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1、本章回顾本章回顾Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.一一 知识结构知识结构Evaluation only.Created with Aspose.Slides for .NET 3
2、.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.二二 方法总结方法总结1.公理的应用公理的应用(1)证明共面问题证明共面问题证明共面问题证明共面问题,一般有两种证法一般有两种证法.一是由某些元素确定一个平一是由某些元素确定一个平面面,再证明其余元素在这个平面内再证明其余元素在这个平面内;二是分别由
3、不同元素确二是分别由不同元素确定若干个平面定若干个平面,再证明这些平面重合再证明这些平面重合.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.(2)证明三点共线问题证明三点共线问题证明空间三点
4、共线问题证明空间三点共线问题,通常证明这些点都在两个面的交线通常证明这些点都在两个面的交线上上,即先确定出某两点在某两个平面的交线上即先确定出某两点在某两个平面的交线上,再证明第三再证明第三点是两个平面的公共点点是两个平面的公共点.当然必在两个平面的交线上当然必在两个平面的交线上.(3)证明三线共点问题证明三线共点问题证明空间三线共点问题证明空间三线共点问题,先证两条直线交于一点先证两条直线交于一点,再证明第三再证明第三条直线经过这点条直线经过这点,把问题转化为证明点在直线上的问题把问题转化为证明点在直线上的问题.Evaluation only.Created with Aspose.Slid
5、es for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.2.判定空间两条直线是异面直线的方法判定空间两条直线是异面直线的方法(1)根据异面直线的定义根据异面直线的定义.(2)反证法反证法.3.求异面直线所成角的方法求异面直线所成角的方法求异面直线所成的角是通过平移直线求异面直线所
6、成的角是通过平移直线,把异面问题转化为共把异面问题转化为共面问题来解决面问题来解决.根据等角定理及推论根据等角定理及推论,异面直线所成的角的异面直线所成的角的大小与顶点位置无关大小与顶点位置无关,将角的顶点取在一些特殊点上将角的顶点取在一些特殊点上(如线如线段端点段端点,中点等中点等),以便于计算以便于计算,具体步骤如下具体步骤如下:Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Cr
7、eated with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.(1)利用定义构造角利用定义构造角;(2)证明所作出的角为异面直线所成的角证明所作出的角为异面直线所成的角;(3)解三角形求角解三角形求角.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Crea
8、ted with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.4.线线平行的判定方法线线平行的判定方法(1)定义定义:同一平面内没有公共点的两条直线是平行直线同一平面内没有公共点的两条直线是平行直线;(2)公理公理4:ab,bc ac;(3)平面几何中判定两直线平行的方法平面几何中判定两直线平行的方法;(4)线面平行的性质线面平行的性质:a,a ,=b ab;(5)线面垂直的性质线面垂直的性质:a,b ab;(6)面面平行的性质面面平行的性质:,=a,=b ab.Eva
9、luation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.5.直线和平面平行的判定方法直线和平面平行的判定方法(1)定义定义:;(2)判定定理判定定理:ab,a ,b a;(3)线面垂直的性质线面垂直的性质:
10、ba,b,a a;(4)面面平行的性质面面平行的性质:,a a.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.6.两个平面平行的判定方法两个平面平行的判定方法(1)依定义采用反证法依定义采用
11、反证法;(2)利用判定定理利用判定定理:a,b,a ,b ,ab=A ;(3)垂直于同一条直线的两个平面平行垂直于同一条直线的两个平面平行:a,a ;(4)平行于同一平面的两个平面平行平行于同一平面的两个平面平行:, .Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0
12、.0.Copyright 2004-2011 Aspose Pty Ltd.7.平行关系的转化平行关系的转化由上面的框图易知三者之间可以进行任意转化由上面的框图易知三者之间可以进行任意转化,因此要判定因此要判定某一平行的过程就是从一平行出发不断转化的过程某一平行的过程就是从一平行出发不断转化的过程,在解在解题时把握这一点题时把握这一点,灵活确定转化的思路和方向灵活确定转化的思路和方向.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose
13、Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.8.线线垂直的判定方法线线垂直的判定方法(1)定义定义:两条直线所成的角为两条直线所成的角为90;(2)平面几何中证明线线垂直的方法平面几何中证明线线垂直的方法;(3)线面垂直的性质线面垂直的性质:a,b ab;(4)线面垂直的性质线面垂直的性质:a,b ab.Evaluation only.Created with Aspose.Slides for
14、.NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.9.线面垂直的判定方法线面垂直的判定方法(1)线面垂直的定义线面垂直的定义:a与与内任意直线垂直内任意直线垂直 a;(3)判定定理判定定理2:ab,a b;(4)面面平行的性质面面平行的性质:,a a;(5)面面垂直的性质面面垂直的性质:
