Monday, May 19, 2008

Elearn Geometry Problem 53

See complete Problem 53
Tangent Circles, angle bisector. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. Valdir Marques FariaOctober 19, 2009 at 5:14 PM

    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.


    Draw 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

  3. Let CED extended meet the common tangent at F. Let BC meet circle A at G.

    Let < FBD = € = < BCD and let < EBD = @

    So < FBE = < FEB = < BGE = €+@
    It follows that < CEG = @ which in turn = < EBG

    Hence < EBG = < EBD

    Sumith Peiris
    Sri Lanka