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. Find $\angle C_{1} B_{1} A_{1}$.
# Solution:

Extend side $A B$ beyond point $\mathrm{B}$, 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 $\angle B A C$, which means it is equidistant from its sides.

We get that $A_{1}$ is equidistant from sides $B_{1} B$ and $B_{1} C$, which means it lies on the bisector of $\angle B B_{1} C$. Similarly, we prove that $B_{1} C_{1}$ is the bisector of $\angle A B_{1} B$.

Therefore, $\angle C_{1} B_{1} A_{1}=90^{\circ}$, as the angle between the bisectors of adjacent angles.

Answer: $\angle C_{1} B_{1} A_{1}=90^{\circ}$.