如圖,在正方體ABCD-A1B1C1D1中,M為CC1的中點,AC交BD於點O,求證:A1O⊥平面MBD.

如圖,在正方體ABCD-A1B1C1D1中,M為CC1的中點,AC交BD於點O,求證:A1O⊥平面MBD.


證明:連接MO.∵DB⊥A1A,DB⊥AC,A1A∩AC=A,∴DB⊥平面A1ACC1.又A1O⊂平面A1ACC1,∴A1O⊥DB.在矩形A1ACC1中,tan∠AA1O=22,tan∠MOC=22,∴∠AA1O=∠MOC,則∠A1OA+∠MOC=90°.∴A1O⊥OM.∵OM∩DB=O,∴A1O⊥平面MBD.



如圖,在正方體ABCD-A1B1C1D1中,M為CC1的中點,AC交BD於點O,求證:A1O⊥平面MBD.


證明:連接MO.∵DB⊥A1A,DB⊥AC,A1A∩AC=A,∴DB⊥平面A1ACC1.又A1O⊂平面A1ACC1,∴A1O⊥DB.在矩形A1ACC1中,tan∠AA1O=22,tan∠MOC=22,∴∠AA1O=∠MOC,則∠A1OA+∠MOC=90°.∴A1O⊥OM.∵OM∩DB=O,∴A1O…