In triangle $A B C$, angle $A$ is equal to $\alpha, A B=A C=b$. A line is drawn through vertex $B$ and the center of the circumscribed circle until it intersects line $A C$ at point $D$. Find $B D$.
Let $O$ be the center of the circumcircle of triangle $ABC$. Then $OA = OB$, $\angle ABD = \angle OAB = \alpha / 2$,

$$
\angle ADB = 180^{\circ} - \angle ABD - \angle BAD = 180^{\circ} - \alpha / 2 - \alpha = 180^{\circ} - 3 \alpha / 2.
$$

Applying the Law of Sines to triangle $ABD$, we get $\frac{BD}{\sin \alpha} = \frac{b}{\sin \left(180^{\circ} - \frac{3 \alpha}{2}\right)}$, which means

$$
BD = \frac{b \sin \alpha}{\sin \frac{3 \alpha}{2}}
$$

## Answer

$$
\frac{b \sin \alpha}{\sin \frac{3 \alpha}{2}}
$$