www.1862.net > (2014?上饶一模)如图,在斜三棱柱ABC

(2014?上饶一模)如图,在斜三棱柱ABC

(1)延长B1E交BC于F,∵△B1EC1∽△FEB,BE=12EC1∴BF=12B1C1=12BC,从而F为BC的中点. (2分)∵G为△ABC的重心,∴A、G、F三点共线,且=FGFA=FEFB1=13,∴GE∥AB1,又GE?侧面AA1B1B,AB1?侧面AA1B1B,∴GE∥侧面AA1B1B (4分)(2)在侧面AA1B1B内,过B...

设三棱柱ABC-A1B1C1的体积为V∵侧棱AA1和BB1上各有一动点P,Q满足A1P=BQ,∴四边形PQBA与四边形PQB1A1的面积相等,∵M是棱CA上的动点,∴M是C时,VM?ABQPVABC?A1B1C1?VM?ABQP 最大又四棱椎C-PQBA的体积等于三棱锥C-ABA1的体积等于13V,∴VM?ABQPVABC?...

证明:(1)∵A1B⊥面ABC,∴A1B⊥AC,------(1分)又AB⊥AC,AB∩A1B=B∴AC⊥面AB1B,------(3分)∵AC?面A1AC,∴平面A1AC⊥平面AB1B;------(4分)(2)如图,以A为原点建立空间直角坐标系,则C(2,0,0),B(02,0),A1(0,2,2),B1(0,4,2...

(1)∵平面A1ACC1⊥平面ABC,AC⊥BC,∴BC⊥平面A1ACC1,∴A1A⊥BC,∵A1B⊥C1C,A1A∥CC1∴A1A⊥A1B,∴A1A⊥平面A1BC,∴A1A⊥A1C;(Ⅱ)建立如图所示的坐标系C-xyz.设AC=BC=2,∵A1A=A1C,则A(2,0,0),B(0,2,0),A1(1,0,1),C(0,0,0).CB=(...

(1)∵CA=CA1=AB=BB1=1,∴ABB1A1,ABB1A1都是菱形,∵面积=1×1×sinB=32,又∠ABB1为锐角,∴∠ABB1=60°,∴△ABB1,△AB1A1,△CAA1均为边长为1的等边三角形. …(3分)∵侧面AA1CC1⊥侧面ABB1A1,设O为AA1的中点,则CO⊥平面ABB1A1,又OB1⊥AA1,∴由三垂线...

∵AC⊥AB,AC⊥BC1,∴AC⊥平面ABC1,AC?平面ABC,∴平面ABC1⊥平面ABC,∴C1在平面ABC上的射影H必在两平面的交线AB上.故答案为:AB

(I)证明:在直三棱柱ABC-A1B1C1中,有AA1⊥平面ABC.∴AA1⊥AC,又AA1=AC,∴A1C⊥AC1. …(2分)又BC1⊥A1C,∴A1C⊥平面ABC1,∵A1C?平面A1C1CA,∴平面ABC1⊥平面A1C1CA. …(4分)(II)解:取AA1中点F,连EF,FD,当E为B1B中点时,EF∥AB,DF∥AC1.即...

(1)证明:∵斜三棱柱ABC-A1B1C1中,点A1在底面ABC上的射影恰好是AB的中点O,底面ABC是正三角形,其重心为G点,D是BC中点,B1D交BC1于E,∴DEEB1=BDB1C1=12,连结AB1,则DEEB1=DGGA=12,∴GE∥AB1,∵GE不包含于侧面AA1B1B,AB1?侧面AA1B1B,∴GE∥...

(1)证明:∵侧面AA1B1B⊥底面ABC,侧棱AA1与底面ABC成60°的角,∴∠A1AB=60°,又AA1=AB=2,取AB的中点O,则AO⊥底面ABC.以O为原点建立空间直角坐标系O-xyz如图,则A(0,-1,0),B(0,1,0),C(3,0,0),A1(0,0,3),B1(0,2,3),C1(3,1...

证明:(1)因为三棱柱ABC-A1B1C1是正三棱柱,所以C1C⊥平面ABC,又AD?平面ABC,所以C1C⊥AD,又点D是棱BC的中点,且△ABC为正三角形,所以AD⊥BC,因为BC∩C1C=C,所以AD⊥平面BCC1B1,又因为DC1?平面BCC1B1,所以AD⊥C1D;(6分)(2)连接A1C交AC1于...

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