如图,在正方体ABCD-A1B1C1D1中,求证A1C⊥平面AB1D1

∵AA'⊥面A‘B’C‘D’
又B'D'∈面A'B'C'D'
∴AA'⊥B‘D’
又B‘D’⊥A‘C’
AA'∩A'C'=面ACC'A'
∴B'D'⊥面ACC'A'
∵A'C∈面ACC'A'
∴B'D'⊥A'C
同理可证AB'⊥A'C
又AB'∩B'D'=面AB'D'
∴A'C⊥面AB'D