Wednesday, June 1, 2011

Problem 613: Altitude of a Triangle, Perpendicular, Collinear Points

Geometry Problem
Click the figure below to see the complete problem 613.

Problem 613: Altitude of a Triangle, Perpendicular, Collinear Points

4 comments:

  1. Triangle Transversal By Menelau
    ABC NEF => AN.CF.BE = NC.FB.EA
    CGB DMF => CM.GD.BF = MG.DB.FC
    BGA EHD => GH.AE.BD = HA.EB.DG
    Multiply 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.DG
    Cancelling common factors:
    AN.CM.GH = NC.MG.HA
    Hence again by Menelau:
    N, H, M are collinear

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  2. 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

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  3. To Pravin:
    Thanks for your comments, in effect, the perpendicularity conditions is not necessary.

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  4. 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.

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