已知三棱柱ABC-A1B1C1中底面边长和侧棱长均为a,侧面A1ACC1⊥底面ABC,A1B=(√6/2)a.求证:A1
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已知三棱柱ABC-A1B1C1中底面边长和侧棱长均为a,侧面A1ACC1⊥底面ABC,A1B=(√6/2)a.求证:A1B⊥平面AB1C
![已知三棱柱ABC-A1B1C1中底面边长和侧棱长均为a,侧面A1ACC1⊥底面ABC,A1B=(√6/2)a.求证:A1](/uploads/image/z/2058992-8-2.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%89%E6%A3%B1%E6%9F%B1ABC-A1B1C1%E4%B8%AD%E5%BA%95%E9%9D%A2%E8%BE%B9%E9%95%BF%E5%92%8C%E4%BE%A7%E6%A3%B1%E9%95%BF%E5%9D%87%E4%B8%BAa%2C%E4%BE%A7%E9%9D%A2A1ACC1%E2%8A%A5%E5%BA%95%E9%9D%A2ABC%2CA1B%3D%28%E2%88%9A6%2F2%EF%BC%89a.%E6%B1%82%E8%AF%81%EF%BC%9AA1)
过点B作BO⊥AC,垂足为点O,则BO⊥侧面ACC1A1,连结A1O,在Rt △A1BO中,A1B=a,BO=a,
∴A1O=a,又AA1=a,AO=.
∴△A1AO为直角三角形,A1O⊥AC,A1O⊥底面ABC.
∵ A1O⊥面ABC,AC⊥BO,
∴ AC⊥A1B,
∴ A1C1⊥A1B.
设A1B与AB1相交于点D,
∵ ABB1A1为菱形,
∴ AB1⊥A1B.
又 A1B⊥AC,
AB1与AC是平面AB1C内两条相交直线,
所以A1B⊥面AB1C.
∴A1O=a,又AA1=a,AO=.
∴△A1AO为直角三角形,A1O⊥AC,A1O⊥底面ABC.
∵ A1O⊥面ABC,AC⊥BO,
∴ AC⊥A1B,
∴ A1C1⊥A1B.
设A1B与AB1相交于点D,
∵ ABB1A1为菱形,
∴ AB1⊥A1B.
又 A1B⊥AC,
AB1与AC是平面AB1C内两条相交直线,
所以A1B⊥面AB1C.
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