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
问题描述：
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
答
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.