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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Let < DHG = x & < CDH = y

ReplyDeleteConsider 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

very nice solution.thank you Sumith

ReplyDeleteThak you for your warm words

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