Sign of the similarity of triangles at two angles

Theorem

If two angles of one triangle are equal to two angles of another triangle, then such triangles are similar.



Evidence.

Let the triangles ABC and A1B1C1 CAB = ∠ C1A1B1, ABC = ∠ A1B1C1. Let us prove that Δ ABC is similar to Δ A1B1C1.
Let k = AB / A1B1. Subject Δ A1B1C1 homothety with coefficient k. Get some Δ A2B2C2.
Δ A2B2C2 = Δ ABC by the second sign of equality of triangles (∠ C2A2B2 = ∠ C1A1B1 = ∠ CAB, ∠ A2B2C2 = ∠ A1B1C1 = ∠ ABC because the similarity transformation preserves angles, A2B2 = k * A1B1 = AB, by condition).
Triangles A1B1C1 and A2B2C2 are homothetic, hence similar. Δ A2B2C2 = Δ ABC, therefore are similar too, which means that triangles A1B1C1 and ABC are similar. The theorem is proved.