With a parallel translation, triangle ABC became triangle A' BʼC'. Find the angles of triangle ABC if triangle AʼBʼC' is isosceles with with bases A'B' and LB' = 20°.

Solution:
1. Given:
- Triangle A'B'C' is isosceles with base A'B' and \angle B' = 20^\circ.
- Triangle A'B'C' is a translation of triangle ABC.

2. Since triangle A'B'C' is isosceles with base A'B', angles at base A' and base C' are equal:
\angle A' = \angle C'

3. The sum of angles in a triangle is 180^\circ. Since A'B'C' is isosceles, we have:
\angle A' + \angle B' + \angle C' = 180^\circ
2\angle A' + 20^\circ = 180^\circ

4. Solve for \angle A':
2\angle A' = 180^\circ - 20^\circ
2\angle A' = 160^\circ
\angle A' = 80^\circ

5. Since triangle A'B'C' is obtained by parallel translation of triangle ABC, angles of ABC are the same as angles of A'B'C'. Therefore, in triangle ABC:
\angle A = \angle A' = 80^\circ
\angle B = \angle B' = 20^\circ
\angle C = \angle C' = 80^\circ