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如图,三棱柱ABC

解答:(1)证明:∵CC1⊥平面ABC,又AD?平面ABC,∴CC1⊥AD∵△ABC是正三角形,D是BC的中点,∴BC⊥AD,又BC∩CC1=C,∴AD⊥平面BB1CC1;(2)证明:如图,连接A1C交AC1于点O,连接OD由题得四边形ACC1A1为矩形,O为A1C的中点,又D为BC的中点,∴A1B∥OD∵OD?...

(Ⅰ)取AB的中点为M,连接EF,EM,CM,∵E是A1B的中点,F是棱CC1中点,∴EM∥AA1,FC∥AA1,EM=FC=12AA1,则四边形EMCF是平行四边形,∴EF∥CM,又∵△ABC为正三角形,侧面AA1C1C是正方形,∴AA1=AB,∴AE⊥A1B,CM⊥AB,∵侧棱AA1⊥平面ABC,∴CM⊥AA1,∴CM⊥...

证明:(1)∵ABC-A1B1C1为三棱柱,D是BC中点,AA1⊥平面ABC,AD?平面ABC,∴AA1⊥AD;又AA1∥BB1,∴AD⊥BB1;又底面ABC为正三角形,D是BC中点,∴AD⊥BC,而BC∩BB1=B,∴AD⊥平面BCC1B1,BC1?平面BCC1B1,∴AD⊥BC1;(2))取C1B1的中点E,连接A1E,ED,则...

解答:解:(1)设AB1与A1B相交于点P,连接PD,则P为AB1中点,∵D为AC中点,∴PD∥B1C.又∵PD∥平面A1BD,∴B1C∥平面A1BD.(2)∵正三棱住ABC-A1B1C1,∴AA1⊥底面ABC.又∵BD⊥AC∴A1D⊥BD∴∠A1DA就是二面角A1-BD-A的平面角.∵AA1=3,AD=12AC=1∴tan∠A1DA=A1...

(1)证明:∵正三棱住ABC-A1B1C1,∴AA1⊥底面ABC,又∵BD⊥AC,A1A∩AC=A,∴BD⊥平面A1ACC1,又∵BD?平面A1BD,∴平面A1BD⊥平面A1ACC1…6分(2)解:作AM⊥A1D,M为垂足,由(1)知AM⊥平面A1DB,设AB1与A1B相交于点P,连接MP,则∠APM就是直线A1B与平面A1B...

证明:(1)取A1C的中点E,取AC的中点F,连接EF,DE,BF.则由条件可得DE∥BF,又面BAC⊥面AA1C1C且交于AC,∴BF⊥AC,BF⊥面AA1C1C,∴DE⊥面AA1C1C而DE?面DA1C,所以平面DA1C⊥平面AA1C1C.(2)延长DA1交AB的延长线于点G,则有CB⊥平面AA1B1C过B作BH⊥A...

(1)三棱柱ABC-A1B1C1中,取C1B1的中点H,连A1H与HC,∵E是BC的中点∴A1H∥AE,∠CA1H是异面直线AE与A1C所成角,∵底面ABC是等腰直角三角形,E是BC的中点,∴AE⊥BC,∴A1H⊥BC,∵侧棱AA′⊥底面ABC,∴侧棱B1B⊥A1H,∴A1H⊥平面BCC1B1,∴A1H⊥HC,在Rt△A1HC中...

解答:解:(1)设正三棱柱ABC-A1B1C1的侧棱长为x.取BC中点E,连AE.∵△ABC是正三角形,∴AE⊥BC.又底面ABC⊥侧面B1C1CB,且交线为BC.∴AE⊥侧面B1C1CB,连ED,则直线AD与侧面B1C1CB所成的角为∠ADE=45°.在Rt△AED中,tan45°=AEED=31+x24,解得x=22...

(1)取BC中点D,连接AD,B1D,由正三棱锥ABC-A1B1C1,得面ABC⊥面BCC1B1.又D为三角形ABC的边BC的中点,故AD⊥BC,于是AD⊥面BCC1B1在矩形BCC1B1中,BC=2,BB1=1,于是Rt△CBC1与Rt△BB1D相似,∠CBC1=∠BB1D,BC1⊥DB1得AB1⊥BC1(2)取BC1的中点D,AC...

解:(Ⅰ)取AB中点E,连接DE,CE因为直棱柱,CC1⊥面ABC,所以CC1⊥AB,又因为△ABC为等腰直角三角形,所以CE⊥AB,所以AB⊥面DEC,即AB⊥DE,所以∠DEC即为二面角D-AB-C的平面角因为CD=1,CE=2,则tan∠DEC=DCCE=12=22(II)连接BC1.因为直棱柱,所...

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