See complete Problem 106
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Geometry Problem 106
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See complete Problem 106
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Construct E on BC such that AE bisects angle BAC, then AE=CE implies triangles ABE and CDE are congruent, thus ABED is cyclic and angle DBC=a, therefore angle ABD=pi-4a.
ReplyDeleteOr
ReplyDeleteextend CA such BA=AE; m(ABE)=m(BEA)=a
BE=BC & DC=AE so AB=BD
x=180-4a
The symmetrical point D about the BC is E, then AB = DC =EC and BD=DE,<BCE=a=<ECB, but <BAC=<ECA=2a and AB=CE then ABEC is isosceles trapezoid (ΒΕ//ΑC ),with <CBE=<ACB=a=<ECB so BE=EC=DC=BD . Then <DBC=<DCB=a, so <BDA=<DBC+<DCB=2a.Therefore x=180-2a-2a=180-4a.
ReplyDeleteIf the perpendicular bisector of BC cuts AC at E, then <AEB=2a, then CE=BE=AB and E=D and x=180-4a
ReplyDelete[For easy typing, I use "a" to replace alpha]
ReplyDelete<BDC=180-x-3a
Consider triangle BDC
sin(180-x-3a)/CD=sina/BD
CD/BD=sin(x+3a)/sina---------(1)
Consider triangle BAD
sin2a/BD=sin(180-x-2a)/AB
AB/BD=sin(x+2a)/sin2a-----------(2)
Since AB=CD, by equating (1) & (2)
sin(x+3a)/sina=sin(x+2a)/sin2a
2cosa=sin(x+2a)/sin(x+3a)
2sin(x+3a)cosa=sin(x+2a)
sin(x+4a)+sin(x+2a)=sin(x+2a)
sin(x+4a)=0
x+4a=0 (rej) or 180
x=180-4a