∴AF=AD,∠AFE=90，

∵ quadrilateral ABCD is a square,

∴AB=AF，

At Rt△ADE and Rt△AFE,

∫AE = AEAF = AD，

∴Rt△ADE≌Rt△AFE(HL)，

[or in Rt△ABG and Rt△AFG,

AG = AGAB = AF，

∴rt△abg≌rt△afg(hl)； ]

(ii)∫CD = 3de, the side length of the square ABCD = 6,

∴de= 13×6=2,ce=cd-de=6-2=4，

∴EF=DE=2，

∵Rt△ABG≌Rt△AFG

∴ let FG=BG=x,

Then EG=x+2, CG=BC-BG=6-x,

In Rt△CEG, EG2=CG2+CE2,

That is (x+2)2=(6-x)2+42,

Finishing, 16x=48,

The solution is x=3,

∴CG=6-x=6-3=3，

∴BG=CG, so ① is correct;

FG = CG = 3，

∴∠GCF=∠GFC，

∵Rt△ABG≌Rt△AFG

∴∠AGB=∠AGF，

According to the exterior angle property of triangle ∠GCF+∠GFC=∠AGB+∠AGF,

∴∠GCF=∠AGB，

∴AG∥CF, so ② is correct;

02CE area = 12CE? CG= 12×4×3=6，

EF = 2，FG=3，

∴S△FGC=33+2×6=3.6, so ③ error;

There are four angles equal to ∠AGB: ∠AGF, ∠GCF, ∠GFC and ∠ GAD * *, so ④ is wrong;

To sum up, the correct conclusion is ① ②.

So the answer is: rt △ ade ≌ rt △ AFE; ①②.