Question

如图,在三棱柱ABC—A1B1C1中,B1B垂直平面A1B1C1,AC=CB=CC1=2,角ACB

Answer 1

令A1B的中点为F.

∵ABC－A1B1C1是三棱柱,且AA1⊥平面A1B1C1,∴BB1＝CC1、BB1∥D1C1.

∵E、F分别是A1B1、A1B的中点,∴EF是△A1B1B的中位线,∴EF＝BB1/2、EF∥BB1.

∵C1D1＝CC1/2、EF＝BB1/2、BB1＝CC1,∴C1D1＝EF.

∵EF∥BB1、BB1∥D1C1,∴C1D1∥EF.

由C1D1＝EF、C1D1∥EF,得：EC1D1F是平行四边形,

∴C1E∥FD1,而FD1在平面A1BD1上,∴C1E∥平面A1BD1.

你把E换成D即可