12.043. Through the vertex $A$ of an isosceles acute triangle $ABC$ and the center of the circumscribed circle around this triangle, a line is drawn, intersecting the side $BC$ at point $D$. Find the length of $AD$, if $AB=BC=b$ and $\angle ABC=\alpha$.
Solution.

Given $A B=B C=b, \angle A B C=\alpha, O$ is the center of the circumscribed circle. Consider $\triangle B E O, \angle B E O=90^{\circ}, B O=\frac{b}{2 \cos \frac{\alpha}{2}}=A O$, $O E=B E \cdot \operatorname{tg} \frac{\alpha}{2}=\frac{b}{2} \operatorname{tg} \frac{\alpha}{2}$ (Fig. 12.45). Since $A O=B O$, we have $\angle B A O=\angle A B O=\frac{\alpha}{2}, \alpha+\frac{\alpha}{2}+\angle B D A=180^{\circ}$, and $\angle B D A=180^{\circ}-\frac{3 \alpha}{2}$. 

$$
\begin{aligned}
& \triangle O E D, \angle O E D=90^{\circ}, O D=\frac{O E}{\sin \angle B D A}=\frac{b}{2} \frac{\operatorname{tg} \frac{\alpha}{2}}{\sin \frac{3 \alpha}{2}} \\
& A D=A O+O D=\frac{b}{2 \cos \frac{\alpha}{2}}+\frac{b \operatorname{tg} \frac{\alpha}{2}}{2 \sin \frac{3 \alpha}{2}}=\frac{b}{2}\left(\frac{\sin \frac{3 \alpha}{2}+\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} \cdot \sin \frac{3 \alpha}{2}}\right)=
\end{aligned}
$$

$$
\begin{aligned}
& =b \cdot \frac{\sin \alpha \cdot \cos \frac{\alpha}{2}}{\cos \frac{\alpha}{2} \cdot \sin \frac{3 \alpha}{2}}=\frac{b \sin \alpha}{\sin \frac{3 \alpha}{2}} \\
& \text { Answer: } \frac{b \sin \alpha}{\sin \frac{3 \alpha}{2}}
\end{aligned}
$$