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如图,在斜三棱柱ABC

∵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

(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)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...

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...

解法一:(Ⅰ)证明:连接AO,∵A1O⊥面ABC,BC?面ABC∴A1O⊥BC∵AO⊥BC,A1O∩AO=O∴BC⊥平面A1OA∵A1A?平面A1OA∴A1A⊥BC.…3分(Ⅱ)解:由(Ⅰ)得∠A1AO=45°由底面是边长为23的正三角形,可知AO=3,∴A1O=3,AA1=32过O作OE⊥AC于E,连接A1E,则∠A1EO为二面角A...

(Ⅰ)证明:∵AB=AC,D是BC的中点,∴AD⊥BC.∵底面ABC⊥平面BB1C1C,∴AD⊥侧面BB1C1C.∴AD⊥CC1.(Ⅱ)解:延长B1A1与BM交于N,连接C1N.∵AM=MA1,∴NA1=A1B1.∵A1B1=A1C1,∴A1C1=A1N=A1B1.∴C1N⊥C1B1.∵截面NB1C1⊥侧面BB1C1C,∴C1N⊥侧面BB1C1C.∴截面...

(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...

(1)证明详见解析;(2)1:1. 试题分析:(1)根据直线与平面垂直的性质可得 ,而已知 ,由直线与平面垂直的判定定理可得 面 ,根据平面与平面垂直的判定定理可得平面 平面 ;(2)由已知可知, =2是三棱锥P ABC的高,△ABC是等腰直角三角形,可...

(Ⅰ)证明:设BC的中点为M.在斜三棱柱ABC-A1B1C1中,点B1在底面ABC上的射影恰好是BC的中点,∴B1M⊥平面ABC.(1分)∵AC?平面ABC,∴B1M⊥AC.(2分)∵∠ACB=90°,∴BC⊥AC.∵B1M∩BC=M,∴AC⊥平面B1C1CB.(4分)∵AC?平面ACC1A1,∴平面ACC1A1⊥平面B1C1C...

(1)证明:∵CA=CB=a,AB=2a,∴AB2=CA2+CB2,∴AC⊥BC∵点B1在底面ABC上的射影落在BC上,∴侧面BCC1B1⊥底面ABC,∵侧面BCC1B1∩底面ABC=BC∴AC⊥平面BCC1B1;(2)解:∵点B1在底面ABC上的射影落在BC上,∴∠B1BC=60°∵AC⊥平面BCC1B1∴BC1⊥AC∵AB1⊥BC1,AB1∩AC...

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