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NMC 2008 Problem 3

  • Answer

Let ABCA B C be a triangle and let DD and EE be points on BCB C and CAC A, respectively, such that ADA D and BEB E are angle bisectors of ABCA B C. Let FF and GG be points on the circumcircle of ABCA B C such that AFA F and DED E are parallel and FGF G and BCB C are parallel. Show that

AGBG=AC+BCAB+CB\frac{A G}{B G}=\frac{A C+B C}{A B+C B}

Dataset text NuminaMath olympiads. Quoted with source.

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\frac{A G}{B G}=\frac{A C+B C}{A B+C B}

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