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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