15、,=l,a ,al a.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.10.两个平面垂直的判定方法两个平面垂直的判定方法(1)利用定义利用定义:两个平面相交两个平面相交,所成的二面角是直二面
16、角所成的二面角是直二面角;(2)判定定理判定定理:a ,a .Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.11.垂直关系的转化垂直关系的转化在证明两平面垂直时一般先从现有直线中寻找平面的垂
17、线在证明两平面垂直时一般先从现有直线中寻找平面的垂线,若这样的直线图中不存在若这样的直线图中不存在,则可通过作辅助线来解决则可通过作辅助线来解决.如有如有平面垂直时平面垂直时,一般要用性质定理一般要用性质定理,在一个平面内作交线的垂在一个平面内作交线的垂线线,使之转化为线面垂直使之转化为线面垂直,然后进一步转化为线线垂直然后进一步转化为线线垂直.故故熟练掌握熟练掌握“线线垂直线线垂直” “面面垂直面面垂直”间的转化条件是解间的转化条件是解决这类问题的关键决这类问题的关键.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client
18、 Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.三三 数学思想数学思想1.转化的思想转化的思想例例1:如下图如下图,点点P是是ABC所在平面外一点所在平面外一点,A B C分别是分别是PBC PCA PAB的重心的重心.(1)求证求证:平面平面ABC平面平面ABC;(2)求求AB:AB.Evaluation onl
19、y.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.解解:(1)如右图如右图,连结并延长连结并延长PA PB PC,其延长线分别交其延长线分别交BC AC AB于于M N Q.A B是是PBC PAC的重心的重心,Evalu
20、ation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.同理由同理由B C是是PAC PAB的重心的重心,可知可知BCQN.ABMN,MN 平面平面ABC,AB 平面平面ABC,AB平面平面ABC,同理同理BC
21、平面平面ABC.又又AB 面面ABC,BC 面面ABC,ABBC=B.平面平面ABC平面平面ABC. Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.(2)平面平面ABC平面平面ABC,平面平
22、面PMN平面平面ABC=MN,平面平面PMN平面平面ABC=AB,ABMN, 2.数形结合思想数形结合思想Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.例例2:某几何体的三视图如下图所示某几
23、何体的三视图如下图所示,P是正方形是正方形ABCD对角线对角线的交点的交点,G是是PB的中点的中点.(1)根据三视图根据三视图,画出该几何体的直观图画出该几何体的直观图;(2)在直观图中在直观图中,证明证明:PD面面AGC;证明证明:面面PBD面面AGC.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Cl
24、ient Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.解解:(1)该几何体是底面为该几何体是底面为2的正方形的正方形,侧面为全等的三角形的侧面为全等的三角形的四棱锥四棱锥,直观图如下图所示直观图如下图所示Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5
25、 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.(2)连结连结AC,BD交于点交于点O,连结连结OG,因为因为G为为PB的中点的中点,O为为BD的中点的中点,所以所以OGPD.又又OG 面面AGC,PD 面面AGC,所以所以PD面面AGC.连结连结PO,由三视图由三视图,PO面面ABCD,所以所以AOPO.又又AOBO,所以所以AO平面平面PBD.AO 平面平面AGC,面面PBD面面AGC.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profi
26、le 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.3.分类讨论思想分类讨论思想Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation o
27、nly.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.例例3:如果二面角如果二面角-l-的平面角是锐角的平面角是锐角,点点P到到 和棱和棱l的距离的距离分别为分别为求二面角的大小求二面角的大小.分析分析:点点P可能在二面角可能在二面角-l-的内部的内部,也可能在外部也可能在外部,应分类应分类解答解答.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2
28、.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.解解:如图所示如图所示,P在二面角在二面角-l-的内部时的内部时,图图(1),点点P在在-l-的的外部时外部时,图图(2).在图在图(1)中中,PA,PAl,Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Prof
29、ile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.ACl,l面面PAC,同理同理,l面面PBC而面而面PAC面面PBC=PC,面面PAC与面与面PBC应重合应重合,即即A C B P在同一平面内在同一平面内,ACB是二面角是二面角-l-的平面角的平面角,在在RtAPC中中, Evaluation only.Created wi
