8. In triangle $A B C$ with $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Segment $A_{1} B_{1}$ intersects the angle bisector $C C_{1}$ at point M. Find $\angle C B M$.

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# Solution:

Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$.

Considering that point $A_{1}$ lies on the bisector of $\angle B A C$, and therefore is equidistant from its sides.

We obtain that $A_{1}$ is equidistant from sides $B_{1} B$ and $B_{1} C$, which means it lies on the bisector of $\angle B B_{1} C$. In triangle $B B_{1} C M$ - point $A_{1}$ is the intersection of the bisectors $B_{1} A_{1}$ and $C C_{1}$, and therefore $B M$ is also the bisector of $\angle B_{1} B C$, hence $\angle B_{1} B M=\angle M B C=30^{\circ}$

Answer: $\angle C B M=30^{\circ}$.