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Geometry ProblemClick the figure below to see the complete problem 613.
Triangle Transversal By MenelauABC NEF => AN.CF.BE = NC.FB.EACGB DMF => CM.GD.BF = MG.DB.FCBGA EHD => GH.AE.BD = HA.EB.DGMultiply out respective sides of the equations, we get AN.CF.BE.CM.GD.BF.GH.AE.BD = NC.FB.EA.MG.DB.FC.HA.EB.DGCancelling common factors:AN.CM.GH = NC.MG.HAHence again by Menelau:N, H, M are collinear
Refer to my solution of Problem 613:Last but one sentence should read as:"Hence again by Menelau applied to triangle AGC"Remark:I haven't used perpendicularity conditions.Please point out flaw, if any, in my solution
To Pravin:Thanks for your comments, in effect, the perpendicularity conditions is not necessary.
By Menelaus in triangle ABC with transversal NEF, NA/NC*FC/BF*BE/EA=1.FC/BF=[DFC]/[DBF] and BE/EA=[DBE]/[DEA].However [DBE]/[DBF]=[DGE]/[DGF].Plugging these into original equation yields NA/NC*[DFC]/[DGF]*[DGE]/[DEA]=1.[DFC]/[DGF]=MC/GM and [DGE]/[DEA]=GH/HA, meaning that NA/NC*MC/GM*GH/HA=1 which is converse of Menelaus in triangle AGC and points N, H, and M.