Tuesday, December 10, 2013

Geometry Problem 939: Circle, Tangent, Secant, Chord, Parallel, Midpoint

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to enlarge the problem 939.

Online Geometry Problem 939: Circle, Tangent, Secant, Chord, Parallel, Midpoint

Posted by Antonio Gutierrez at 9:14 AM
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4 comments:

  1. Ivan BazarovDecember 10, 2013 at 5:12 PM

    <EDC=<ECA (tangent angle theorem) and <EDC=<EAF (parallel). That makes circumcircle of EAC tangent to FA and then by powerpoint theorem, FA^2=FE*FC. But FB^2=FE*FC, so FB^2=FA^2

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  2. nakaDecember 10, 2013 at 5:17 PM

    Construct circle circumscribe AEC
    AngleBAE=angleEDC=angleECA
    So AB is tangent to circle O and also to new constructed circle.
    Further as FEC is the common chord of both circles it will bisect the tangent segment
    QED

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  3. Sailendra TDecember 21, 2017 at 10:51 AM

    Triangles AEF,DEC & CAF are similar (Apply alternate segment theorem)
    Considering AEF & CAF
    => AF/FE = CF/FA
    => AF^2 = FC.FE = BF^2

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  4. Sumith PeirisNovember 7, 2018 at 11:14 AM

    FB is tangent to circle BCE
    FA is tangent to circle ACE

    Follows that AF = BF

    Sumith Peiris
    Moratuwa
    Sri Lanka

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