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

Posted by Antonio Gutierrez at 4:12 PM
Labels: , , , ,

3 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

    ReplyDelete

  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

    ReplyDelete

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

    ReplyDelete

Subscribe to: Post Comments (Atom)