Sign of similarity of triangles on three sides

Theorem

If the sides of one triangle are proportional to the sides of another triangle, then such triangles are similar.



Evidence.

Let the triangles ABC and A1B1C1 ∠ CBA = ∠ C1B1A1, AB = k * A1B1, BC = k * B1C1, AC = k * A1C1. Let us prove that Δ ABC is similar to Δ A1B1C1.
Subject Δ A1B1C1 homothety with coefficient k. Get some Δ A2B2C2.
Δ A2B2C2 = Δ ABC on the third sign of equality of triangles (A2С2 = k * A1С1 = AC, A2B2 = k * A1B1 = AB, B2С2 = k * B1С1 = BS, 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.