题干

如图,在四棱锥E﹣ABCD中,底面ABCD为正方形,AE⊥平面CDE,已知AE=DE=2,F为线段DF的中点.

(I)求证:BE∥平面ACF;

(II)求平面BCF与平面BEF所成锐二面角的余弦角.

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答案(点此获取答案解析)

解:(1)连接BD和AC交于点O,连接OF,因为四边形ABCD为正方形,所以O为BD的中点.

因为F为DE的中点,所以OF∥BE.

因为BE⊄平面ACF,OF⊂平面AFC,

所以BE∥平面ACF.

(II)因为AE⊥平面CDE,CD⊂平面CDE,

所以AE⊥CD.

因为ABCD为正方形,所以CD⊥AD.

因为AE∩AD=A,AD,AE⊂平面DAE,

所以CD⊥平面DAE.

因为DE⊂平面DAE,所以DE

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