Saturday, September 30, 2017

Geometry Problem 1348: Three Concentric Circles, Equilateral Triangle, Perpendicular Bisector, Congruence

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1348: Three Concentric Circles, Equilateral Triangle, Perpendicular Bisector, Congruence.

Posted by Antonio Gutierrez at 7:43 PM
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4 comments:

  1. AnonymousSeptember 30, 2017 at 9:17 PM

    Drop perpendiculars from O to EA & B to DA
    Since BA=OA => m(BAD) = m(OAE)
    Hence triangles BAD & OAE are congruent
    => DA = EA
    m(DAE)=m(DAO)+m(OAE)
    =>m(DAE)=60-m(BAD)+m(OAE)
    =>m(DAE)=60
    Hence DAE is equilateral

    ReplyDelete
    Replies
    1. Antonio GutierrezOctober 2, 2017 at 6:57 AM

      To Anonymous: Please explain your second line. Thanks

      Delete
    2. Stan FulgerOctober 3, 2017 at 9:03 AM

      It is just construction of an equilateral triangle with vertices onto 3 concentric circles

      Delete
  2. APOSTOLIS MANOLOUDISOctober 2, 2017 at 5:45 AM


    Let a circle (A,AD) that intersects the circle C_3 at the point Z.Then triangle DBA=triangle ZOA (AD=AZ,AB=AO,DB=ZO).So triangle DAZ is
    equilateral.If ZN perpendicular of AD,then AN=ND or N,M=coincide.Therefore and
    the points Z,E= coincide.So triangle DAE is equilateral .
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete

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