题目内容
如图,在正三棱柱ABC-A1B1C1中,E∈BB1,F是AC的中点,截面A1EC⊥侧面AC1.求证:BF∥平面A1EC.
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过E向平面AC1作垂线,∵截面A1EC⊥侧面AC1,截面A1EC∩侧面AC1=A1C
∴垂足必落在A1C上,设为D,则ED⊥平面AC1,
∵F是AC的中点,三棱柱ABC-A1B1C1为正三棱锥,
∴BF⊥AC,BF⊥AA1,∴BF⊥平面AC1,
∴BF∥ED,
∵ED?平面A1EC.BF?平面A1EC.
∴BF∥平面A1EC.
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∴垂足必落在A1C上,设为D,则ED⊥平面AC1,
∵F是AC的中点,三棱柱ABC-A1B1C1为正三棱锥,
∴BF⊥AC,BF⊥AA1,∴BF⊥平面AC1,
∴BF∥ED,
∵ED?平面A1EC.BF?平面A1EC.
∴BF∥平面A1EC.
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