平行线中作辅助线的方法.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,*,初中数学知识点精讲课程,平行线中作辅助线的方法,授课教师 朱 永,授课班级 初一（,4,）班,知识储备,平行线的判定方法有哪几种？,1,、同位角相等，两直线平行。,2,、内错角相等，两直线平行。,3,、同旁内角互补，两直线平行。,4,、平行于同一条直线的两条直线互相平行。,5,、在同一平面内，垂直于同一条直线的两条直线互相平行。,知识储备,平行线的性质有哪些？,1,、两直线平行，同位角相等。,2,、两直线平行，内错角相等。,3,、两直线平行，同旁内角互补。,A,B,C,D,E,如图：,AB,EF,，,CD,EF,，,A=120,，,C=145,，求,AEC,的度数？,F,解：,AB,EF,，,A=120,A+,AEF=180,AEF=60,CD,EF,，,C=145,C+,CEF=180,CEF=35,AEC=,AEF+,CEF,AEC=60+35=95,练 练 手,自主探究,F,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,F,右线,左线,自主探究,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,F,F,F,F,F,F,图,1,图,2,图,3,图,4,图,5,图,6,F,F,F,F,延长,1,延长,2,连接,1,连接,2,连接,3,典例精讲,如图,AB,CD,1=50,2=110,则,3=_.,类型一：含一个拐点的平行线问题,1,2,E,A,D,C,H,B,G,3,典例精讲,如图,AB,CD,1=50,2=110,则,3=_.,1,2,E,A,D,C,H,B,G,3,F,方法一：过点,E,作,EFAB.,1=HEF=50,EF,AB,，,AB,CD,EF,CD,2+FEG=180,FEG=70,3=,180,-,FEG-,HEF,=,60,60,典例精讲,如图,AB,CD,1=50,2=110,则,3=_.,1,2,E,A,D,C,H,B,G,3,F,方法二：延长,HF,与,CD,交于点,F.,AB,CD,1=HFG=50,2=110,EG,F,=180,-110,=,70,3=G,EF,3=,180,-,EG,F,-,EFG,=180,-,70,-50,=,60,60,典例精讲,如图所示,AB,CD,，,则,A+,E+,F+,C,等于,_,类型二：含两（或多）个拐点的平行线问题,典例精讲,如图所示,AB,CD,则,A+,E+,F+,C,等于,_,方法一,：分别过点,E,和点,F,作,EM/AB,，,FN/AB,AB/EM/FN/CD,A+,AEM=180,MEF+,EFN=180,NFC+,C=180,A+,AEM+,MEF+,EFN+,NFC+,C,=,A+,AEF+,EFC+,C=540,M,N,540,方法二：,连接,AF,，,AC,AB/CD,BAC+,ACD=180,又,AEF+,EAF+,AFE=180,AFC+,FAC+,ACF=180,BAC+,ACD+,AEF+,EAF+,AFE+,AFC+,FAC+,ACF,=,A+,AEF+,EFC+,C=540,典例精讲,如图所示,AB,CD,则,A+,E+,F+,C,等于,_,540,如图：,AB,CD,，,ABF=,DCE,，,求证：,BFE=,FEC,。,A,B,C,D,E,F,典例精讲,如图：,AB,CD,，,ABF=,DCE,，求证：,BFE=,FEC,。,A,B,C,D,E,F,典例精讲,方法一,：分别过点,E,和点,F,作,EM/CD,，,FN/AB,C=,CEM,，,B=,BFN,ABF=,DCE,CEM=,BFN,AB/CD,，,EM/CD,，,FN/AB,EM/FN,MEF=,NFE,CEM+,MEF=,BFN+,NFE,FEC=,BFE,即,BFE=,FEC,M,N,A,B,C,D,E,F,典例精讲,方法二,：把,EF,向两边延长，分别,与,BA,、,CD,交于,G,、,H,点。,AB/CD,G=,H,ABF=,DCE,180,ABF,G=180,DCE,H,BFG=,CEH,BFE=,FEC,H,G,如图：,AB,CD,，,ABF=,DCE,，求证：,BFE=,FEC,。,A,B,C,D,E,F,典例精讲,方法三,：分别延长,BF,、,DC,，交于,G,点。,AB/CD,B=,G,ABF=,DCE,G=,DCE,BG/CE,BFE=,FEC,G,如图：,AB,CD,，,ABF=,DCE,，求证：,BFE=,FEC,。,典例精讲,方法四,：连接,BC,，交,EF,于,G,点。,AB/CD,ABC=,BCD,ABF=,DCE,ABC,ABF=,BCD,DCE,FBC=,ECB,BF/CE,BFE=,FEC,A,B,C,D,E,F,G,如图：,AB,CD,，,ABF=,DCE,，求证：,BFE=,FEC,。,理解平行线中添加辅助线的方法，会利用平行线的性质与判定进行解答是解决问题的关键。,1,、原题的难点在于平行线间没有截线或截线不明显；,2,、添加辅助线的目的是构造截线或构造新的平行线；,3,、处理平行线间拐点的问题，,过所有拐点作平行线,是一种通法；,4,、,加截线（连结两点、延长线段相交）,构造三角形，应用三角形内角和定理，也是一种“转化”的数学思想。