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

Labels:
angle bisector,
circle,
tangent

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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