Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 707.
Tuesday, December 20, 2011
Problem 707: Triangle, Excenters, Angle Bisectors, Angle, Measurement
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http://img84.imageshack.us/img84/8180/problem707.png
ReplyDeleteMake 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
∠A is trisected by AD,AE
ReplyDelete∠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°