In triangle $ABC$, the angle at vertex $B$ is $120^o$. Let $A_1, B_1, C_1$ be points on the sides $BC, CA, AB$ respectively  such that $AA_1, BB_1, CC_1$ are bisectors of the angles of triangle $ABC$. Determine the angle $\angle A_1B_1C_1$.
1. Given that $\angle B = 120^\circ$ in $\triangle ABC$, we need to determine the angle $\angle A_1B_1C_1$ where $A_1, B_1, C_1$ are points on the sides $BC, CA, AB$ respectively such that $AA_1, BB_1, CC_1$ are the angle bisectors of $\triangle ABC$.

2. First, recall that the angle bisectors of a triangle intersect at the incenter, which is the center of the circle inscribed in the triangle. The incenter is equidistant from all three sides of the triangle.

3. Since $\angle B = 120^\circ$, the angle bisectors $AA_1, BB_1, CC_1$ divide $\angle B$ into two angles of $60^\circ$ each. Therefore, $\angle AB_1B = 60^\circ$ and $\angle CB_1B = 60^\circ$.

4. The incenter $I$ of $\triangle ABC$ is the intersection of the angle bisectors. Since $I$ is equidistant from all sides, it forms right angles with the points of tangency on each side of the triangle.

5. To find $\angle A_1B_1C_1$, we need to consider the properties of the triangle formed by the points of tangency of the incircle with the sides of $\triangle ABC$. These points of tangency form a right triangle because the incenter is equidistant from all sides and forms right angles with the points of tangency.

6. Specifically, in $\triangle ABC$, the points $A_1, B_1, C_1$ are the points where the incircle touches the sides $BC, CA, AB$ respectively. The angles at these points are right angles because the radius of the incircle is perpendicular to the sides of the triangle at the points of tangency.

7. Therefore, $\angle A_1B_1C_1$ is a right angle because it is formed by the intersection of the angle bisectors at the incenter, which creates right angles with the sides of the triangle.

8. Hence, $\angle A_1B_1C_1 = 90^\circ$.

The final answer is $\boxed{90^\circ}$.