Challenging Geometry Puzzle: Problem 1596. 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 P be the circumscribing circle of △DEF, the line segment AC, and a point other than point F.
ReplyDeleteBy the circular angle theorem,
∠DFE＝∠DPE＝α , ∠FDE＝∠FPE＝θ
∠DAP＝∠DPA＝180°－(α＋θ)
Therefore, △DAP and △EPC are isosceles triangles with DA＝DP and EP＝EA, respectively.
In other words,
GP=GA=5, HP=HC=8,
∴ GH=GP+HP=5+8=13
Wonderful solution. The auxiliary construction was key
ReplyDelete