Wednesday, June 25, 2008

Elearn Geometry Problem 130



See complete Problem 130
Triangle, Concurrent Cevians, Sum of Ratios. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 11:23 AM
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4 comments:

  1. c .t . e. oDecember 14, 2009 at 12:09 PM

    can I know what is a cevian, please

    ReplyDelete
  2. Antonio GutierrezDecember 16, 2009 at 9:38 AM

    A Cevian is any line segment which joins a vertex of a triangle with a point on the opposite side (or its extension).

    ReplyDelete
  3. c .t . e. oFebruary 10, 2010 at 10:13 AM

    from an cevian theorem

    PD/AD + PE/BE + PF/FC = 1
    =>
    (AD - PA)/AD + (BE - PB)/BE + (FC - PC)/FC = 1

    AD/AD - PA/AD + BE/BE - PB/BE + FC/FC - PC/FC = 1

    1 - PA/AD + 1 - PB/BE + 1 - PC/FC = 1

    PA/AD + PB/BE + PC/FC = 2
    -------------------------------------------

    ReplyDelete
  4. rv.littlemanMay 24, 2020 at 1:32 PM

    PA/AD=PBA/ABD=PCA/ACD=(PBA+PCA)/(ABD+ACD)=(PBA+PCA)/ABC
    PB/BE=PBC/BEC=PBA/BEA=(PBC+PBA)/(BEC+BEA)=(PBC+PBA)/ABC
    PC/CF=PBC/CFB=PAC/CAF=(PBC+PAC)/(CFB+CAF)=(PBC+PAC)/ABC
    Therefore
    PA/AD+PB/BE+PC/CF=(PBA+PCA)/ABC+(PBC+PBA)/ABC+(PBC+PAC)/ABC
    =(2PBA+2PCA+2PBC)/ABC=2(PBA+PCA+PBC)/ABC=2ABC/ABC=2

    ReplyDelete

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