高中数学2.2.3直线与平面平行的性质.ppt

2.2.3 直线与平面平行的性质Evaluation only. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 1.掌握直线线与平面平行的性质质定理. 2.体会直线线与平面平行的性质质定理的应应用. 3.通过线线过线线 平行与线线面平行的转转化,培养学生的学习兴习兴 趣. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 直线线与平面平行的性质质定理 (1)文字语语言：一条直线线与一个平面平行,则过这则过这 条直线线的任 一平面与此平面的交线线与该该直线线平行.简简述为为“若线线面平行, 则则_”. (2)符号语言： 线线线线 平行 a _ _ ab =b Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (3)图形语言： Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 1.“判一判”理清知识识的疑惑点(正确的打“”,错误错误 的打 “”). (1)如果直线线a,b和平面满满足ab,a,b,那么b.( ) (2)如果m,n,m,n共面,那么mn.( ) (3)如果m,n,m,n共面,那么mn.( ) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 提示：(1)正确.由ab,a,直线b与平面平行或者在平 面内,又b ,则b. (2)正确.利用直线与平面平行的性质定理可得. (3)错误.m,n可以是同一个平面内的相交直线. 答案：(1) (2) (3) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2.“练练一练练”尝试尝试 知识识的应应用点(请请把正确的答案写在横线线 上). (1)直线线a与平面平行,则则在平面内有 条直线线 和直线线a平行. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (2)如图图,在直四棱柱ABCD-A1B1C1D1中,E,F分别别是AB1,BC1的中 点,和EF平行的面是 . (3)直线线a,b都与平面平行,则则a,b的位置关系是 . Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【解析】(1)直线a与平面平行,过直线a可作无数个平面和 平面相交,直线a与交线平行,故有无数条直线和直线a平行. (2)若连接BA1,则EF是BA1C1的中位线,故EF平面A1B1C1D1,同 理EF平面ABCD,和其他平面都不平行. (3)a,b的位置关系可以是平行、相交或异面. 答案：(1)无数 (2)平面A1B1C1D1,平面ABCD (3)平行、相交或异面 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 直线与平面平行的性质定理 根据直线与平面平行的性质定理及符号表示 和图形表示探究以下问题 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 探究1：如果一条直线线与平面平行,那么这这条直线线是否与这这个 平面内的所有直线线都平行?这这条直线线与这这个平面内多少直线线 平行? 提示：如果一条直线与平面平行,这条直线不会与这个平面内 的所有直线都平行,但在这个平面内有无数条直线与这条直线 平行. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 探究2：如果一条直线与一个平面平行，那么这条直线与这个 平面内的直线有哪些位置关系？ 提示：由直线与平面平行的定义可知， 如果一条直线a与平面平行，那么a 与平面无公共点，即a上的点都不在 平面内，平面内的任何直线与a都无公共点，这样， 平面内的直线与平面外的直线a只能是异面直线或平行 直线. 探究提示：从直 线与平面没有公 共点考虑. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 探究3：如果一条直线a与平面平行，在什么条件下直线a与 平面内的直线平行？ 提示：由于直线a与平面内的任何直线无公共点，所以过直 线a的某一平面，若与平面相交，则直线a就平行于这条交 线. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 探究4：怎样利用性质定理画一条与已知直线平行的直线？ 提示：如果已知直线平行于一个平 面，要想在该平面内画一条与已知 直线平行的直线，只需要过已知直 线作一个与已知平面相交的平面， 那么交线就是要画的与已知直线平行的直线. 探究提示：利用性 质定理确定一个过 此直线的平面即可. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【拓展延伸】线面平行的其他性质 (1)平面外的两条平行线中的一条平行于这个平面，则另一条 也平行于这个平面. (2)直线和平面平行，平面内有无数条直线和该直线平行，但 不一定与平面内任意一条直线平行. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【探究提升】对直线与平面平行的性质定理的理解 (1)该性质定理可以看作直线与直线平行的判定定理.可简述 为“若线面平行,则线线平行”. (2)用该定理判断直线a与b平行时,必须具备三个条件： 直线a和平面平行,即a; 平面和相交,即=b; 直线a在平面内,即a .