Challenging Geometry Puzzle: Problem 1560. 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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ReplyDeleteImage of the solution.
ADH is isosceles and AGD is isosceles so AHDG is a deltoid. Let HG meets BC at point M, HG meets AD at point L. < HLD is a right angle.
Let GD and BC extended meets at point N.
< BNG = 65
Let AE = 2a, AB = 2k
Because of LF = a and FD = 2a then MC = k and CN = 2k, so CDN is isosceles.
< CDN = < CND = 65
Then < HDA = 180 - (< CDN + < ADG) = 50 and GHD = 40
< AHD = < LHD + < AHL
< AHD = 80
Let CB, GA extended meet at P
ReplyDeleteSince PC//AF and AE = EF, it follows that PB = BC = BA
Hence < PAC = 90 and < CAD = 90 - 65 = 25
Since ABCD is an isosceles trapezoid it is also concyclic
Hence < CAB = < CAD = 25
So < BAD = 50 = < HBC since BC//AD
Similarly < BCH = 50
Therefore < AHC = 180 - 50 - 50 = 80
Sumith Peiris
Moratuwa
Sri Lanka