如图所示,已知正四棱柱ABCD-A 证明:(Ⅰ)连结AD1.∵ABCD-A1B1C1D1 是正四棱柱,∴AA1⊥平面ABCD.∴平面ADD1A1⊥平面ABCD.又AB⊥AD,∴AB⊥平面ADD1A1.∴AB⊥AD1.由已知AD=22,DD1=4,∴AD1=AD2+DD12=26.而AE=2,∴tan∠ADE1=AD1AE=2.

(2014?江西一模)如图,在正四棱柱ABCD-A (1)证明:连结AC交BD于点O,连结C1O,PO∵正四棱柱ABCD-A1B1C1D1,∴C1C⊥平面ABCD且O为BD、AC中点,∴C1C⊥CD,C1C⊥BC又∵正四棱柱ABCD-A1B1C1D1,∴CD=CB,∴C1D=C1B,∴C1O⊥BD又C1O=(2)2+62=38,PO=OA2+PA2.

如图,已知正四棱柱ABCD-A (I)如图,以D为原点,DA、DC、DD1所在直线分别为x、y、z轴,建立空间直角坐标系D-xyz如图所示,可得D(0,0,0),A(2,0,0),B(2,2,0),C(0,2,0),A1(2,0,4),B1(2,2,4),C1(0,2,4),D1.