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°