Tuesday, August 15, 2023

Geometry Problem 1551: Unraveling Angle Relations in Cyclic Quadrilaterals: Solving for Angle GEJ

Challenging Geometry Puzzle: Problem 1551. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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Geometry Problem 1551: Unraveling Angle Relations in Cyclic Quadrilaterals: Solving for Angle GEJ

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4 comments:

  1. < GEJ = 64, proof to follow

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. < FBE = < FAE = < CAD = < FGE = a (say)
    < JHE = < HEJ = < BDH = < BDA = b (say)
    < FEB = < FAB = < FGB = u (say)
    < HEC = < HJC = < HDC = v (say)
    < ABG = < AFG = < AEG = t (say)
    < JCD = < JHD = < JED = y (say)

    a + b = 52.........(1) (Triangle BCE)
    u + v + 52 + 46 = 180 (angles on Point E)
    So u +v = 82 .......(2)

    u + 2a + t = 180 - 52 = 128...(3) (Angles on Point G)
    v + 2b + y = 180 - 52 = 128 ...(4) (Angles on Point J)

    (3)+(4) ==> u + v + 2a + 2b + t +y = 256

    So 82 + 104 + t + y = 256
    t + y = 70

    < GEJ = 180 - (a+t) - (b+y) = 180 - (a+b) - (t+y) = 180 - 52 - 70
    < GEJ = 58

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Because < BDC= < BAC;
    So, < HFE= < FHE
    Same for <EGJ = EJG
    So < FHE =90-46/2 = 67 degree
    So < EDC = 67 degree
    So < ECD =180 -52-67 = 61 degree
    So < EJG =61 degree
    So <GEJ = 2*(90-61)= 58 degree

    ReplyDelete
  4. Antonio - a new problem arises from this problem

    Prove that FHJG is a cyclic quadrilateral

    ReplyDelete