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在斜三棱柱A1B1C1獵中,底面是等腰三角形,AB=AC...

(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)过A1作A1H⊥平面ABC,垂足为H.连接AH,并延长交BC于G,于是∠A1AH为A1A与底面ABC所成的角.∵∠A1AB=∠A1AC,∴AG为∠BAC的平分线.又∵AB=AC,∴AG⊥BC,且G为BC的中点.因此,由三垂线定理A1A⊥BC.∵A1A∥B1B,且EG∥B1B,∴EG⊥BC.于是∠AGE为二面...

证明:(1)取BC中点M,连接FM,C1M,在△ABC中,因为F,M分别为BA、BC的中点,所以FM∥.12AC,因为E为A1C1的中点,AC∥.A1C1,所以EF∥EC1,又FM∥A1C1从而四边形EFMC1为平行四边形,所以EF∥C1M,又因为C1M?平面BB1C1C,EF?平面BB1C1C,EF∥平面BB1C1...

(1)证明:取BC中点M,连结FM,C1M.在△ABC中,∵F,M分别为BA,BC的中点,∴FM∥AC,FM=12AC.∵E为A1C1的中点,AC∥A1C1∴FM∥EC1且FM=EC1,∴四边形EFMC1为平行四边形∴EF∥C1M.∵C1M?平面BB1C1C,EF?平面BB1C1C,∴EF∥平面BB1C1C.(2)证明:连接A1C...

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°;--...

∵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) 过O作OF//AB交BC于F 过F做FE//CC1交BC1于E OF//AB,FE//CC1//AA1 所以平面OFE//平面ABA1,即OE//平面A1AB (2) 过A作AG⊥A1B于G 过G作GH//A1C1交BC1于H 角AGH即所求二面角 角AGH=角AGO+角OGH =arccos(OG/AG)+π/2 =arccos(((21)^0.5)/7)+π/...

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⊥...

过点B作BM⊥AA1于M,连结CM,在△ABM和△ACM中,∵AB=AC,∠MAB=∠MAC=45°,MA为公用边,∴△ABM≌△ACM,∴∠AMC=∠AMB=90°,∴AA1⊥面BHC,即平面BMC为直截面,又BM=CM=ABsin45°=22a,∴BMC周长为2x22a+a=(1+2)a,且棱长为b,∴S侧=(1+2)ab.

(Ⅰ)B1C与AC1垂直.证明如下:连接BC1,由题意知B1C⊥BC1作B1D⊥AB,由条件知B1D⊥面ABC,又侧棱与底面所成的角为60°,∴D为AB的中点,∴CD⊥AB,CD为CB1在底面ABC上的射影又B1C⊥AB,∴B1C⊥面ABC1,∴B1C⊥AC1(Ⅱ)由题意知cos∠B1BC=cos∠B1BA?cos∠CBA=...

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