题目内容：

已知正方体ABCD-A1B1C1D1,O是底ABCD对角线的交点,求证A1C⊥面AB1D1(已证C1O‖面AB1D1)

优质解答

证明：连接A1C1;∵正方体ABCD-A1B1C1D1,A1C1,B1D1是面A1B1C1D1的对角线；∴A1C1⊥B1D1,又CC1⊥面A1B1C1D1,∴CC1⊥B1D1又A1C1∩CC1=C1∴B1D1⊥面A1CC1∴B1D1⊥A1C同理连接A1B,可证AB1⊥A1C这样B1D1∩AB1=B1∴A1C⊥面A...