## Friday, August 27, 2010

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

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

Level: High School, SAT Prep, College geometry

1. Draw CE perpendicular to BD and cut AB at E. ED cut BC at F.
Since 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)
Peter Tran

2. 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.
Ajit

3. Problem 513
You 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