题目内容
如图,在六面体ABCD-A1B1C1D1中,AA1∥CC1,A1B=A1D,AB=AD.
求证:
(1)AA1⊥BD;
(2)BB1∥DD1.
求证:
(1)AA1⊥BD;
(2)BB1∥DD1.
(1)取BD中点E,连接AE、A1E
∵△ABD中,AB=AD,E为BD中点
∴AE⊥BD,同理可得A1E⊥BD,
∵AE、A1E?平面A1AE,AE∩A1E=E
∴BD⊥平面A1AE,
∵AA1?平面A1AE,∴AA1⊥BD;
(2)∵AA1∥CC1,AA1?平面AA1B1B,CC1?平面AA1B1B,
∴CC1∥平面AA1B1B
∵CC1?平面CC1B1B,平面CC1B1B∩平面AA1B1B=BB1
∴BB1∥CC1,同理可得DD1∥CC1,
∴BB1∥DD1.
∵△ABD中,AB=AD,E为BD中点
∴AE⊥BD,同理可得A1E⊥BD,
∵AE、A1E?平面A1AE,AE∩A1E=E
∴BD⊥平面A1AE,
∵AA1?平面A1AE,∴AA1⊥BD;
(2)∵AA1∥CC1,AA1?平面AA1B1B,CC1?平面AA1B1B,
∴CC1∥平面AA1B1B
∵CC1?平面CC1B1B,平面CC1B1B∩平面AA1B1B=BB1
∴BB1∥CC1,同理可得DD1∥CC1,
∴BB1∥DD1.
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