Wednesday, June 25, 2008

Elearn Geometry Problem 129



See complete Problem 129
Triangle, Concurrent Cevians, Sum of Ratios. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

2 comments:

  1. PROOF of PROBLEM 129

    BB' – altitude of ∆ABC
    PE' -– altitude of ∆APC
    PE/BE=(PE^')/(BB^' )
    In the same way:
    PD/AD=(PD^')/AA',PF/CF=(PF^')/CC'
    PD/AD+PE/BE+PF/CF=(PD^')/(AA^' )+(PE^')/(BB^' )+(PF^')/(CC^' )
    S∆ABC=(PD^'∙BC+PE^'∙AC+PF^'∙AB)/2=(AA^'∙BC)/2=(BB^'∙AC)/2=(CC^'∙AB)/2→
    (PD'∙BC)/(AA'∙BC)+(PE'∙AC)/(BB'∙AC)+(PF'∙AB)/(CC'∙AB)=S∆ABC/S∆ABC=1

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  2. solution of 129 :
    you can see that : PD\AD = [PBD]/[ABD] = [PCD]/[ACD] = ([PBD]+[PCD])/([ABD]+[ACD])
    = [BPC]/[ABC]
    and after sum the other we have : \sum PD\AD = [ABC]/[ABC] .

    adil azrou .

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