∴EC=ED=2a,CD=2a
∴EC⊥ED, ... (2 points)
∵BC⊥ plane CC 1D 1D, DE? Plane CC 1D 1D,
∴BC⊥DE, ... (4 points)
∫BC∩EC = C
∴DE⊥ Aircraft EBC, …(7 points)
(2) solution: ∫AD∨BC,
∴∠EBC is the included angle (or remaining angle) of straight lines on different planes, …(9 points)
Take BC⊥ plane CC 1D 1D, EC? BC⊥EC Plane CC 1D 1D, …( 1 1)
That is, △EBC is a right triangle,
In the right angle △EBC, EC=2a, BC=a,
∴tan∠EBC=ECBC=2
∴∠ EBC = Arctan2...( 14 points)