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.
Wednesday, June 25, 2008
Elearn Geometry Problem 129
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PROOF of PROBLEM 129
ReplyDeleteBB' – 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
solution of 129 :
ReplyDeleteyou 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 .