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.
See also:
Complete Problem 448
Level: High School, SAT Prep, College geometry

2 comments:

  1. 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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  2. 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

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