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