Sunday, October 22, 2023

Geometry Problem 1557: Calculate the Angle DHG in Right Triangle ABC.

Challenging Geometry Puzzle: Problem 1557. 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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Illustration of Geometry Problem 1557: Calculate the Angle DHG in Right Triangle ABC.

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Posted by Antonio Gutierrez at 4:16 PM
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4 comments:

  1. Sumith PeirisOctober 23, 2023 at 11:19 PM

    Let < DHG = x & < CDH = y

    Consider Triangles GED & DFH
    < GED = < DFH = 30, DE = FH & EG = FD
    Hence Triangles GED & DFH are congruent SAS
    So GD = DH & < DGH = < DHG =x

    Further, since Triangles GED & DFH are congruent, < EGD = < FDH = A + y and so
    < GDA = 180 - (60 - A) - (60 + A + y) = 60 - y (by considering the angles of Triangle AGD)
    Hence < GDH = 180 - y - (60 - y) = 120

    So in Isosceles Triangle GDH, x + x + 120 = 180
    and x = 30 = < DHG

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. AnonymousFebruary 1, 2024 at 4:11 AM

    very nice solution.thank you Sumith

    ReplyDelete
  3. Stan FULGEROctober 25, 2024 at 4:35 AM

    Remark: If, instead we construct the equilateral triangles BEG and BHF, then triangle DHG is equilateral (proof quite similar with the one above!).

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