B_{1}=\angle B_{1} B C$ (Fig. 3.7). We have $\angle A B C=\angle A C B=\frac{\pi}{2}-\frac{\alpha}{2}, \angle B_{1} B C=\frac{\pi}{4}-\frac{\alpha}{4}$,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-501.jpg?height=362&width=633&top_left_y=1088&top_left_x=276)

Fig. 3.7

$\angle B B_{1} C=\pi--\left(\frac{\pi}{2}-\frac{\alpha}{2}+\frac{\pi}{4}-\frac{\alpha}{4}\right)=\frac{\pi}{4}+\frac{3 \alpha}{4}$. From $\triangle B B_{1} C$ by the sine rule we find

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