七年级数学下册第五章-相交线与平行线-复习课件(1)人教版教学内容.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,我的课堂我做主,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,5.1.1,相交线,第五章 相交线与平行线,全章复习,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,相交线与平行线,了解补角、余角、对顶角，知道等角的余角相等、等角的补角相等、对顶角相等,.,了解垂线、垂线段等概念，了解垂线段最短的性质，体会点到直线距离的意义,.,知道过一点有且仅有一条直线垂直于已知直线，会用三角尺或量角器过一点画一条直线的垂线,.,一课标链接,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,了解线段垂直平分线及其性质,.,知道两直线平行同位角相等，进一步探索平行线的性质,.,知道过直线外一点有且仅有一条直线平行于已知直线,会用三角尺和直尺过已知直线外一点画这条直线的平行线,.,体会两条平行线之间距离的意义，会度量两条平行线之间的距离,.,一,.,课标链接,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,（,1,）对顶角的概念及应用,.,（,2,）相交线、垂线的定义,.,（,3,）垂线的性质：,经过直线外一点有且只有一条直线与已知直线垂直,.,垂线段最短,.,（,4,）点到直线的距离：从直线外一点向已知直线作垂线，这一点和垂足之间线段的长度叫做点到直线的距离,.,二,.,复习目标及知识要点,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,（,5,）三线八角形成的相关角；同位角、内错角、同旁内角,.,（,6,）平行线的性质（特征）：,公理：两直线平行，同位角相等,.,两直线平行，内错角相等,.,两直线平行，同旁内角互补,.,二,.,复习目标及知识要点,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,（,7,）平行线的判别（判定）,公理：同位角相等，两直线平行,.,内错角相等，两直线平行,.,同旁内角互补，两直线平行,.,如果两条直线都和第三条直线平行，那么这两条直线也互相平行,.,（,8,）两条平行线间的距离及其应用,.,二,.,复习目标及知识要点,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,例,1,（,06,大连西岗）如图,1,，直线,AB,、,CD,相交于点,O,，,OM,AB,，若,COB,=135,，,则,MOD,等于,(),A.45 B.35,C.25 D.15,三,.,典型例题,图1,M,O,D,C,B,A,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,分析：,DOB,与,COB,互为邻补角，,则,DOB,=180-,COB=,45,，又,OM,AB,，,则,MOD,与,DOB,互余，故有,MOB,90,DOB,45.,还可以这样思考：,DOA,是,COB,的对顶角，则,DOA,=,COB,135,，,又,OM,AB,，则,MOA,90,，,故有,MOB,135,MOA,45.,三,.,典型例题,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,例,1,（,06,大连西岗）如图,1,，直线,AB,、,CD,相交于点,O,，,OM,AB,，若,COB,=135,，,则,MOD,等于,(),A.45 B.35,C.25 D.15,三,.,典型例题,图1,M,O,D,C,B,A,解：选,A.,点评：解答本题首先要认真阅读题目，明确题目的要求，其次通过图形关系及已知条件，找出所需的对顶角、邻补角、垂线，最后结合其性质建立等量关系求出结果,.,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,例,2,（,05,安徽）如图,2,所示，,AB,/,CD,，,EF,分别交,AB,、,CD,于,M,、,N,，,MG,平分，,MG,交,CD,于,G,.,求的度数,.,三,.,典型例题,解析：方法一：由于,AB,/,CD,，,可得,1=,BMG,；又,MG,平分,则,BMG,=,GMN,；又,BNE,与,BMN,互补，则,BMF,=130,，得,1=,BMG,=,方法二：（提示结合角平分线、平行线,的性质和三角形内角和的知识求得角的度数）,，,BMF,=65.,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,例,3,（,06,十堰）如图,3,，已知,则,_,三,.,典型例题,，,，,（图,3,）,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,例,3,（,06,十堰）如图,3,，已知,则,_,三,.,典型例题,，,，,（图,3,）,解析：过点,P,作,PE,CD,，由,AB,CD,得,PE,CD,，故,APE,=,A,=55,CPE,=,C,=20,所以,AP,C=,APE,-,CPE,=,A,-,C,=35.,点评：过点,P,作原平行线的平行,直线是解本题的关键,.,注意此问题,中的转化的思想的应用,.,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,一选择题：,1,下面四个图形中（如图,5,），,1,与,2,是对顶角的图形有,(),A,0,个,B.1,个,C.2,个,D.3,个,四,.,能力训练,2,1,1,2,1,2,1,2,图5,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,2,如图,6,，直线,AB,、,CD,相交于,O,，,OE,AB,于点,O,，,OF,平分,AOE,，,1=1530,，,则下列结论中不正确的是（）,A.2=45,B.1=3,C.,AOD,与,1,互为补角,D.1,的余角等于,7530