Saturday, May 8, 2010

Problem 448: Tangent Circles, Diameter, Perpendicular, Chord

Geometry Problem
Click the figure below to see the complete problem 448 about Tangent Circles, Diameter, Perpendicular, Chord.

Problem 448: Tangent Circles, Diameter, Perpendicular, Chord.
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Complete Problem 448
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 7:20 PM
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4 comments:

  1. c .t . e. oMay 9, 2010 at 6:02 AM

    DCE = DAB = 1/2 DO'E = DOB ( DO'E, DOB isoc, ODB comm )
    DEC = DCA ( have same arc DC )
    =>
    ▲ADC ~ ▲DCE (1)
    DC common for ADC, DCE (2)
    (1) & (2)
    ▲ADC = ▲DCE
    =>
    AD = DC = DE = x (3)

    ▲DGE ,
    GDE = 90 - ADG = 90 - B
    GED = 90 - B (ADC isoceles => DG bisector )
    =>
    DGE isoceles =>
    ▲DGE ~ ▲ADC
    =>
    x/a = b/x

    x² = ab
    ------------------------------------------

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    Replies
    1. Sumith PeirisApril 30, 2019 at 2:28 AM

      Tr.s ADC and DCE are not necessarily congruent. This will be so only in the special case of each triangle being isoceles.
      Could u elaborate?

      Delete
  2. JankonyexMay 9, 2010 at 9:56 AM

    consider common tangent, mBAD=mECD
    hence mADC=mBCE=mCDE=pi/4, mEGD=pi/4 as well
    draw DH//AB with GCEH collinear, then mDHC=pi/4 and HD=b
    consider similar triangles DCA and HDC,
    DC/CA=HD/DC <=> x^2=a.b

    ReplyDelete
  3. Sumith PeirisMay 2, 2019 at 11:05 AM

    Extend GCE to P such that DC = DP = x

    Let < O'DC = < O'CD = < GDC = p
    CD bisects < ADB and so Tr.s ADC and GDP are similar having angles 45, 45+p and 90-p

    Hence x/a = b/x and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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