,作业：,基础训练,42,43,页,下课,!,F,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,解：,A+,AEC+,C=360,理由：过,E,点作,EF,AB,A+,AEF=180,AB,CD,，,EF,AB,EF,CD,C+,CEF=180,A+,AEF+,C+,CEF=360,A+,AEC+,C=360,F,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,解：,A+,AEC+,C=360,理由：过,E,点作,EF,AB,A=,AEF,AB,CD,，,EF,AB,EF,CD,C=,CEF,AEF+,AEC+,CEF=360,A+,AEC+,C=360,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,解：,AEC=,A+,C,理由：过,E,点作,EF,AB,A=,AEF,AB,CD,，,EF,AB,EF,CD,C=,CEF,AEC=,AEF+,CEF,AEC=,A+,C,A,B,C,D,E,F,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,解：,AEC=,A+,C,理由：过,E,点作,EF,AB,A+,AEF=180,AB,CD,，,EF,AB,EF,CD,C+,CEF=180,A+,AEF+,C+,CEF=360,AEF+,AEC+,CEF=360,AEC=,A+,C,A,B,C,D,E,F,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,解：,AEC=,C,A,理由：过,E,点作,EF,AB,A=,AEF,AB,CD,，,EF,AB,EF,CD,C=,CEF,AEC=,CEF,AEF,AEC=,C,A,A,B,C,D,E,F,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,解：,AEC=,A,C,理由：过,E,点作,EF,AB,A=,AEF,AB,CD,，,EF,AB,EF,CD,C=,CEF,AEC=,AEF,CEF,AEC=,A,C,A,B,C,D,E,F,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,解：,AEC=,A,C,理由：过,E,点作,EF,AB,A=,AEF,AB,CD,，,EF,AB,EF,CD,C=,CEF,AEC=,AEF,CEF,AEC=,A,C,A,B,C,D,E,F,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,解：,AEC=,C,A,理由：过,E,点作,EF,AB,A=,AEF,AB,CD,，,EF,AB,EF,CD,C=,CEF,AEC=,CEF,AEF,AEC=,C,A,A,B,C,D,E,F,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,解：,A+,AEC+,C=360,理由：分别延长,AE,和,DC,，交于,F,点。,AB,CD,A+,F=180,AEC+,CEF=180,，,ECF+,ECD=180,A+,F+,AEC+,CEF+,ECF+,ECD=1803,F+,CEF+,ECF=180,A+,AEC+,ECD=360,A,B,C,D,E,F,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,解：,A+,AEC+,C=360,理由：分别延长,BA,和,CE,，交于,F,点。,AB,CD,F+,C=180,AEC+,AEF=180,，,EAF+,EAB=180,F+,C+,AEC+,AEF+,EAF+,EAB=1803,F+,AEF+,EAF=180,EAB+,AEC+,C=360,A,B,C,D,E,F,A,B,C,D,E,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,解：,A+,AEC+,C=360,理由：连接,AC,AB,CD,BAC+,ACD=180,EAC+,AEC+,ACE=180,BAC+,ACD+,EAC+,AEC+,ACE=360,A+,AEC+,C=360,A,B,C,D,E,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,解：,A+,AEC+,C=360,理由：在,CD,上任取一点,F,，连接,AF,AB,CD,DAF=,AFC,EAF+,AEC+,ECF+,AFC=360,EAF+,AEC+,ECF+,DAF=360,DAE+,AEC+,C=360,F,A,B,C,D,E,如图：,AB,CD,，猜想,AEC,与,A,、,C,的关系，并说明理由。,解：,A+,AEC+,C=360,理由：在,AB,、,CD,上分别任取一点,F,、,H,，连接,FH,。,AB,CD,AFH+,CHF=180,A+,AEC+,C+,CHF+,AFH=540,A+,AEC+,C+,CHF+,AFH,AFH,CHF=540,180,A+,AEC+,C=360,H,F,