6. In triangle $A B C$ with angle $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Find the degree measure of angle $C_{1} B_{1} A_{1}$.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_9fc9f23f10a928994848g-02.jpg?height=460&width=936&top_left_y=1849&top_left_x=640)

Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B B_{1} B K$, 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 is therefore equidistant from its sides, we get that $A_{1}$ is equidistant from sides $B_{1} B$ and $B_{1} C$, and thus lies on the bisector of $\angle B B_{1} C$. Therefore, $B_{1} A_{1}$ is 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, $\angle C_{1} B_{1} A_{1}=90^{0}$, as the angle between the bisectors of adjacent angles.

Answer: 90.