在正方体ABCD－A1B1C1D1中,求证：BD1⊥平面ACB1.

连结A1B,在正方体ABCD-A1B1C1D1中,面A1B1BA是正方形,对角线A1B⊥AB1,又,在正方体ABCD-A1B1C1D1中,D1A1⊥面A1B1BA,AB1在面A1B1BA上,∴D1A1⊥AB1∵AB1⊥A1B,AB1⊥D1A1,A1B和D1A1是面A1BD1内的相交直线∴AB1⊥面A1BD1,又BD1在面A1BD上,∴AB1⊥BD1同理,D1D⊥面ABCD,AC在面ABCD上,D1D⊥AC,在正方形ABCD中对角线AC⊥BD∵AC⊥D1D,AC⊥BD,D1D和BD是面BDD1内的相交直线∴AC⊥面BDD1,又BD1在面BDD1上,∴AC⊥BD1∵BD1⊥AB1,BD1⊥AC,AB1和AC是面ACB1内的相交直线∴BD1⊥面ACB1