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Circle + triangle = hexagon ==> two parallel lines.

Source: Tuymaada 2002, day 1, problem 3. - Author : S. Berlov.

May 3, 2007
geometrycircumcircleincentergeometry proposed

Problem Statement

A circle having common centre with the circumcircle of triangle ABCABC meets the sides of the triangle at six points forming convex hexagon A1A2B1B2C1C2A_{1}A_{2}B_{1}B_{2}C_{1}C_{2} (A1A_{1} and A2A_{2} lie on BCBC, B1B_{1} and B2B_{2} lie on ACAC, C1C_{1} and C2C_{2} lie on ABAB). If A1B1A_{1}B_{1} is parallel to the bisector of angle BB, prove that A2C2A_{2}C_{2} is parallel to the bisector of angle CC.
Proposed by S. Berlov