AC=CD, ∴△ ACD is an equilateral triangle,

And df∨BC, ∠ ACB = 90, ∴DF⊥AC,

∴AE and DF are equilateral triangles ACD, ∴ DE = the height of AE, and ∠ GAD = ∠ GDA = 30.

∴GA=GD，

∴GE=GF。

⑵ Connect CG, gf = ge, CG=CG, ∴ RT δ CGF ≌ RT δ CGE (HL),

∴∠GCE=30 =∠BCE，

ce = ce，∠ceg =∠CEB = 90 ,∴δceg≌δceb，

∴EG=EB, that is, CD bisects BG vertically, ∴BD=DG= 1=AG,

In rt δ afg, ∠ a = 30, ∴FG= 1/2AG= 1/2,

∴DF= 1+ 1/2=3/2。