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

[ Criteria and properties of an isosceles triangle. ]

Given an isosceles triangle $A B C(A B=A C)$. On the extension of side $A C$ beyond point $C$, segment $C D$ is laid out, equal to $B C$. It turns out that $B D=A B$.

Find the angles of triangle $A B C$.
Let $\angle B=\angle C=\alpha$. By the exterior angle theorem of a triangle, $\angle C B D=\angle C D B=\alpha / 2$.

The base angles of the isosceles triangle $A B D$ are equal, so $\angle B A C=\angle B A D=\angle A D B=\alpha / 2$.

Applying the theorem of the sum of angles in a triangle to triangle $A B C$, we get the equation $\alpha+\alpha+\alpha / 2=180^{\circ}$, from which $\alpha=72^{\circ}$.

## Otвет

$36^{\circ}, 72^{\circ}, 72^{\circ}$.