[ Law of Sines ]

In triangle $ABC$, it is known that $\angle A=\alpha, \angle C=\beta, AB=a$; $AD$ is the angle bisector. Find $BD$.

#
Use the Law of Sines.

## Solution

Angle $B A D$ is the exterior angle of triangle $A D C$, so

$$
\angle A D B=\angle D A C+\angle A C B=\alpha / 2+\beta .
$$

By the Law of Sines in triangle $A D B$, we find that

$$
A B / \sin \angle A D B=B D / \sin \angle B A D, \text{ or } a / \sin (\alpha / 2+\beta)=B D / \sin \alpha / 2 \text{, }
$$

from which

$$
B D=a \cdot \sin \alpha / 2 / \sin (\alpha / 2+\beta)
$$

## Answer

$a \cdot \sin \alpha / 2 / \sin (\alpha / 2+\beta)$