Geometry Problem

Click the figure below to see the complete problem 513 about Triangle, Double Angle, Exterior Angle Bisector, Measure.

See also:

Complete Problem 513

Level: High School, SAT Prep, College geometry

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## Friday, August 27, 2010

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Problem 513: Triangle, Double Angle, Exterior Angle Bisector, Measure

See also:

Complete Problem 513

Level: High School, SAT Prep, College geometry

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Geometry Problem

Click the figure below to see the complete problem 513 about Triangle, Double Angle, Exterior Angle Bisector, Measure.

See also:

Complete Problem 513

Level: High School, SAT Prep, College geometry

Labels:
angle bisector,
double angle,
exterior,
measurement,
triangle

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Draw CE perpendicular to BD and cut AB at E. ED cut BC at F.

ReplyDeleteSince BD is angle bisector of angle (CBE) so triangle EBC and CDE are isosceles and BD become symmetrical line of triangles EBC.and CDE

We have angle (BEF)=angle(BCA) =alpha (property of symmetric)

Angle (ADE)= (BAC)-(AED)= alpha

And triangle EAD is isosceles. ( angle AED=angle ADE= alpha)

AD=AE=BE+BA=BC+BA= a+c

Peter Tran

Another way: Extend CB to E such that BE=c. Now triangles EBD & ABD are SAS congruent. /_DEB=/_DAB=180-2α and thus /_EDC=180-(180-2α)-α=α which means AD=ED=EC or d=a+c.

ReplyDeleteAjit

Problem 513

ReplyDeleteYou get the point E on the side DA such as AE = AB = c,( the point E is between the D, A) then <AEB=<EBA=α,so BE=BC=a.

Is 2β=2α+α or β=3α/2, but < DBE=β-α=3α/2-α=α/2 and <EDB=<BAC-<DBA=2α-β=

2α-3α/2=α/2=<DBE. Then DE=EB=a.Therefore d=DA=DE+EA=a+c.

TOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE