Wednesday, July 2, 2008

Elearn Geometry Problem 133



See complete Problem 133
Triangle, Angle Bisectors, Collinear Points. 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 5:35 PM

    AE is an interior bisector of angle A
    BE/EC = AB/AC
    CD is an interior bisector of angle C
    AD/DB = AC/BC
    AE is an exterior bisector of angle B
    FA/FC = AB/BC
    therefore
    BD/DA * AF/FC * CE/EB = BC/AC * AB/BC * AC/BC = 1
    therefore
    D, E, F are collinear points
    Magdy Essafty

    ReplyDelete
  2. EditorSeptember 17, 2013 at 3:32 PM

    - P is on AC such that BP bisects <B.
    - It is known that P is the harmonic conjugate of F w/r AC.
    In conclusion, F-E-D must be colinear.

    ReplyDelete
  3. AnonymousMarch 16, 2021 at 3:06 PM

    Considering Pappu's hexagon Theorem A,D,B and A,C,F are two sets of collinear points. As B,E,C are collinear D,E,F should be collinear to form the pappus line AE.

    ReplyDelete

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