∴AD⊥ surface DC 1,

D 1F again? Surface DC 1,

∴AD⊥D 1F

(2) take the midpoint g of AB and connect A 1G, FG,

F is the midpoint of CD.

∴GF

∥ ...AD and a 1d 1∩ ... advertisement

∴GF

∥ ... a1d1∴ gfd1a1is a parallelogram ∴A 1G∥D 1F Let a/kloc.

∠AHA 1 is the angle formed by AE and d1f.

E is the midpoint of bb1∴ rt △ a1ag ≌ rt △ Abe.

∴∠ ga1a = ∠ gah ∴∠ a1ha = 90, that is, the angle formed by the straight line AE and D 1F is a right angle.