初中数学全等三角形辅助线技巧

专业资料分享专业资料分享完美完美DOC格式整理D，CEABC1=BEF1+ABCBDBDBA，CEBE=BEF=ABDABDBAC=EBD=2CEBEF=BEC，BEF=EC3+F=ACF1=ACFBD=CFBEF

BEC=90°，CF=2CE1=33AB=ACBBD=2CE。BECBAD=B，DBD=2CECAF=ABCACBD2 ABCADBAC ADBCABC1）题意分析：本题考查全等三角形常见辅助线的知识2）解题思路：在证明三角形的问题中特别要注意题目中出现ADBCAB=ACAD.EADEDE=ADBEADBC BD=DCBDE=CDABEDCADEB=ACE=2ADBAC1=21=EAB=EBAB=ACABC例3：已ADC=180ACBADCD=CBAB>AD B+CBADACBADCBADCEABECFADFACBADCE=CF。RtCBERtCDFCE=CF，CB=CD，RtCBERtCDFB=CDFCDF+ADC=180

B+ADC=180（4）过图形上某一点作特定的平行线，构造全等三角形，全等变换中的“平移”或“翻转折叠”4 ABCAB=ACEABFACEFBCDEB=CFDE=DF1）题意分析：本题考查全等三角形常见辅助线的知识：作平行线。2 DEDF DEBDFC EB=CFEEG//CFBAC=60C=40APBACBCBAC=60C=40APBACBCEEG//ACBCGEGB=ACBAB=ACB=ACBB=EGBEGD=DCFEB二EG=CFEDB=CDFDGEDCFDE=DF5ABCBQABCACQAB+BP=BQ+AQBCADOBD=ODAB+BP=BQ+AQAQOOD=OQAD=AQ60 40=8060 40=80ADO=ABC=180AQO=C+QBC=80ADO=AQODAO=QAOOA=AOADOAQOOD=OQAD=AQODBPPBO=DOBPBO=DBODBO=DOBBD=ODBPA=C+PAC=70BOP=OBA+BAO=70BOP=BPOBP=OBAB+BP=AD+DB+BP=AQ+OQ+BO=AQ+BQABAD=AQOD2OODBCACDADOABOMAado^Aaqo,MAado^Aaqo,②如图（3）,过口作DE/7BC交AB于D,交AC于E,△AB□丝△AE□从而得以解决口则则△/口丝△apc从而③如图（4）,过P作PDZ/BQ交AB的延长线于D,得以解决.ABPADPADBCABADE=CDEDCE=ECBCD=AD+BCCD=AD+BCCDCF=CBDF=DACD CF=BC2=1ADBCADC+BCD=180DCE+CDE=902+3=90 1+4=903=4FDEADE2FDE=ZADE<DE=DEZ3=Z4FDEADEASADF=DACD=DF+CFCD=AD+BC180DDEBA EDFBCF1-2RtADERtCDF(HL)DAE=DCFBAD+DAE=180BAD+DCF=180BAD+BCD=1802 1 180BCP=EAPPPEBA E2-2S 口C图2-2且皿LEG：.PE=PD,在用△ 与舟也EP口中,\PE=PD[BP=BP，及△BPE^^1XBPIX.HE),.,.BE=BDa'：AB+BC=2BD,.'.AB+BI^rDC=BI^rBE,AB+DC=B^DC=BE-AB=AEa在用△工F区与用色CFD中，‘FE=PD<ZPSA=ZPDCAE=DCRtAPERtCPD(SAS)PAE=PCDBAP+PAE=180BAP+BCP=180ACECE=CD AB AF=ACACE DC=CE CDECED3-2.'.ZACB=2ZSf'：ZACB=2ZB,二/E=/国在&4班与&4班中，21=/2<ZB=ZEAD=AD:.Aabd^Aasd(.AAS),.'.AB=AEa又A£=AC+CE=AC+DC,.'.AB=AC+DCa方法二(截长法)在/E上截取力F=HC，如图3TDF=DCAFDACDACBBFDBBFD=FBAB=AF+FB=AC+FDAB=AC+CD4DE ABACMNAMNAM+AN>MD+DE+NEBDMMB+MD>BDCENCN+NE>CEAM+AN+MB+MD+CN+NE>MD+DE+NE+BD+CEAB+AC>BD+DE+EC4-2BDACGFCGDEAB+AF>BD+DG+GFGF+FC>GE+CEDG+GE>DE++AB+AF+GF+FC+DG+GE>BD+DG+GF+GE+CE+DEAB+AC>BD+DE+EC5 AB+AC>2AD AB+BD>ADAC+CD>ADAB+AC+BD+CD>AD+AD=2AD BD+CD2AD 2AD

图5-2证明：延长AD至E,使DE=AD,连接BE,CE;AD为色抽。的中线（已知）.,.BD=CD（中线定义）在△ACD和色如。中^BD=CD（已证），Z1=/2（对顶角相等）AD=ED（辅助线作法）■-ACDEBDSASBE=CAABEAB+BE>AEAB+AC>2AD6 AC=BFACBFACBF ACBFACACBFADCACACBFBFH

ADCHDBADHDH=ADBHADCHDBAC=BHH=BFH BF=BHAA证明：延长AD到H,使得DH=AD,连接BHD为BC中点/.BD=DC在△ADC和△HDB中。AADC=£BDHBD=CDADCHDB(SAS)AC=BHH=HACEA=EFHAE=AFEBFH=AFEBH=BFBF=ACBBHACAD HADCHDBBFD BFACBFFDHDH=FDHCCDHBDFBUJ//CmJ证明：延长FD至H,使得