Thursday, December 8, 2011

Problem 701: Intersecting Circles, Diameter, Perpendicular, Angles, Congruence

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 701.

Online Geometry Problem 701: Intersecting Circles, Diameter, Perpendicular, Angles, Congruence.

2 comments:

  1. ∠ACB = 90° and CD ⊥ AB
    Also OD ⊥ EF and bisects it (DE = DF)
    ∴ CD² = AD.DB = DE.DF = DE² = DF²
    So ∠ECF = 90° = ∠ACB
    Subtracting common ∠FCB from both sides,
    We get ∠ACF = ∠BCE

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  2. Note that m(ACB)=90 and triangle EOF is isosceles . OD is an altitude of tri. EOF ,so DE=DF
    In Circle O, power from point D is DA.DB=DE.DF=DE^2 (1)
    In circle O’, power from point D is DA.DB=DC^2 (2)
    From (1) and (2) we get DC=DE=DF and m( ECF)=90
    Both angles( ACF) and ( BCE) supplement to angle (BCF) so m(ACF)=m(BCE)

    Peter Tran

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