## Task 3.

In triangle $A B C$, the angle at vertex $B$ is $120^{\circ}$. Let $A_{1}, B_{1}, C_{1}$ be points on the sides $\overline{B C}$, $\overline{C A}$, $\overline{A B}$, respectively, such that $A A_{1}$, $B B_{1}$, $C C_{1}$ are the angle bisectors of triangle $A B C$. Determine the angle $\varangle A_{1} B_{1} C_{1}$.
## Solution.

Let $X$ be any point on the extension of side $\overline{A B}$ beyond vertex $B$. Notice that $\varangle A B B_{1} = \varangle B_{1} B C = \varangle C B X = 60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_eef7af868a6d1038a7ecg-09.jpg?height=706&width=1328&top_left_y=1229&top_left_x=364)

This means that point $A_{1}$ lies on the angle bisector of $\varangle B_{1} B X$, and it also lies on the angle bisector of $\varangle B A C$.

From this, it follows that point $A_{1}$ is the excenter of triangle $A B B_{1}$ opposite vertex $A$, and thus it lies on the angle bisector of $\varangle B B_{1} C$.

This proves that line $B_{1} A_{1}$ is the angle bisector of $\varangle B B_{1} C$. Similarly, it can be shown that line $B_{1} C_{1}$ is the angle bisector of $\varangle A B_{1} B$.

Therefore, we have

$$
\varangle A_{1} B_{1} C_{1} = \varangle A_{1} B_{1} B + \varangle B B_{1} C_{1} = \frac{1}{2} \varangle C B_{1} B + \frac{1}{2} \varangle B B_{1} A = \frac{1}{2} \cdot 180^{\circ} = 90^{\circ}
$$