是证明A1C⊥平面AB1D1吧?
连A1C1
∵CC1⊥平面A1B1C1D1
∴CC1⊥B1D1
∵B1D1⊥A1C1,A1C1∩CC1=平面A1C1C
∴B1D1⊥平面A1C1C
∴B1D1⊥A1C
同理可证AB1⊥A1C
∵AB1∩B1D1=平面AB1D1
∴A1C⊥平面AB1D1
是证明A1C⊥平面AB1D1吧?
连A1C1
∵CC1⊥平面A1B1C1D1
∴CC1⊥B1D1
∵B1D1⊥A1C1,A1C1∩CC1=平面A1C1C
∴B1D1⊥平面A1C1C
∴B1D1⊥A1C
同理可证AB1⊥A1C
∵AB1∩B1D1=平面AB1D1
∴A1C⊥平面AB1D1