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SHARYGIN 2010 Problem 23

    A cyclic hexagon ABCDEFABCDEF is such that AB⋅CF=2BC⋅FA,CD⋅EB=2DE⋅BCAB \cdot CF = 2BC \cdot FA, CD \cdot EB = 2DE \cdot BC, and EF⋅AD=2FA⋅DEEF \cdot AD = 2FA \cdot DE. Prove that the lines AD,BEAD, BE and CFCF concur.

    Official text Sharygin Geometry Olympiad 2010 official paper, page 3. Quoted with source.

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