以上三个条件缺一不可. (3)在应用这个定理时,要防止出现“一条直线平行于一个平 面,就平行于这个平面内的一切直线”的错误. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 类型 一 对对直线线与平面平行的性质质定理的理解 试试着完成下列各题题,总结线总结线 面平行的性质质定理的实质实质 、 作用及应应用时时的注意点. 1.下列判断正确的是( ) A.a,b,则则ab B.a=P,b,则则a与b不平行 C.a,则则a D.a,b,则则ab Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2.求证证：如果一条直线线和两个相交平面平行,那么这这条直线线 和它们们的交线线平行. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【解题指南】1.利用直线与平面的位置关系考虑线线、线面 是否平行. 2.(1)用数学符号语言描述上述命题,写出已知和求证. (2)用图形语言描述上述命题,即画出相应图形. (3)综合利用线面平行的性质定理与判定定理解答本题. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【解析】1.选B.对于A,直线a,b可能平行也可能异面 ;a=P, b ,则a与b相交或异面,故不平行,B正确;对于C,直线a可能 和平面相交;对于D,直线a,b不一定平行. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2.已知：a,a,=l,求证：al. 证明：如图, Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 过a作平面,使得=c,=d,那么有 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【技法点拨】线面平行的性质定理的实质、作用及应用时的 注意点 (1)实质：线面平行的性质定理是由线面平行推出线线平行, 定理内容揭示了“直线与平面平行之后它们具有什么样的性 质”. (2)作用：证明直线与直线平行的一种方法. (3)注意点：在应用定理时要避免出现“一条直线平行于一个 平面就平行于这个平面内的一切直线”的错误认识. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 类型 二 线线面平行的性质质定理的简单应简单应 用 尝试尝试 完成下列试题试题 ,体会性质质定理的应应用,并归纳应归纳应 用 定理证证明线线线线 平行的步骤骤. 1.直线线a,b,c及平面,使ab成立的条件是( ) A.a,b B.a,b C.ac,bc D.a,=b Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2.如图图,E,H分别别是空间间四边边形ABCD的边边AB,AD的中点,平面 过过EH分别别交BC,CD于F,G.求证证：EHFG. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【解题指南】1.考虑是否满足线面平行的性质定理,可以排除 A,B,D. 2.先证明EH平面BCD,再利用直线与平面平行的性质定理推 出线线平行. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【解析】1.选C.a,b ,则ab或a,b异面,所以A错误 ;a,b,则ab或a,b异面或a,b相交,所以B错误 ;a,=b,则ab或a,b异面,所以D错误;ac,bc, 则ab,这是公理4,所以C正确. 2.因为E,H分别是AB,AD的中点,所以EHBD. 又BD 平面BCD,EH 平面BCD,所以EH平面BCD. 又EH ,平面BCD=FG,所以EHFG. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【互动动探究】题题2中若已知EFGH是平行四边边形,如何证证明： BD平面EFGH,AC平面EFGH. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【证明】因为EFGH是平行四边形 同理可证AC平面EFGH. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【技法点拨】利用线面平行的性质定理证明线线平行的四个 步骤 (1)在已知图形中确定(或寻找)一条直线平行于一个平面. (2)作出(或寻找)过这条直线且与这个平面相交的平面. (3)得出交线. (4)根据线面平行的性质定理得出结论. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 类型 三 线线面平行的判定定理与性质质定理的综综合应应用 试试着完成下列各题题,总结应总结应 用线线面平行判定和性质质定理的 应应用方法. 1.如图图,四棱锥锥S-ABCD的所有的棱长长 都等于2,E是SA的中点,过过C,D,E三点 的平面与SB交于点F,则则四边边形DEFC 的周长为长为 ( ) A.2+ B.3+ C.3+2 D.2+2 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2.如图图,三棱柱ABC-ABC中,D是BC上一点,且满满足AB 平面ACD,则则D是BC的 . Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【解题指南】1.根据线面平行的判定定理得到CD平面SAB, 再根据线面平行的性质定理得到平行关系,进而得到线段 EF,FC的长,可求四边形DEFC的周长. 2.利用线面平行的性质定理,得出平面ABC与平面ACD的交 线与AB平行,进而利用中位线性质,得出D是BC的中点. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【解析】1.选C.因为AB=BC=CD=AD=2, 所以四边形ABCD为菱形,所以CDAB. 又CD 平面SAB,AB 平面SAB, 所以CD平面SAB. 又CD 平面CDEF, 平面CDEF平面SAB=EF, 所以CDEF.所以EFAB. 又因为E为SA的中点,所以EF= AB=1. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 又因为SAD和SBC都是等边三角形, 所以DE=CF=2sin60= , 所以四边形DEFC的周长为 CD+DE+EF+FC=2+ +1+ =3+2 . Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2.连接AC交AC于E,连接DE,在平行四边形AACC中, AC与AC互相平分. 所以AE=EC. 又因为AB平面ACD, 平面ABC平面ACD=DE, 所以ABDE. 在ABC中,AE=EC,ABDE, 所以BD=DC,所以D是BC的中点. 答案：中点 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【技法点拨】利用线面平行的判定定理和性质定理的关注点 (1)利用性质定理证明的关键是过直线作平面与已知平面相交 . (2)证明过程,实际上就是不断交替使用线面平行的判定定理 、性质定理及公理4的过程.这是证明线线平行的一种典型的 思路. (3)若题目中含有线线平行的条件,可考虑线面平行的判定定 理的条件;若含有线面平行的条件,可考虑线面平行的性质定 理得线线平行. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【变变式训练训练 】如图图所示,在长长方体ABCD-ABCD中,点 PBB(不与B,B重合).PABA=M,PCBC=N,求证证： MN平面BAC. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【证明】连接AC,AC,BA,BC, 因为ABCD-ABCD是长方体,所以ACAC. 又AC 平面BAC,AC 平面BAC, 所以AC平面BAC. 又因为平面PAC过AC与平面BAC交于MN, 所以MNAC. 因为MN 平面BAC,所以MN平面BAC. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 1.对对于直线线m,n和平面,下面说说法中正确的是( ) A.如果m,n,m,n是异面直线线,那么n B.如果m,n,m,n是异面直线线,那么n与相交 C.如果m,n,m,n共面,那么mn D.如果m,n,m,n共面,那么mn 【解析】选C.如果m ,n,m,n共面,根据线面平行的性 质定理,则mn,故选项C正确.在选项A中,n与可能相交.在 选项B中,n与可能平行.在选项D中,m与n可能相交. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2.直线线ab,且a与平面相交,那么b与平面的关系是( ) A.必相交 B.可能平行 C.相交或平行 D.相交或在平面内 【解析】选A.两条平行线,其中一条与平面相交,另一条也与平 面相交. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 3.下列结论结论 中,正确的个数是( ) (1)若直线线l上有无数个点不在平面内,则则l. (2)若直线线l平行于平面内的无数条直线线,则则l. (3)若直线线l与平面平行,则则l与平面内的任一直线线平行. (4)若直线线l在平面外,则则l. A.0个 B.1个 C.2个 D.3个 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【解析】选A.(1)直线l上有无数个点不在平面内,并没有说 是所有点都不在平面内,因而直线可能与平面平行,亦有可 能与平面相交.解题时要注意“无数”并非“所有”.(2)直线l 虽与内无数条直线平行,但l有可能在平面内,所以直线l不 一定平行于.(3)当l时,若m 且ml,则在平面内, 除了与m平行的直线以外的每一条直线与l都是异面直线.(4)直 线l在平面外,应包括两种情况：l和l与相交,所以l与 不一定平行.故选A. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 4.已知m,n为为两条不同的直线线,为为两个不同的平面,则则下 列结论结论 中正确的是( ) A.m,mnn B.m,nmn C.m,m,=nmn D.m,nmn 【解析】选C.A中,n还有可能在平面内;B中m,n可能相交、 平行、异面;由线面平行的性质定理可得C正确.D中m,n可能异 面. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 5.如图图所示的三棱柱ABC-A1B1C1中,过过A1B1的 平面与平面AB