Sunday, February 12, 2023

Geometry Problem 1504: Squares, Equilateral Triangle, Parallel, Angle

Geometry Problem 1504. 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 1504: Squares, Equilateral Triangle, Parallel, Angle

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4 comments:

  1. Tr. CEB is isoceles with angles 15,15,150
    Tr.s CEI & BEF are congruent SAS
    If < CEI = < BEF = p then considering the angles in Tr. BEC, 15+15+p + p + 90-p = 180
    So p = 60 and < IEF = 30 and so < GJH = 150

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. 2 interesting results

      1 - FC is perpendicular to AC
      2 - Tr. BIF and Tr. JFH have angles 15,30 and 135

      Delete
  2. A simpler solution

    Since < BEC = 150 and < FEC = 90, < BEF = 60 so < EFI = 30

    Hence considering cyclic quadrilateral EIJF, < GJH = 150

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. https://photos.app.goo.gl/poMiXSFHd2rRzTxS6

    Construct equilateral triangle CBP as shown
    Note that triangle CBP is the translation of triangle DAE with EP=DC and DE=CP
    Since EP is an angle bisector of ∠(CPB) > ∠(CPE)=30
    Since EP=CP -> CEP is isoceles - > ∠(CEP)=∠(ECP)= 75 and ∠(ECB)= 15
    So ∠(CEB)= ∠(GJH)= 150

    ReplyDelete