Wednesday, June 25, 2008

Elearn Geometry Problem 128



See complete Problem 128
Incenter of a Triangle, Angle Bisectors, Sum of Ratios. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

5 comments:

  1. BI is a bisector of angle B
    AI/ID = AB/BD
    CI is a bisector of angle C
    AI/ID = AC/CD
    AI/ID = AB/BD = AC/CD
    AI/ID = ( AB + AC )/( BD + CD )
    AI/ID = ( AB + AC )/BC
    ( AI + ID )/ID = ( AB + BC + CA )/BC
    ID/AD = BC/( AB + BC + CA )
    Similarly
    IE/BE = AC/( AB + BC + CA )
    IF/CF = AB/( AB + BC + CA )
    Therefore by addition
    ID/AD + IE/BE + IF/CF = 1
    Magdy Essafty

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  2. using theorem of van aubel we have :
    AI\ID = AF\FB + AE\EC = AC\BC + AB\BC = (AC+AB)\BC
    now AI\ID + 1 = AD\ID = s\BC (and s = AB + AC + BC)
    so : ID\AD = BC\s
    and similar after sum we get : ID\AD + IE\BE + IF\FC = BC\s + AC\s + AB\s = s\s = 1 .
    so we are done !!
    adil azrou from morocco .

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  3. To Anonymous:
    As far as I know the theorem of Van Aubel, I can't see how it is applyed in your solution posted in April 13, 2012. Could you please explain it?
    Thanks.

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  4. There is a simpler proof using areas:
    ID/AD
    = (IDB)/(ADB)= (IDC)/(ADC)
    = [(IDB)+(IDC)]/[(ADB)+(ADC)]=(IBC)/(ABC)
    = (IBC)/(ABC)
    Similarly
    IE/BE = (IAC)/(ABC) and
    IF/CF = (IAB)/(ABC)
    Hence
    ID/AD + IE/BE + IF/CF
    =[(IBC)+(IAC)+(IAB)]/(ABC)=ABC)/(ABC)
    = 1
    Note:In fact PD/AD + PE/BE + PF/CF = 1
    holds for any three concurrent cevians APD, BPE, CPF

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  5. To Anonymous: My previous question is already solved. I've learned more about Van Aubel. Thanks.

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