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triangle sequence, centers of squares on sides of previous triangle

Source: Bulgaria 1970 P4

June 22, 2021
geometryTriangles

Problem Statement

Let δ0=△A0B0C0\delta_0=\triangle A_0B_0C_0 be a triangle. On each of the sides B0C0B_0C_0, C0A0C_0A_0, A0B0A_0B_0, there are constructed squares in the halfplane, not containing the respective vertex A0,B0,C0A_0,B_0,C_0 and A1,B1,C1A_1,B_1,C_1 are the centers of the constructed squares. If we use the triangle δ1=△A1B1C1\delta_1=\triangle A_1B_1C_1 in the same way we may construct the triangle δ2=△A2B2C2\delta_2=\triangle A_2B_2C_2; from δ2=△A2B2C2\delta_2=\triangle A_2B_2C_2 we may construct δ3=△A3B3C3\delta_3=\triangle A_3B_3C_3 and etc. Prove that:
(a) segments A0A1,B0B1,C0C1A_0A_1,B_0B_1,C_0C_1 are respectively equal and perpendicular to B1C1,C1A1,A1B1B_1C_1,C_1A_1,A_1B_1; (b) vertices A1,B1,C1A_1,B_1,C_1 of the triangle δ1\delta_1 lies respectively over the segments A0A3,B0B3,C0C3A_0A_3,B_0B_3,C_0C_3 (defined by the vertices of δ0\delta_0 and δ1\delta_1) and divide them in ratio 2:12:1.
K. Dochev
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