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.

Gain comprehensive insights! Click below to reveal the complete details.

Click for additional details.

Share your solution by clicking 'Comment' below the post or entering your solution or comment in the 'Enter Comment' field and pressing 'Publish'.

Share your solution by clicking 'Comment' below the post or entering your solution or comment in the 'Enter Comment' field and pressing 'Publish'.

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