30、th Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.在在RtBCP中中, BCP=45ACB=30+45=75在图在图(2)中中,应有应有ACB=45-30=15.Evaluation only.Created with Aspose.Slides
31、 for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.4.函数的思想函数的思想Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Asp
32、ose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.例例4:在长方体在长方体A1B1C1D1ABCD中中, 点点M在在AB1上移动上移动,点点N在在BC1上移动上移动,求点求点M和点和点N的最短距的最短距离离.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2
33、011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.解解:如右图所示如右图所示,在在BB1上取动点上取动点P,作作PMAB1于于M,PNBC交交BC1于于N,连结连结MN,因为因为BC垂直于平面垂直于平面A1ABB1内的所有内的所有直线直线,所以所以BCBB1,BCPM.又又PNBC,PNBB1,PNPM.设设BP=x, Evaluation only.Created with As
34、pose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspo
35、se Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.四四 数学方法数学方法1.利用平移求异面直线所成的角利用平移求异面直线所成的角例例5:如右图如右图,在空间四边形在空间四边形ABCD中中,E F分别是分别是AB CD的中的中点点,且且AD=BC,ADBC,求求EF与与BC所成的角所成的角.Evaluation only.Created with Aspose.Slides for .NET 3.5
36、 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.解解:连结连结EF,取取BD的中点的中点G,连结连结GE GF.E F分别是分别是AB CD的中点的中点,EG和和FG分别是分别是ABD和和DBC的中位线的中位线.GE .故故EFG(或其补角或其补角)就是就是EF和和BC所成的角所成的角.AD=BC,G
37、E=GF.又又ADBC,EGF=90.EFG=45.故故EF与与BC所成的角为所成的角为45.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.2.反证法反证法例例6:求证求证:过直线外一点有且
38、只有一条直线和这条直线平行过直线外一点有且只有一条直线和这条直线平行.已知已知:点点P 直线直线a.求证求证:过点过点P和直线和直线a平行的直线平行的直线b有且只有一条有且只有一条.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2
39、011 Aspose Pty Ltd.证明证明:存在性存在性:如下图如下图,P a,点点P和直线和直线a确定一个平面确定一个平面,在平面在平面内过点内过点P作直线作直线b与直与直线线a平行平行,故这样的直线故这样的直线b存在存在.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile
40、5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.唯一性唯一性:假设过点假设过点P还有一条直线还有一条直线c与与a平行平行.ab,ac,bc,这与这与b c共点于共点于P矛盾矛盾.故假设不成立故假设不成立,因此直线因此直线b唯一唯一.过直线外一点有且只有一条直线和这条直线平行过直线外一点有且只有一条直线和这条直线平行.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluati
41、on only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd. 规律技巧规律技巧:对于对于“有且唯一有且唯一”性命题的证明性命题的证明,既要证明既要证明“有有”即存在性即存在性,又要证明唯一性又要证明唯一性,其中唯一性的证明大多采用反其中唯一性的证明大多采用反证法证法.反证法的证明过程是反证法的证明过程是:从否定结论入手从否定结论入手,进行推理进行推理,直直至推出矛盾至推出矛盾(可以与假设相矛盾可以与假设相矛盾,也可以与已知也可以与已知 定理定理
42、 公理公理相矛盾相矛盾)矛盾产生的原因是假设不成立矛盾产生的原因是假设不成立,故原命题成立故原命题成立.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.3.同一法同一法例例7:已知平面已知平面
43、平面平面,直线直线AB过过内一点内一点A,且且AB平面平面.求证求证:AB .分析分析:此题直接证明此题直接证明,不易表达不易表达,可用同一法证明可用同一法证明.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose
44、 Pty Ltd.证明证明:在平面在平面内过点内过点A作直线作直线AC垂直于平面垂直于平面 的交线的交线OE.如下图如下图,则则AC平面平面.因为因为AB和和AC都过点都过点A,且垂直于平且垂直于平面面,所以所以AB AC重合重合.也就是也就是AB 平面平面.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.
