Let △ABC be an equilateral triangle. Let C1 and C2 be on AB, B1 and B2 on AC and A1 and A2 on BC such that A1A2=B1B2=C1C2. Let A2B1 and B2C1, B2C1 and C2A1, C2A1 and A2B1 intersect at E, F, G correspondently. Prove that the triangle formed by the segments B1A2, A1C2 and C1B2 is similar to △EFG.
Solution
Let us denote the triangle formed by the segments B1A2, A1C2 and C1B2 with △A3B3C3. Let P be a point of the interior of the triangle △EFG such that C1C2PB2 is a parallelogram. Then △B2PB1 is equilateral, hence PA1A2B1 is a parallelogram. From the above observations we get that PC2∥EF, PA1∥EG. Now it's obvious that △PC2A1∼△EFG and because △PC2A1≅△A3B3C3 we conclude that △A3B3C3∼△EFG.
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