5.5. In triangle $A B C$ with an angle of $120^{\circ}$ at vertex $A$, the angle bisectors $A A_{1}, B B_{1}$ and $C C_{1}$ are drawn. Find the angle $C_{1} A_{1} B_{1}$.
5.5. Answer. $90^{\circ}$.

Let the length of side $AB$ be denoted by $c$, $AC$ by $b$, and $BC$ by $a$. First, we will prove that $AA_1 = \frac{bc}{b+c}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_20eb9f6ef2c568605a63g-45.jpg?height=304&width=556&top_left_y=1207&top_left_x=358)

Fig. 5.5.

Extend $AA_1$ and draw a line $BK$ parallel to $AC$ (see the figure). In triangle $ABK$, angles $A$ and $K$ are both $60^{\circ}$, so it is equilateral and thus $AK = BK = c$. Now, from the similarity of triangles $BA_1K$ and $AA_1C$, we get $\frac{c - AA_1}{AA_1} = \frac{c}{b}$, from which it follows that $AA_1 = \frac{bc}{b+c}$.

Since the bisector of an internal angle of a triangle divides the opposite side into segments proportional to the adjacent sides, we have $\frac{BA_1}{a - BA_1} = \frac{c}{b}$, from which $BA_1 = \frac{ac}{b+c}$. Therefore, $AA_1 : BA_1 = b : a$. The base $C_1$ of the bisector $CC_1$ divides side $BA$ in the same ratio. Hence, $A_1C_1$ is the bisector of angle $A_1$ in triangle $BA_1A$. Similarly, it can be shown that $A_1B_1$ is the bisector of angle $AA_1C$. Therefore, $\angle C_1A_1B_1 = \left(\angle BA_1A + \angle AA_1C\right) / 2 = 90^{\circ}$. 5.6. Answer. $36^{\circ}, 36^{\circ}, 108^{\circ}$.

It is clear that this triangle $ABC$ is isosceles. Let $AB$ be its base, and $O$ be the center of the circumscribed circle. Prove that $\angle ACO = \angle CAO = \frac{3}{2} \angle CAB$.