8. 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. Segment $A_{1} B_{1}$ intersects the angle bisector $C C_{1}$ at point M. Find the degree measure of angle $B_{1} M C_{1}$.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_b77a4ddf94595f81d054g-04.jpg?height=457&width=905&top_left_y=251&top_left_x=610)

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 thus is equidistant from its sides, we get that $A_{1}$ is equidistant from sides $B_{1} B$ and $B_{1} C$, and therefore 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.

In triangle $B B_{1} C$, point $M$ is the intersection of the bisectors $B_{1} A_{1}$ and $C C_{1}$, and thus $B M$ is also the bisector of $\angle B_{1} B C$, so $\angle B_{1} B M=\angle M B C=30^{\circ}$. Further, $\angle A B M=\angle C_{1} B B_{1}+\angle B_{1} B M=60^{\circ}+30^{\circ}=90^{\circ}$, which means a circle can be circumscribed around quadrilateral $B M B_{1} C_{1}$. Therefore, $\angle B_{1} M C_{1}=\angle B_{1} B C_{1}=60^{\circ}$, as they subtend the same arc.

Answer: 60.