∴∠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