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,典型例题突破,图形的性质之四边形,典型例题突破,【,例,1,】,如图,在矩形,ABCD,中,E,是,AD,的中点,延长,CE,BA,交于点,F,连接,AC,DF.,(1),求证,:,四边形,ACDF,是平行四边形,;,(2),当,CF,平分,BCD,时,写出,BC,与,CD,的数量关系,并说明理由,.,【,解析,】,(1),四边形,ABCD,是矩形,ABCD,FAE=CDE,E,是,AD,的中点,AE=DE,又,FEA=CED,FAECDE(ASA),CD=FA,又,CDAF,四边形,ACDF,是平行四边形,.,(2)BC=2CD.,理由如下,:CF,平分,BCD,DCE=45,CDE=90,CDE,是等腰直角三角形,CD=DE,E,是,AD,的中点,AD=2CD,AD=BC,BC=2CD.,【,例,2,】,如图,ABCD,的对角线,AC,BD,相交于点,O.E,F,是,AC,上的两点,并且,AE=CF,连接,DE,BF.,(1),求证,:DOEBOF;,(2),若,BD=EF,连接,BE,DF.,判断四边形,EBFD,的形状,并说明理由,.,【,解析,】,(1),四边形,ABCD,是平行四边形,OA=OC,OB=OD,AE=CF,OE=OF,在,DEO,和,BFO,中,DOEBOF.,(2),四边形,EBFD,是矩形,.,理由,:OD=OB,OE=OF,四边形,EBFD,是平行四边形,BD=EF,四边形,EBFD,是矩形,.,【,例,3,】,如图,菱形,ABCD,的对角线,AC,BD,相交于点,O,分别延长,OA,OC,到点,E,F,使,AE=CF,依次连接,B,F,D,E,各点,.,(1),求证,:BAEBCF;,(2),若,ABC=50,则当,EBA=_,时,四边形,BFDE,是正方形,.,【,解析,】,(1),菱形,ABCD,的对角线,AC,BD,相交于点,O,AB=BC,BAC=BCA,BAE=BCF,在,BAE,与,BCF,中,BA=BC,BAE=BCF,AE=CF,BAEBCF(SAS).,(2),四边形,BFDE,对角线互相垂直平分,只要,EBF=90,即得四边形,BFDE,是正方形,BAEBCF,EBA=FBC,又,ABC=50,EBA+FBC=40,EBA=,40,=20,.,答案,:,20,【,例,4,】,如图,在,ABCD,中,点,E,F,分别在边,CB,AD,的延长线上,且,BE=DF,EF,分别与,AB,CD,交于点,G,H,求证,:AG=CH.,【,证明,】,四边形,ABCD,是平行四边形,ADBC,AD=BC,A=C,E=F,又,BE=DF,AD+DF=CB+BE,即,AF=CE,在,CEH,和,AFG,中,CEHAFG(ASA),CH=AG.,【,例,5,】,如图,在平行四边形,ABCD,中,E,为,BC,的中点,连接,AE,并延长交,DC,的延长线于点,F.,(1),求证,:AB=CF;,(2),当,BC,与,AF,满足什么数量关系时,四边形,ABFC,是矩形,并说明理由,.,【,解析,】,(1),四边形,ABCD,是平行四边形,ABCD,AB=CD,BAE=CFE,ABE=FCE.,E,为,BC,的中点,BE=EC,ABEFCE(AAS),AB=CF.,(2),当,BC=AF,时,四边形,ABFC,是矩形,.,理由如下,:,ABCF,AB=CF,四边形,ABFC,是平行四边形,BC=AF,四边形,ABFC,是矩形,.,【,例,6,】,如图,将平行四边形纸片,ABCD,沿一条直线折叠,使点,A,与点,C,重合,点,D,落在点,G,处,折痕为,EF.,求证,:,(1)ECB=FCG;,(2)EBCFGC.,【,证明,】,(1),四边形,ABCD,是平行四边形,A=BCD,由折叠可得,A=ECG,BCD=ECG,BCD-ECF=ECG-ECF,ECB=FCG.,(2),四边形,ABCD,是平行四边形,D=B,AD=BC,由折叠可得,D=G,AD=CG,B=G,BC=CG.,又,ECB=FCG,EBCFGC(ASA).,【,例,7,】,如图,在四边形,ABCD,中,ADBC,点,E,F,分别在,AD,BC,上,AE=CF,过点,A,C,分别作,EF,的垂线,垂足为,G,H.,(1),求证,:AGECHF;,(2),连接,AC,线段,GH,与,AC,是否互相平分,?,请说明理由,.,【,解析,】,(1)AGEF,CHEF,G=H=90,AGCH,ADBC,DEF=BFE,AEG=DEF,CFH=BFE,AEG=CFH,在,AGE,和,CHF,中,AGECHF(AAS).,(2),线段,GH,与,AC,互相平分,理由如下,:,连接,AH,CG,如图所示,:,由,(1),得,:AGECHF,AG=CH,AGCH,四边形,AHCG,是平行四边形,线段,GH,与,AC,互相平分,.,
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