Sunday, October 25, 2009

Problem 372: Circles, Common Tangent, Angles

Proposed Problem
Click the figure below to see the complete problem 372 about Circles, Common Internal and External Tangent, Angles.

Problem 372: Circles, Common Internal and External Tangent, Angles.
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Complete Geometry Problem 372
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 7:48 AM
Labels: , , , ,

4 comments:

  1. c .t . e. oJanuary 30, 2010 at 10:18 AM

    draw BC' perpendicular to CE
    extend GB to G' ( on circle)
    draw G'E' tangent on G'
    => E'G'//EG ( both perpendicular to GG')
    =>ang G'C'E' = FCE
    => C'G'//HM
    => HC' = MG' ( HMG'C' trapezoid )
    => ang MG'G'' = α ( G" tg E'G'G")
    => ang MBG' = 2α
    => x + 2α = 180

    x = 180 - 2α
    --------------------------------------------

    ReplyDelete
  2. JankonyexMay 2, 2010 at 2:57 AM

    EB bisect mDEG and mECF=mCFE =>
    HM//EG⊥GD
    α+x/2=mHMD+mMDG=90° <=>
    x=180°-2α

    ReplyDelete
  3. Sumith PeirisOctober 15, 2016 at 8:05 AM

    If < FGH = € then < GDH = €

    Now < GHM = x/2 hence < FCE = x/2 - €

    Considering the angles of Tr. CMD,
    x/2 -€ + @ + x/2 + € +@ = 180

    Therefore x = 180-2@

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Sailendra TMarch 29, 2021 at 2:16 PM

    Let m(BED) = a and read @ as alpha
    EDBG is cyclic and Tr.EGB & Tr.EBD are congruent (SSS). Hence m(GBD)=180-2a---(1)
    Since CE=CF=> m(ECF)=a and CM||EB.
    Hence m(MHD)=a+@=>m(MBD)=2a+2@----(2)
    =>x=360-(1)-(2) = 180-2@

    ReplyDelete

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