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*,*,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,*,*,*,5.2.2,平行线,的判定,(,一,),核心目标,.,课堂导学,.,1,课前预习,.,2,3,课后巩固,.,4,培优学案,.,5,核心目标,掌握利用同位角相等判定两条直线平行的方法,能运用判定方法对两条直线的位置关系进行判定,课前预习,1,如图,1,,若,1,2,,则直线,a,b,,理由是,_,同位角相等,两直线平行,图,1,2,如图,2.(1),若,1,2,,则,_,;,(2),若,1,3,,则,_,BC,AB,AD,CD,图,2,课堂导学,知识点,1,:同位角相等,两直线平行,【,例题,】,如图,下列说法正确的是,(,),A,若,1,2,,则,cd,B,若,1,3,,则,ab,C,若,1,4,,则,cd,D,若,1,3,,则,cd,【,解析,】,根据同位角相等,两直线平行判断即可,【,答案,】D,【,点拔,】,正确识别“三线八角”中的同位角是正确,解题的关键,课堂导学,对点训练,1,如图,直线,a,,,b,被,c,所截,,1,60,,当,2,_,时,,ab.,60,第,1,题图,2,如图,,(1),若,1,2,,则,_,;,(2),若,1,3,,则,_,a,c,b,d,第,2,题图,课堂导学,对点训练,3,如图,,(1),若,DAB,CBE,时,_,;,(2),若,DAB,_,时,,ABCD.,AD,第,3,题图,BC,CDF,课堂导学,对点训练,4,如图,完成下列推理:,(1)A,3,,,_,;,(2)A,_,,,ACDE,;,(3)1,_,,,DEAC,;,(4)2,B,,,_,第,4,题图,DF,C,4,AB,DF,AB,课堂导学,对点训练,5,如图,直线,a,、,b,被直线,c,所截,且,1,3,,,求证:,ab.,2,3,,,1,3,,,1,2,,,a,b.,课后巩固,6,如图,直线,a,,,b,被直线,c,所截,现给出下列四个条,件,:,1,5,,,1,7,,,1,3,,,4,8,,其中能判定,ab,的条件的序号是,(,),A,B,C,D,C,课后巩固,7,如图,,1,2,,则下列结论正确的是,(,),A,ADBC,B,ABCD,C,ADEF,D,EFBC,C,第,7,题图,课后巩固,8,如图,下列说法正确的是,(,),A,若,1,2,,那么,ab,B,若,1,3,,那么,cd,C,若,1,4,,那么,ab,D,若,1,2,,那么,cd,D,第,8,题图,课后巩固,9,如图,能判定,EBAC,的条件是,(,),A,C,1,B,A,2,C,C,3,D,2,C,D,第,9,题图,课后巩固,10,如图,直线,a,、,b,被直线,c,所截,,1,50,,,3,50,,求证:,ab.,3,50,,,2,3,50,,,1,50,,,1,2,,,a,b.,课后巩固,11,如图,直线,a,、,b,被直线,c,所截,,1,50,,,2,130,,求证:,ab.,2,130,,,3,180,2,50,,,1,50,,,1,3,,,a,b.,课后巩固,12,如图,已知,1,2,,,ac,,那么,bc,吗?说明你,的理由,b,c,,理由:,1,2,,,2,3,,,1,3,,,a,b,,,a,c,,,b,c.,课后巩固,13,如图,已知,BEMN,,,DFMN,,垂足分别为,B,,,D,,,且,1,2,,那么,ABCD,吗?说明你的理由,AB,CD,,理由:,BE,MN,,,DF,MN,,,EBM,FDM,90,,,1,2,,,EBM,1,FDM,2,即,ABM,CDM,,,AB,CD.,培优学案,14,如图,,AD,平分,BAC,,,EF,平分,DEC,,且,1,2.,求证:,(1)ADEF,;,(2)ABED.,(1),AD,平分,BAC,,,EF,平分,DEC,,,DAC,1,,,FEC,2,,,1,2,,,DAC,FEC,,,AD,EF.,(2),AD,平分,BAC,,,EF,平分,DEC,,,BAC,2,1,,,DEC,2,2,,,1,2,,,BAC,DEC,,,AB,ED.,感谢聆听,
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