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BALKAN-SL 2018 G1

  • Answer

In an acute triangle ABCA B C, the midpoint of the side BCB C is MM and the centers of the excircles relative to MM of the triangles AMBA M B and AMCA M C are DD and EE respectively. The circumcircle of the triangle ABDA B D meets line BCB C at BB and FF. The circumcircle of the triangle ACEA C E meets line BCB C at CC and GG. Prove that BF=CGB F=C G. (Romania)

Dataset text NuminaMath olympiads. Quoted with source.

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B F=C G

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