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
Friday, August 27, 2010
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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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