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

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# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_b77a4ddf94595f81d054g-07.jpg?height=514&width=1037&top_left_y=1870&top_left_x=538)

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 $\angle B B_{1} C$. Thus, $B_{1} A_{1}$ is the bisector of $\angle B B_{1} C$.

In triangle $B B_{1} C$, point $M$ is the intersection of the bisectors $B_{1} A_{1}$ and $C C_{1}$, and therefore $B M$ is also the bisector of $\angle B_{1} B C$, so $\angle B_{1} B M = \angle M B C = 30^{\circ}$.

Answer: 30.