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20 17 tai' an one-mode mathematics examination questions
Solution: Solution: (i) When F is the midpoint of BE, the plane ADE of CF ∨ ... (1point).

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).