
This geometry puzzle is from mr. Omid Motahed, math teacher in Beijing.
Given are:
Equilateral triangle ABC.
M is the center of BC and also center of a circle arc through B and C.
Arcs CF, FG and GB are equal in length.
Prove that :
CD = DE = EB.
Proof.
Because of the symmetry: CD = EB.
Extend AB and FG, their intersection is H.
Draw lines FM, MG, GB.
LFMG = 60^{0}.....one sixth of 360^{0}
FM = GM, so LMFG = LMGF = 60^{0}
Similar:
LGMB=LMGB=LGBM=60^{0}.
LBGH=180^{0}60^{0}60^{0}= 60^{0}
Conclusion:
FHBC
Triangles FMG and BMG are equilateral and have a common edge MG.
Triangles BMG and BHG are equilateral and have a common edge BG.
So:
FG = GH.......(1)
Triangles ADE and AFG are similar so:
DE / FG = AE / AG .........(2)
Triangles AEB and AGH are similar so:
EB / GH = AE / AG .........(3)
Combining (2) and (3):
DE / FG = EB / GH now combine with (1):
DE = EB

