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已知斜三棱柱ABC

解答:(1)证明:取AC中点P,则BP⊥AC∵平面A1ACC1⊥平面ABC,平面A1ACC1∩平面ABC=AC,∴BP⊥平面A1ACC1,∵A1C?平面A1ACC1,∴A1C⊥BP∵A1C⊥AC1,AC1∥PM∴A1C⊥PM∵BP∩PM=P∴A1C⊥面BPM∵BM?面BPM∴A1C⊥BM;(2)解:作PQ⊥A1A于Q,连接BQ∵BP⊥平面A1ACC1,∴A1A⊥...

(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...

解:(I)取AC中点D,连接A1D,则A1D⊥AC.又∵侧面ACC1A1与底面ABC垂直,交线为AC,∵A1D⊥面ABC(2分)∴A1D⊥BC.假设AA1与平面A1BC垂直,则A1A⊥BC.又A1D⊥BC,由线面垂直的判定定理,BC⊥面A1AC,所以BC⊥AC,这样在△ABC中有两个直角,与三角形内角...

如图,取A1B1的中点E,连结C1E,AE,由正三棱柱性质得面A1B1C1⊥面A1B1BA,交线是A1B1.又C1E⊥A1B1,∴C1E⊥面A1B1BA.∴∠C1AE为所求.∵AB=a,C1C=2a,∴Rt△C1EA中,C1E=3a2,AE=32a.∴tan∠C1AE=C1EAE=33.∴∠C1AE=30°.∴AC1与面ABB1A1所成的角为30°....

解答:(本小题满分12分)解:(Ⅰ)∵△ABC为正三角形,D为AC中点,∴BD⊥AC,由AB=6可知,CD=3,BD=33,∴S△BCD=12?CD?BD=932.又∵A1A⊥底面ABC,且A1A=AB=6,∴C1C⊥底面ABC,且C1C=6,∴VC1?BCD=13?S△BCD?C1C=93. …(4分)(Ⅱ)∵A1A⊥底面ABC,∴...

(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∥...

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...

(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...

(1)∵侧面ABB1A1是菱形,∠A1AB=60°,M是A1B1的中点,∴△BA1B1是等边三角形,BM⊥A1B1 .再由面ABB1A1垂直于底面,可得BM⊥面 A1B1C1 .故BM⊥面ABC,∴BM⊥AC.(2)作MN⊥B1C1 ,由三垂线定理可得BN⊥B1C1 ,故∠MNB为二面角B-B1C1-A1的平面角.MN=BMsi...

解答:证明:(1)∵△ABC为正三角形,D是BC的中点∴BC⊥AD,…(1分)∵AA1⊥平面ABC,BC?平面ABC,∴BC⊥AA1 …(3分)∵AD,AA1是平面DAA1内的两条相交直线,∴BC⊥平面DAA1 …(5分)∵A1D?平面DAA1∴BC⊥A1D …(6分)(2)∵D,E,F分别为BC,B1C1,A1B1的中...

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