## [ The sum of the angles of a triangle. The exterior angle theorem.]

The angle bisectors of angles $A$ and $B$ of triangle $A B C$ are equally inclined to the sides $B C$ and $A C$. Find the relationship between angles $A$ and $B$.
Consider two cases. Use the exterior angle theorem of a triangle.

## Solution

Let $A K$ and $B M$ be the angle bisectors of triangle $A B C, P$ be their intersection point, $\angle A=\alpha, \angle B=\beta$.

From the condition, it follows that $\angle A K C=\angle B M C$ or $\angle A K C+\angle B M C=180^{\circ}$. In the first case, $\angle A K C=\angle K A B + \angle K B A=\alpha / 2+\beta$,

$\angle B M C=\angle M B A+\angle M A B=\alpha+\beta / 2$.

Then $\alpha / 2+\beta=\alpha+\beta / 2$. Therefore, $\alpha=\beta$.

In the second case, similarly, we get that $\alpha+\beta=120^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-43.jpg?height=371&width=878&top_left_y=1498&top_left_x=590)

Answer

$\angle A=\angle B$ or $\angle A+\angle B=120^{\circ}$.