Sunday, August 8, 2010

Problem 495: Triangle, 120 Degrees, Angle Bisectors

Geometry Problem
Click the figure below to see the complete problem 495 about Triangle, 120 Degrees, Angle Bisectors.

Problem 495. Triangle, 120 Degrees, Angle Bisectors.
See also:
Complete Problem 495

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 9:48 AM
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6 comments:

  1. VihaanAugust 8, 2010 at 6:53 PM

    From Tr. DAC Ang.BDF = (60 + C)/2 = 30 + C/2 = x + C/2 from Tr. DFC. Hence x = 30 deg.
    Vihaan, Dubai

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  2. Peter TranAugust 8, 2010 at 9:36 PM

    To Vihaan

    You mention that “From Tr. DAC Ang.BDF = (60 + C)/2 “ . I am not sure how do you get this
    I understand that DF is not angle bisector of angle BDA per problem statement.


    Peter Tran

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  3. VihaanAugust 8, 2010 at 10:25 PM

    BF is the external bisector of Ang. DAC and F is the intersection of BF & CF which is internal bisector of ang. C. Hence DF is the external bisector of ang. ADC.
    Would that stand to reason?
    Vihaan

    ReplyDelete
  4. Peter TranAugust 8, 2010 at 11:55 PM

    Thanks Vihaan

    Peter Tran

    ReplyDelete
  5. AnonymousNovember 26, 2013 at 9:52 AM

    Angle Bisector of <DAC hits FC at point P which is the incenter of triangle DAC. If <FCA is a, then <AFC is 60-a. <ADP=<CDP=(180-(60+2a))/2=60-a making FDAP an circumscribed quadrilateral. Therefore <CFD=<DAP=30

    Ivan Bazarov

    ReplyDelete
  6. Sumith PeirisJuly 13, 2015 at 2:42 AM

    The result follows upon observing that FD is indeed the external bisector in Tr. ADC

    ReplyDelete

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