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已知斜三棱柱ABC

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...

∵A1ABB1的面积=|A1A||AB|sin∠A1AB,A1ACC1的面积=|A1A||AC|sin∠A1AC, A1ABB1的面积=A1AC1的面积,|AB|=|AC|, ∴sin∠A1AB=sin∠A1AC, ∴∠A1AB=∠A1AC(都是锐角) 作A1D⊥底面ABC,D为垂足, ∴A1A在底面ABC上的射影AD平分∠BAC, 又∵AB=AC,∴AD⊥BC,...

证明:(Ⅰ)因为∠ACB=90°,所以AC⊥CB,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)连接A1B,交AB1于O点,连接MO,在△A1BN中,O,M分别为A1B,BN的中点,所以OM∥A1N又...

解答:解:取AC的中点E,连接BE,C1E,∵正三棱柱ABC-A1B1C1中,∴BE⊥面ACC1A1,∴∠BC1E就是BC1与侧面ACC1A1所成的角,BC1=3,BE=32,∴sinθ=12,θ=30°.故答案为30°.

取AC的中点O,连结BO,ON. 由题意知 BO⊥AC, 又∵平面A1ACC1⊥平面ABC, ∴BO⊥平面A1ACC1. ∵A1C?平面A1ACC1 ∴所以BO⊥A1C ∴四边形A1ACC1为菱形, ∴A1C⊥AC1 又∵ON∥AC1,所以 A1C⊥ON ∴A1C⊥平面BON,又 BN?平面BON ∴A1C⊥BN.

答案是:四分之根号六

呵呵,这个题我做过2次 解题步骤:取AC中点E,作EF⊥AC1于F,连接DF,连接A1C交AC1于O ∵侧面ABB1A1,ACC1A1均为正方形 ∴AA1⊥AB,AA1⊥AC ∴AA1⊥面ABC ∵∠BAC=90° ∴BA⊥AC ∴BA⊥面ACC1A1 DE是BC,AC中点 ∴DE//BA ∴DE⊥面ACC1A1 ∴DE⊥AC1 ∵EF⊥AC1 ∴AC1⊥面DE...

(I)证明:由已知得AA1⊥平面A1B1C1,∴侧面BCC1B1⊥平面A1B1C1,又A1B1=A1C1,∴A1D1⊥B1C1.∴A1D1⊥平面BB1C1C.…(4分)(Ⅱ)证明:∵D1、D分别是棱B1C1、BC的中点,∴B1D∥CD1,∴CD1∥平面AB1D.又ADD1A1为矩形,∴A1D1∥AD,∴A1D1∥平面AB1D.∵AD∩DB1=D...

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