Sunday, September 13, 2009

Problem 355: Circles, Common external tangent, Concyclic points

Proposed Problem
Click the figure below to see the complete problem 355 about Circles, Common external tangent, Concyclic points.

 Problem 355. Circles, Common external tangent, Concyclic points.
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Complete Problem 355
Level: High School, SAT Prep, College geometry

2 comments:

  1. join H to D, E to A
    ang CDN = ang NHD, ang NDH = ang GHN (same arc ND, NH)

    => ang D = ang H (1)

    at the same way

    ang A = ang E (2)

    (1) & (2) => DH//AE

    ang PNG = ang CND, ang QNC = GHN ( vertical ang )
    ang BMQ = ang MEF, ang FMP = BAM

    => ang BMF + ang GNC = BMQ + M + FMP + QNC + N + PNG
    => (BMQ + FMP) + M + (QNC + PNG) + N
    => (BMQ + FMP + QNC + PNG) + M + N
    => ( A + D ) + M + N = 360
    => 180 + M + N = 360 ( A + D = 180, DH//AE )

    => M + N = 180

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  2. Let < AEM = < CAM = < AMB = < FMP = p
    Let < FME = < FEM = q
    Let < GNH = < GHN = < QNC = < HND = r
    Let < CND = < CDN = < DHN = s

    Let HE and DA be extended and meet at X. Then AEX and DHX are both isosceles triangles. Hence ADHE is an isosceles trapezoid.

    Hence < AEH + < ADH = (p+q) + (r+s) = 180

    But < EMP = p+q and < QND = r+s

    So < EMP + < QND = 180
    Therefore < QMP + QNP = 180

    Hence M,P,N,Q are concyclic.

    Sumith Peiris
    Moratuwa
    Sri Lanka

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