[ Auxiliary area. The area helps to solve the task, $]

Angles between bisectors

The sides of the triangle are 3 and 6, and the angle between them is $60^{\circ}$. Find the bisector of the triangle, drawn from the vertex of this angle.

#
Let $AD$ be the bisector of triangle $ABC$, where $AB=6, AC=3, \angle BAC=60^{\circ}$.

## First method.

Let $AD=x$. Then

$$
S_{\triangle ABC}=\frac{1}{2} AB \cdot AC \sin 60^{\circ}=\frac{1}{2} \cdot 6 \cdot 3 \cdot \frac{\sqrt{3}}{2}=\frac{9 \sqrt{3}}{2}
$$

On the other hand,

$$
S_{\triangle ABC}=S_{\triangle ABD}+S_{\triangle ACD}=\frac{1}{2} AB \cdot AD \sin 30^{\circ}+\frac{1}{2} AC \cdot AD \sin 30^{\circ}=\frac{1}{2} \cdot 6 \cdot x \cdot \frac{1}{2}+\frac{1}{2} \cdot 3 \cdot x \cdot \frac{1}{2}=\frac{9}{4} x
$$

From the equation $\frac{9}{4} x=\frac{9 \sqrt{3}}{2}$, we find that $x=2 \sqrt{3}$.

## Second method.

Notice that triangle $ABC$ is a right triangle. Then triangle $ACD$ is also a right triangle, and $\angle CAD=30^{\circ}$. Therefore,

$$
AD=\frac{AC}{\cos \angle CAD}=\frac{3}{\cos 30^{\circ}}=2 \sqrt{3}
$$

## Answer

## Problem