解答:(Ⅰ)证明:∵∠VAB=∠VAC=90°,∴VA⊥AB,VA⊥AC,又AB∩AC=A,
∴VA⊥平面ABC.∴VA⊥BC.
∠ABC=90°,∴AB⊥BC,VA∩VB=V,
∴BC⊥平面VBA.又BC?平面VBC,
∴平面VBA⊥平面VBC.
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(Ⅱ)过点E作EF⊥AC于点F,连接BF,则EF∥VA.
VA⊥平面ABC,∴EF⊥平面ABC,
∴∠EBF为BE与平面ABC所成的角.
∵点E为VC的中点,∴点F为AC的中点.
∴BF=
AC,EF=VA.
在Rt△EFB中,由tan
∠EBF===,∴∠EBF=60°,
∴直线BE与平面ABC所成的角为60°.