摘要:18.解: 1)过点D作DE ⊥ A1 C 于E点.取AC的中点F.连BF ?EF.∵面DA1 C⊥面AA1C1C且相交于A1 C.面DA1 C内的直线DE ⊥ A1 C∴直线DE⊥面AA1C1C ---3分又∵面BA C⊥面AA1C1C且相交于AC.易知BF⊥AC.∴BF⊥面AA1C1C由此知:DE∥BF .从而有D.E.F.B共面.又易知BB1∥面AA1C1C.故有DB∥EF .从而有EF∥AA1.

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