Sunday, May 29, 2011

Problem 609: Triangle, Perpendiculars, 90 Degrees, Collinearity

Geometry Problem
Click the figure below to see the complete problem 609.

Problem 609: Triangle, Perpendiculars, 90 Degrees, Collinearity.

Posted by Antonio Gutierrez at 7:49 AM
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1 comment:

  1. PravinMay 31, 2011 at 8:37 PM

    Let(XYZ) denote the area of triagle XYZ.

    AG/GC
    =(ADG)/(GDC)=(AD.GD.sin ADG)/(GD.CD.sin GDC)
    =(AD/CD)(sin ADG /sin GDC)

    CF/FB
    =(CDF)/(FDB)=(CD.FD.sin CDF)/(FD.BD.sin FDB)
    =(CD/BD)(sin CDF / sin FDB)

    BE/EA
    =(BDE)/(EDA)=(BD.ED.sin BDE)/(ED.AD.sin EDA)
    =(BD/AD)(sin BDE / sin EDA)

    From perpendicularity conditions,
    sin ADG = sin FDB, sin GDC = sin BDE, and
    sin CDF = sin EDA

    Follows
    (AG/GC)(CF/FB)(BE/EA)= 1 numercally

    Applying Converse of Menelau's Theorem,
    E,F,G are collinear

    ReplyDelete

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