Question

棱长为2的正方体ABCD-A1B1C1D1中,E是BC的中点,F是DD1的中点.(1)求证:CF‖平面A1DE谢谢了,

Answer 1

设O是A1D中点.G是AD中点.H是OD中点.

⑴OF‖＝EC∴OECF是平行四边形,CF‖EOEO∈平面A1DE.∴CF‖平面A1DE.

⑵EG⊥AA1D,GH⊥A1D（∵GH‖AO中位线）∠EHG是二面角E-A1D-A的平面角.

GH＝AO/2＝√2/2.HE＝√[EG²+GH²]＝3/√2cos∠EHG＝GH/HE＝1/3