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

[ Area of a triangle (using two sides and the angle between them).]

In triangle $ABC$, it is known that $AB=a, AC=b, \angle BAC=120^{\circ}$. Find the bisector $AM$.
Since $S_{\triangle A B C}=S_{\triangle A B M}+S_{\triangle A C M}$, then

$$
\frac{1}{2} A B \cdot A C \sin \angle B A C=\frac{1}{2} A B \cdot A M \sin \angle B A M+\frac{1}{2} A M \cdot A C \sin \angle C A M
$$

or

$\frac{1}{2} a b \sin 120^{\circ}=\frac{1}{2} a \cdot A M \sin 60^{\circ}+\frac{1}{2} b \cdot A M \sin 60^{\circ}$,

$\frac{1}{2} a b \cdot \frac{\sqrt{3}}{2}=\frac{1}{2} a \cdot A M \cdot \frac{\sqrt{3}}{2}+\frac{1}{2} b \cdot A M \cdot \frac{\sqrt{3}}{2}, a b=A M(a+b)$.

Therefore, $A M=\frac{a b}{a+b}$.

## Answer

$\frac{a b}{a+b}$.

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