Solution: As shown in the figure, extend F? The extension lines of D ′ and AB intersect at G point. Because the quadrilateral ABCD is a diamond, we can get ∠ bgd ′ = 90. Let the intersection of FD' and CB be H, the length of CF be 1, DG = X, and ∠ C = 60 in the right triangle FCH, so ∠ Ch. D' f = 120, so ∠ BD' g = 60, ∠ D' BG = 30, and because ∠BG? D' = 90, what about BD' = 2x? D' h = 2x, BH = 2 √ 3 in the right triangle BGH? X, because CD = BC and DF = FD', 1+√ 3+2x = 2+2 √ 3x, x = 1/2, 2x = 1, that is, FD = FD' = √ 3+ 1. So cf/FD =1/(√ 3+1) = (√ 3-1)/2.