(1)∵AC1是正方体 ∴AD⊥面DC1, 又D1F⊂面DC1, ∴AD⊥D1F (2)取AB中点G,连接A1G,FG, ∵F是CD中点 ∴GFAD又A1D1AD ∴GFA1D1∴GFD1A1是平行四边形∴A1G∥D1F设A1G∩AE=H 则∠AHA1是AE与D1F所成的角 ∵E是BB1的中点∴Rt△A1AG≌Rt△ABE ∴∠GA1A=∠GAH∴∠A1HA=90°即直线AE与D1F所成角是直角 (3)∵AD⊥D1F((1)中已证) AE⊥D1F,又AD∩AE=A,∴D1F⊥面AED,又∵D1F⊂面A1FD1, ∴面AED⊥面A1FD1 |