4. In $\triangle A B C$ with $\angle B=120^{0}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn.

Segment $A_{1} B_{1}$ intersects the bisector $C C_{1}$ at point M. Find the angle $\Delta B_{1} C_{1} M$.

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# Solution.

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 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 $\triangle B B_{1} C M$, the intersection point 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 $\angle B_{1} B M=\angle M B C=30^{\circ}$.

$\angle A B M=\angle C_{1} B B_{1}+\angle B_{1} B M=60^{\circ}+30^{\circ}=90^{\circ}$, and therefore a circle can be circumscribed around quadrilateral $B M B_{1} C_{1}$. Thus, $\angle B_{1} M C_{1}=\angle B_{1} B C_{1}=60^{\circ}$, as they subtend the same arc.

Answer: 60.

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