www.1862.net > (2013?广州三模)斜三棱柱A1B1C1獵中,侧面AA1C...

(2013?广州三模)斜三棱柱A1B1C1獵中,侧面AA1C...

(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...

(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)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:设AA1中点为D,连BD,CD,C1D,AC1.因为A1B=AB,所以BD⊥AA1.--------------------------2分因为侧面ABB1A1⊥AA1C1C,所以BD⊥面AA1C1C.----------4分又△ACC1为正三角形,AC1=C1A1,所以C1D⊥AA1.------6分所以AA1⊥面BDC1,所以AA1⊥B...

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) 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:(Ⅰ)证明:因为∠ACB=90°,所以 AC⊥BC,又侧面ACC1A1⊥平面ABC,且平面ACC1A1∩平面ABC=AC,BC?平面ABC,所以 BC⊥平面ACC1A1,又AA1?平面ACC1A1,所以 BC⊥AA1.(Ⅱ)证明:设A1B与AB1的交点为O,连接OD,在△A1BC中,O,D分别为A1B,BC的中点...

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⊥...

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...

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...

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