Friday, October 14, 2022

Geometry Problem 1499: Triangle, Angles, Cevian, Congruence, Isosceles, Equilateral, Auxiliary Construction

Geometry Problem 1499. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1499: Triangle, Angles, Cevian, Congruence, Isosceles, Equilateral, Auxiliary Construction

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6 comments:

  1. Extend AB to D such that AE = ED

    Now < CBD = 60, <BDE = 60-2x and so < BED = 60+x = < EBD
    Hence DE = BD = AE = BC and so Tr. BCD is equilateral

    Therefore, D is the circumcentre of Tr. BCE and so <BCE = 1/2.(<BDE) = 1/2. (60-2x) = 30 - x
    But < BCE = 2x and hence 2x = 30 - x from which x = 10

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. Replies
    1. Consider Tr. ABC

      <CBD = < A + < C = (60-2x) + 2x = 60

      Rgds

      Sumith Peiris

      Delete
  3. My solution may be found on my blog, https://stanfulger.blogspot.com/2022/11/problem-1499-go-geometry.html

    ReplyDelete