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

Labels:
cevian,
concurrent,
ratio,
triangle

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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 .