## [ Sum of angles in a triangle. Theorem about the exterior angle.]

[ Criteria and properties of an isosceles triangle.]

Given triangle $A B C$. On the extension of side $A C$ beyond point $A$, segment $A D=A B$ is laid out, and beyond point $C$, segment $C E=C B$ is laid out.

Find the angles of triangle $D B E$, knowing the angles of triangle $A B C$.
Apply the exterior angle theorem of a triangle.

## Solution

Angle $B A C$ is the exterior angle of isosceles triangle $D A B$. Therefore, $\angle B E D = 1 / 2 \angle A$. Similarly, $\angle B E D = 1 / 2 \angle C$. Consequently,

$\angle D B E = \angle D B A + \angle A B C + \angle C B E = 1 / 2 \angle A + \angle B + 1 / 2 \angle C = 1 / 2 (\angle A + \angle B + \angle C) + 1 / 2 \angle B = 90^{\circ} + 1 / 2 \angle B$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4fce0705b02fa9e106d4g-11.jpg?height=312&width=1378&top_left_y=1759&top_left_x=340)

## Answer

$1 / 2 \angle A, 1 / 2 \angle C, 90^{\circ} + 1 / 2 \angle B$