Proposed Problem
Click the figure below to see the complete problem 372 about Circles, Common Internal and External Tangent, Angles.
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Complete Geometry Problem 372
Level: High School, SAT Prep, College geometry
Sunday, October 25, 2009
Problem 372: Circles, Common Tangent, Angles
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draw BC' perpendicular to CE
ReplyDeleteextend 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
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EB bisect mDEG and mECF=mCFE =>
ReplyDeleteHM//EG⊥GD
α+x/2=mHMD+mMDG=90° <=>
x=180°-2α