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.
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Planometry
Terms: Planometry