Saturday, October 1, 2016

Geometry Problem 1269 Triangle, Incircle, Incenter, Inscribed Circle, Tangency Points, Perpendicular, 90 Degrees, Angle Bisector

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

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Geometry Problem 1269 Triangle, Incircle, Incenter, Inscribed Circle, Tangency Points, Perpendicular, 90 Degrees, Angle Bisector.

Posted by Antonio Gutierrez at 10:00 AM
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4 comments:

  1. AnonymousOctober 1, 2016 at 10:44 AM

    connect OC, <BDA1 = <COA1, so OCDA1 is on the same circle. <ODC = 90 = <OA1C =<AA1C

    ReplyDelete
  2. Sumith PeirisOctober 1, 2016 at 10:54 AM

    < DEB = 90-B/2 and < A1AE = A/2 so
    < EA1A = 90-B/2 - A/2 = C/2.

    But < OCD = C/2 hence OCDA1 is concyclic

    Therefore <OA1C = <ODC = 90

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Sumith PeirisOctober 1, 2016 at 11:00 AM

    It follows that < OCA1 = B/2

    ReplyDelete

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