Wednesday, June 25, 2008

Elearn Geometry Problem 131



See complete Problem 131
Van Aubel Theorem, Concurrent Cevians, Sum of Ratios. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.
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3 comments:

  1. مجدى الصفتىJuly 5, 2008 at 5:15 PM

    [align=center][img]http://www.al3ez.net/upload/d/essafty_1466.jpg[/img][/align]

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  2. AnonymousSeptember 3, 2008 at 4:18 AM

    BP/PE=[ABCP]/[APC]=[ABP]/[APC]+[PBC]/[APC]=BD/DC+BF/FA

    ReplyDelete
  3. Nilton LapaJune 14, 2012 at 8:13 AM

    Solution to problem 131.
    Let’s use Menelaus’ theorem to triangle ABE, with the transversal FPC, and triangle CBE, with transversal DPA, getting
    (AF/FB).(BP/PE).(EC/CA) = 1
    and (CD/DB).(BP/PE).(EC/CA) = 1.
    Thus BF/FA = (BP/PE).(EC/CA)
    nd BC/DC = (BP/PE).(EA/CA).
    Adding the last equalities,
    BF/FA + BC/DC = (BP/PE).(EC/CA) + (BP/PE).(EA/CA) = BP(EC+EA)/(PE.CA) = (BP.CA)/(PE.CA) = BP/PE.

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