Problem 2. The bisector of the exterior angle at the base of an isosceles triangle intersects the bisector of the exterior angle at the vertex of the same triangle at an angle of $80^{\circ}$. Determine the angles of the triangle.
Solution. From the properties of interior and exterior angles in a triangle, it follows

![](https://cdn.mathpix.com/cropped/2024_06_05_1d413d45628348221ff6g-3.jpg?height=648&width=772&top_left_y=1154&top_left_x=477)

From $\triangle C B D$ it follows:

$$
\begin{gathered}
\frac{\gamma_{1}}{2}+\frac{\beta_{1}}{2}+80^{\circ}=180^{\circ} \\
\alpha+90^{\circ}-\frac{\alpha}{2}=100^{\circ} \\
\frac{\alpha}{2}=10^{\circ} ; \alpha=20^{\circ}
\end{gathered}
$$

and $\gamma=180^{\circ}-2 \alpha=140^{\circ}$.