已知四棱柱ABCD-A1B1C1D1中,AA1⊥平面ABCD,底面ABCD为棱形
解答:(1)证明:在四棱柱ABCD-A1B1C1D1中,∵AA1⊥平面ABCD,∴AA1⊥BD,∵平行四边形ABCD中,AB=BC,∴AC⊥BD,∵AA1∩AC=A,∴BD⊥平面ACC1A1.(2)解:连结A1O,EO,A1E,∵在四棱柱中,底面ABCD是棱形,且E是棱CC1的中点,∴A1B=A1D,EB=ED,又∵O是BD的中点,∴A1B=A1D,EB=ED,又∵O是BD的中点,∴A1O⊥BD,EO⊥BD,∴∠A1OE是二面角A1-BD-E的平面角,∵四棱柱中,AA1⊥平面ABCD,AB=BC=2,AA1=4,∴A1O=17,EO=5,A1E=22,∴在△A1OE中,cos∠A1OE=A1O2+EO2?A1E22A1O?EO=17+5?8217?5=78585,∴二面角A1-BD-E的平面角的余弦值为78585.