See complete Problem 53
Tangent Circles, angle bisector. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Elearn Geometry Problem 53
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See complete Problem 53
Tangent Circles, angle bisector. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
A reta BD intercepta a circunferência menor no ponto F e BC intercepta no ponto G. Ligando EF e EG, teremos que o angulo FED = (1/2)(arcEF) = angulo DBE. Angulo EGC = (1/2)(arcEG)= angulo CBE. Traçando a reta (r) tangente em B temos que ang(FBr)=(1/2).(arcFB) = angBGF e ang(DBr) = (1/2).(arcBD) = ang(BCD). Então GF é paralelo a CD o que se conclui que o arco(EG) = arco(EF) então angulo FED = angulo(GEC). Assim, BE é bissetriz do ângulo B.
ReplyDeletehttp://img59.imageshack.us/img59/7384/problem53.png
ReplyDeleteDraw OM perpendicular to chord CD . M is on circle O and M is midpoint of arc CD.
We will prove that M, E, B are collinear.
1. Tri. AEB and MOB are isosceles with common angle OEB
OM // AE and m(MOB)=m(EAB) ( corresponding angles)
2. m(OMB)=90- ½ m(MOB)
m(AEB)=90- ½ m(EAB)
so m(OMB)=m(AEB)
3. M,E,B are collinear and BE is the angle bisector of angle CBD
Peter Tran
Let CED extended meet the common tangent at F. Let BC meet circle A at G.
ReplyDeleteLet < FBD = € = < BCD and let < EBD = @
So < FBE = < FEB = < BGE = €+@
It follows that < CEG = @ which in turn = < EBG
Hence < EBG = < EBD
Sumith Peiris
Moratuwa
Sri Lanka