题目内容
如图,在正三棱柱(底面为正三角形的直棱柱)ABC-A1B1C1中,F是A1C1的中点.
(1)求证:BC1∥平面AFB1;
(2)求证:平面AFB1⊥平面ACC1A1.
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(1)求证:BC1∥平面AFB1;
(2)求证:平面AFB1⊥平面ACC1A1.
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证明:(1)连接A1B与AB1交于点E,连接EF.在正
三棱柱ABC-A1B1C1中,可得四边形ABB1A1是矩形,∴A1E=EB.
又A1F=FC1,∴EF∥BC1.
∵EF?平面AB1F,BC1?平面AB1F,
∴BC1∥平面AFB1;
(2)由正三棱柱ABC-A1B1C1中,可得AA1⊥底面A1B1C1,∴AA1⊥B1F.
由F是正△A1B1C1的A1C1的中点,∴B1F⊥A1C1.
又A1A∩A1C1=A1,∴B1F⊥平面ACC1A1,
∴平面AFB1⊥平面ACC1A1.
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又A1F=FC1,∴EF∥BC1.
∵EF?平面AB1F,BC1?平面AB1F,
∴BC1∥平面AFB1;
(2)由正三棱柱ABC-A1B1C1中,可得AA1⊥底面A1B1C1,∴AA1⊥B1F.
由F是正△A1B1C1的A1C1的中点,∴B1F⊥A1C1.
又A1A∩A1C1=A1,∴B1F⊥平面ACC1A1,
∴平面AFB1⊥平面ACC1A1.
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