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3

Part of 2009 JBMO TST - Macedonia

Problems(1)

Macedonian JBMO TST 2009 - Problem 3

Source: Macedonian JBMO TST

9/23/2012
Let △ABC \triangle ABC be equilateral. On the side AB AB points C1 C_{1} and C2 C_{2} , on the side AC AC points B1 B_{1} and B2 B_{2} are chosen, and on the side BC BC points A1 A_{1} and A2 A_{2} are chosen. The following condition is given : A1A2 A_{1}A_{2} = B1B2 B_{1}B_{2} = C1C2 C_{1}C_{2} . Let the intersection lines A2B1 A_{2}B_{1} and B2C1 B_{2}C_{1} , B2C1 B_{2}C_{1} and C2A1 C_{2}A_{1} and C2A1 C_{2}A_{1} and A2B1 A_{2}B_{1} are E E , F F , and G G respectively. Show that the triangle formed by B1A2 B_{1}A_{2} , A1C2 A_{1}C_{2} and C1B2 C_{1}B_{2} is similar to △EFG \triangle EFG .
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