Monday, May 19, 2008

Elearn Geometry Problem 57



See complete Problem 57
Tangent Circles, Angle Bisector, Cyclic Quadrilateral. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.
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2 comments:

  1. Valdir Marques Faria,Colégio VisãoOctober 20, 2009 at 12:09 PM

    Tracemos a reta r tangente ao círculo de centro O no ponto B. Sabemos do problema 53 que se prolongarmos BE até encontrar o círculo de centro O no ponto I, teremos que o arco CE = arco ID. Seja arc(ID) =  e arc(EB) = x. Assim, ang(EBr) = (1/2)arc(IB) = (1/2)arc(EB). Teremos que e ang(DEB) = (1/2) arc(EB). Como o prolongamento de GH também encontro o círculo de centro O no ponto I, temos Ang(IGB) é inscritro do círculo de centro O e Ang(IGB) = (1/2).arc(IB). Concluimos que Ang(IGB) = Ang(DEB). Assim, Ang(IGB) + Ang(HEB) = 180º, o que concluímos que o quadrilátero BEHG é cíclico.

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  2. Peter TranJanuary 17, 2011 at 12:37 AM

    http://img695.imageshack.us/img695/8219/problem57.png

    Per result of problem 54
    GH will intersect EB at midpoint M of arc CD
    Triangle MED similar to tri. MDB ( case AA)
    Ratio of sides of 2 triangles will give ME.MB=MD^2
    Similarly Triangles MHC similar to tri. MCG and MH.MG=MC^2=MD^2
    So ME.MB=MH.MG and quadrilateral BEHG is cyclic.

    Peter Tran

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