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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⊥...

取AA1的中点E,连接CE、BE。易知CB垂直于面AA1C1C,所以AA1垂直于CB,结合AA1垂直于CE知,AA1垂直于面CEB,在直角三角形CEB中斜边BE上的高即为所求。

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...

(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)过B1点作B1O⊥BA于点O.∵侧面ABB1A1⊥底面ABC,侧面ABB1A1∩底面ABC=AB,∴B1O⊥面ABC,∴∠B1BA是侧面BB1与底面ABC所成的角,可得∠B1BO=π3在Rt△B1OB中,BB1=2,∴BO=BB1cosπ3=1又∵BB1=AB=2,∴BO=12AB∴O是AB的中点,可得点B1在平面ABC上的射影O为AB...

解:(Ⅰ)过A1作A1H⊥平面ABC,垂足为H.过H作HD⊥AB,连A1D则A1D⊥AB作HF⊥AC,连结A1F则A1F⊥AC,又∠A1AB=∠A1AC=45°∴Rt△DAA1≌Rt△FAA1,∴AD=AF∴Rt△ADH≌Rt△FAH所以H在∠CAB平分线AE上,由△ABC为正三角形,∴BC⊥AE?BC⊥AA1异面直线AA1与BC所成角为90°;--...

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°....

(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...

(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...

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