Consider an isosceles triangle. Let's draw the angle bisectors of the exterior angles of this triangle. What are the angles of the original triangle if the vertex angle of the isosceles triangle formed by the angle bisectors is $88^{\circ}$?
If $A B C$ is the original triangle, and $A_{1} B_{1} C_{1}$ is the triangle formed by the angle bisectors, then

$$
\begin{gathered}
\angle B A C = \angle A B C = 180^{\circ} - \angle A B C_{1} - \angle C B A_{1} = 180^{\circ} - 2 \angle A B C_{1} = \\
= 180^{\circ} - 2\left(90^{\circ} - \frac{\angle A_{1} B_{1} C_{1}}{2}\right) = 88^{\circ}
\end{gathered}
$$

and thus

$$
\angle A C B = 4^{\circ}
$$

(Kálmán Fekete, Arad.)