[X版高考数学人教A版·数学文全程复习方略配套课件7.4直线、平面平行的判定及其性质.ppt

【解题指南】(1)证明四边形EFGH为平行四边形即可；(2)利用面面平行的判定定理，转化为线面平行来证明． 【规范解答】(1)∵E、H分别是AB、DA的中点， ∴EH BD. 同理，FG BD，∴FGEH． ∴四边形EFGH是平行四边形， ∴E、F、G、H共面． 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. (2)平面ABD和平面α有一个公共点A， 设两平面交于过点A的直线AD′． ∵α∥β，∴AD′∥BD. 又∵BD∥EH，∴EH∥BD∥AD′. ∴EH∥平面α，同理，EF∥平面α， 又EH∩EF＝E，EH?平面EFGH， EF?平面EFGH，∴平面EFGH∥平面α． 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. 【反思·感悟】1.线面、面面平行的判定和性质常常结合在一起进行考查，解题中要注意性质和判定交替应用． 2.利用判定或性质解题时，应注意解题过程的规范性，即要准确地使用数学语言及符号来表示出定理的有关内容． 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. 【变式训练】在正方体ABCD－A1B1C1D1中，M、N、P分别是C1C、B1C1、C1D1的中点，求证：平面MNP∥平面A1BD. 【证明】如图，连接B1D1、B1C． ∵P、N分别是D1C1、B1C1的中点， ∴PN∥B1D1. 又B1D1∥BD，∴PN∥BD． 又PN平面A1BD，∴PN∥平面A1BD． 同理MN∥平面A1BD， 又PN∩MN＝N，∴平面MNP∥平面A1BD． 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. 平行关系中的计算问题 【方法点睛】求解平行关系中范围问题的数学思想 解答立体几何中的有关最值或范围问题，常用函数思想解决，通过设出适当的变量、建立函数关系，转化为求函数的最值(或值域)的问题．解题时要弄清哪些是定值，哪些是变量，如何根据题意建立函数关系，如何求函数的最值等． 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3】(1)如图，已知平面α∥平面β∥平面γ，且β位于α 与γ之间.点A、D∈α，C、F∈γ，AC∩β=B，DF∩β=E.设AF 交β于M，AC与DF不平行，α与β间距离为h′，α与γ间距离 为h，则当△BEM