证明:连结A1C1,由于AC∥A1C1,EF⊥AC,
∴EF⊥A1C1,
又EF⊥A1D,A1D∩A1C1=A1,
∴EF⊥平面A1C1D, ①
∵BB1⊥平面A1B1C1D1,A1C1平面A1B1C1D1,
∴BB1⊥A1C1,
又A1B1C1D1为正方体,
∴A1C1⊥B1D1,
∵BB1∩B1D1=B1,
∴A1C1⊥平面BB1D1D,
而BD1平面BB1D1D,
∴BD1⊥A1C1,
同理,DC1⊥BD1,DC1∩A1C1=C1,
∴BD1⊥平面A1C1D, ②
由①②,可知EF∥BD1。
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