Find the angles of an isosceles triangle if it is known that the angle between the bisector drawn from its vertex and the bisector drawn to its lateral side is equal to the angle at the vertex.
Let the angle at the base $BC$ of the isosceles triangle $ABC$ be $\alpha$, and the angle bisectors $AD$ and $CE$ intersect at point $M$. Then $\angle A=180^{\circ}-2 \alpha$.

Consider the case when $\angle AMC=\angle A$. Since $\angle AMC=90^{\circ}+1 / 2 \angle B$, then $90^{\circ}+\alpha / 2=180^{\circ}-2 \alpha$, from which $\alpha$ $=36^{\circ}$.

If $\angle AME=\angle A$, then $90^{\circ}-\alpha / 2=180^{\circ}-2 \alpha$, from which $\alpha=60^{\circ}$.

In the first case, the angles of triangle $ABC$ are $108^{\circ}, 36^{\circ}, 36^{\circ}$. In the second case, triangle $ABC$ is equilateral.

## Answer

$36^{\circ}, 36^{\circ}, 108^{\circ}$ or $60^{\circ}, 60^{\circ}, 60^{\circ}$.