∴AG=BG and FG=AG,

∴FG=AG=BG, that is, FG= 12AB,

∴∠AFB=90，

Therefore, option ① is correct;

FG = AG，

∴∠GFA=∠GAF，

And EF⊥FD,

∴∠EFG=∠EAG=90，

∴∠EFG-∠GFA=∠EAG-∠GAF, that is ∠EFA=∠EAF,

And the bisector whose EC is ∠DEF,

∴∠DEC=∠FEC，

∫∠DEF is the outer corner of △EAF,

∴∠def=∠dec+∠fec=2∠fec=∠efa+∠eaf=2∠efa，

∴∠FEC=∠EFA，

∴AF∥EC，

Therefore, option ② is correct;

△EHD and△ △BGF are not necessarily similar, so option ③ is wrong;

∫AF∑EC，

∴DHFH=DEEA，

∠∠EFD =∠GAD = 90，∠EDF=∠GDA，

∴△EFD∽△GAD，

∴DEEF=DGAG，

∠∠EFA =∠EAF，

∴AE=EF and AG=FG,

∴DEEA=DGFG，

∴DHFH=DGFG, that is, DH? FG=FH？ DG，

Therefore, option ④ is correct,

To sum up, the correct option is ① ② ④.

So choose B.