∴∠DAE=∠AEB

∫AB∨CD

∴∠BAE=∠CFE

∫AE segmentation ∠ bad

∴∠BAE=∠DAE

∴∠CEF=∠AEB=∠CFE

∴CE=CF

⑵45

Even BG, CG

∫BE = AB = DC

EG=CG

∠BEG= 135 =∠DCG

∴△BEG≌△DCG,BG=DG

∴∠BGE=∠DGC

∴∠BGD=∠EGC=90

∴△BDG is an isosceles right triangle.

∴∠BDG=45

(3) Even BG, CG

It is proved that the quadrilateral CEGF is a diamond.

∠ ABC = 120。

∴EG=CG

And ∠ beg = 120 = ∠ DCG, BE=AB=DC.

∴△BEG≌△DCG

∴BG=DG,∠BGE=∠DGC

∴∠BGD=∠EGC=60

∴△BGD is an equilateral triangle

∴∠BDG=60