,四,.,能力训练,图6,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,3,如图,7,，将直角三角尺的直角顶点靠在直尺上，且斜边与这根直尺平行，那么，在形成的这个图中与,互余的角共有（）,A.4,个,B.3,个,C.2,个,D.1,个,四,.,能力训练,图7,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,4,如图,8,，给出了过直线外一点作已知直线的平行线的方法，其依据是（）,A.,同位角相等，两直线平行,B.,内错角相等，两直线平行,C.,同旁内角互补，两直线平行,D.,两直线平行，同位角相等,四,.,能力训练,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,5,已知：如图,9,，,l,1,l,2,，,1=50,则,2,的度数是,A,135 B,130C,50 D,40,四,.,能力训练,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,6,如图,10,，,AD,BC,，点,E,在,BD,的延长线上，若,ADE,=155,，则,DBC,的度数为（）,A.,155,B.,50,C.,45,D.,25,四,.,能力训练,A,B,C,D,E,图10,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,7,如图,11,，直线,a,b,，点,B,在直线,b,上，且,AB,BC,1,二,55,，则,2,的度数为,:,A.35 B.45 C.55 D.125,四,.,能力训练,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,8,如图,12,，,AB,CD,，,直线,EF,分别交,AB,，,CD,于,E,，,F,两点，,BEF,的平分线交,CD,于点,G,，若,EFG,72,，则,EGF,等于,(),A,36B,54C,72 D,108,四,.,能力训练,F,G,D,C,A,E,B,图12,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,9,已知：如图,13,，,A0B,的两边,0A,、,0B,均为平面反光镜，,A0B=,在,0B,上有一点,P,从,P,点射出一束光线经,0A,上的,Q,点反射后,反射光线,QR,恰好与,0B,平行,则,QPB,的度数是,(),A,60 B,80 C,100 D,120,四,.,能力训练,A,B,P,Q,R,图13,o,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,二填空题,10,若 的补角是,150,，则 ,，,cos,=,.,11,如图,14,，直线,AB,、,CD,相交于点,O,，若,1=26,则,2=,.,四,.,能力训练,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,13.,如图,16,，在,ABC,中，,ABC,=90,，,A,=50,，,BD,AC,，则,CBD,的度数是,.,四,.,能力训练,12,如图,15,，已知,AB,CD,，若,1=45,，则,BAC,=_,度,.,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,14.,如图,17,，已知,AB,CD,，,CE,、,AE,分别平分,ACD,、,CAB,，则,1+2=,四,.,能力训练,C,D,2,1,E,B,A,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,15,如图,18,，,AB,CD,，,B,=68,，,E,=20,，则,D,的度数为,.,四,.,能力训练,B,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,16.,如图,19,，已知,AB,CD,，,直线,EF,分别交,AB,、,CD,于点,E,，,F,，,EG,平分,BEF,交,CD,于点,G,，,如果,1=50,，那么,2,的度数是,_,度,.,四,.,能力训练,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.,三解答题,18,将直尺与三角尺按如图所示的方式叠放在一起在图（如图,21,）中标记的角中，写出所有与,1,与互余的角,四,.,能力训练,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,19,画图,:,如图,22,，作出线段的中点,.B,（,要求：用尺规作图，保留作图痕迹，写出作法，不用证明）,.,四,.,能力训练,图22,A,B,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,20,已知：如图,23,，直线,AB,CD,，,直线,EF,分别交,AB,，,CD,于点,E,，,F,，,BEF,的平分线与,DFE,的平分线相交于点,P,求证：,P,=90,四,.,能力训练,P,A,B,C,D,E,F,图23,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,五,.,参考答案,参考答案：,1,B 2.D 3.C 4.A 5.B,6.D 7.A 8.B 9.B,10.30 11.26 12.135,13.40 14.9015.48 16.65,17.65 18.2,3,4,19.,略,20.,略,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,感谢各位老师莅临指导，,谢谢！,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,数学来源于生活,再见,Evaluation only.,Created with Aspose.Slides for.NET 3.5 Client Profile 5.2.0.0.,Copyright 2004-2011 Aspose Pty Ltd.,