It is proved that the midpoint f of BE and the midpoint g of AE are connected to GD, GD and CF.
∴GF= 12AB,GF∥AB
∫DC = 12AB，CD∨AB。
∴CD∥GF,CD=GF
∴CFGD is a parallelogram ... (3 points)
∴CF∥GD
∴CF∥ Aircraft appearance ... (4 points)
(ⅱ)∵cf⊥bf,cf⊥ab
∴CF⊥ Aircraft Abe
∫CF∨DG
∴DG⊥ Aircraft Abe ... (6 points)
∵DG？ Aircraft Abe
∴ Aircraft ABE⊥ Aircraft Ade ... (7 points)
(ⅲ)∵AB = BE
∴AE⊥BG
∴BG⊥ Aircraft appearance
G is GM⊥DE, BM is connected, and then BM⊥DE.
Then ∠BMG is the plane angle of dihedral angle A-DE-B …(9 points).
Let AB=BC=2CD=2, then
BG=2，GE=2
In Rt△DCE, CD= 1 and CE=2.
∴DE=5
And DG=CF=3.
Youde? GM=DG？ EG gets GM = 305...( 1 1)
∴tan∠BMG=BGGM= 153
The ∴ tangent of the surface angle A-DE-B is 153...( 12 minutes).