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

If < FGH = € then < GDH = €

ReplyDeleteNow < 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