Monday, May 30, 2011

Problem 610: Intersecting Circles, Radius, Parallel, Common Tangent, Angle, 90 Degrees

Geometry Problem
Click the figure below to see the complete problem 610.

Problem 610: Intersecting Circles, Radius, Parallel, Common Tangent, Angle, 90 Degrees.

Posted by Antonio Gutierrez at 10:05 AM
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1 comment:

  1. PravinMay 31, 2011 at 6:59 AM

    AG/GB =AE/BF=AC/BC
    So CG bisects angle ACB internally
    AH parallel BM
    AN/BN = AH/BM =AC/BC
    So CH bisects angle ACB externally
    Therefore angle GCH = 90 degrees

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