Wednesday, December 29, 2010

Problem 559: Triangle, Cevian, Incenters, Angle, 90 Degrees

Geometry Problem
Click the figure below to see the complete problem 559.

Problem 559: Triangle, Cevian, Incenters, Angle, 90 Degrees.
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Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 9:13 AM
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4 comments:

  1. EditorDecember 29, 2010 at 11:20 AM

    el angulo entre las bisectrices de dos angulos suplementarios es 90° y aquí, por llegar al incentro, tenemos bisectrices.

    ReplyDelete
  2. c .t . e. oDecember 29, 2010 at 11:40 AM

    ADB/2 = (B1 + C)/2 and BDC/2 = (180 - B1 - C)/2
    =>
    ADB/2 + BDC/2 = 90

    I have sent yesterday solution for 558 (delete this )

    ReplyDelete
  3. sourav mishraFebruary 12, 2011 at 4:06 AM

    angle EDB = (1/2)*angle BDA and angle BDF = (1/2)* angle BDC. adding the results we get angle EBD + angle BDF = angle EDF = (1/2)*(angle BDA + angle BDC) = (1/2)*angle ADC = (1/2)*180 degrees = 90 degrees. ADC is a straight line so angle ADC is 180 degrees. therefore angle EDF = 90 degrees.
    Q. E. D.

    ReplyDelete
  4. Sumith PeirisJune 24, 2016 at 8:08 AM

    Since DF bisects < BFA and EF bisects < CFA the result is obvious

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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