Tuesday, December 20, 2011

Problem 707: Triangle, Excenters, Angle Bisectors, Angle, Measurement

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 707.

Online Geometry Problem 707: Triangle, Excenters, Angle Bisectors, Angle, Measurement.
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2 comments:

  1. Peter TranDecember 20, 2011 at 4:29 PM

    http://img84.imageshack.us/img84/8180/problem707.png

    Make projections of D,E and A over BC ( see picture)
    Since m(ADE)+m(AED)=142 so m(DAE)=180-142=38
    Since AD and AE are angle bisectors of ( BAE) and (DAC) so m(BAD)=m(DAE)=m(EAC)=38
    Note that m(CEN)=1/2m(ACG) and m(BDM)=1/2m(ABF)
    And m(ACG)+m(ABF)=180-3*38=66
    . m(ADB)+m(AEC)=m(BDM)+m(CEN)+m(MDF)+m(GEN)
    = .5*66+38=71

    Peter Tran

    ReplyDelete
  2. PravinDecember 20, 2011 at 8:13 PM

    ∠A is trisected by AD,AE
    ∠DAE=A/3=38°(supplement of ∠ADE+∠AED=142°)
    ∠ABD is 90°+B/2 and ∠ACE is 90°+C/2 and their sum is 270°-A/2=213°
    ∠BAD+∠CAE is 38°+38°=76°
    The sum of all the angles in ∆les ABD,ACE is 360°
    Required sum=360°-213°-76°=71°

    ReplyDelete

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