初中數(shù)學(xué)全等三角形輔助線技巧

專業(yè)資料分享專業(yè)資料分享完美完美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）題意分析：本題考查全等三角形常見(jiàn)輔助線的知識(shí)2）解題思路：在證明三角形的問(wèn)題中特別要注意題目中出現(xiàn)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）過(guò)圖形上某一點(diǎn)作特定的平行線，構(gòu)造全等三角形，全等變換中的“平移”或“翻轉(zhuǎn)折疊”4 ABCAB=ACEABFACEFBCDEB=CFDE=DF1）題意分析：本題考查全等三角形常見(jiàn)輔助線的知識(shí)：作平行線。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如圖（3）,過(guò)口作DE/7BC交AB于D,交AC于E,△AB□絲△AE□從而得以解決口則則△/口絲△apc從而③如圖（4）,過(guò)P作PDZ/BQ交AB的延長(zhǎng)線于D,得以解決.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圖2-2且皿LEG：.PE=PD,在用△ 與舟也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區(qū)與用色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=/國(guó)在&4班與&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(截長(zhǎng)法)在/E上截取力F=HC，如圖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

圖5-2證明：延長(zhǎng)AD至E,使DE=AD,連接BE,CE;AD為色抽。的中線（已知）.,.BD=CD（中線定義）在△ACD和色如。中^BD=CD（已證），Z1=/2（對(duì)頂角相等）AD=ED（輔助線作法）■-ACDEBDSASBE=CAABEAB+BE>AEAB+AC>2AD6 AC=BFACBFACBF ACBFACACBFADCACACBFBFH

ADCHDBADHDH=ADBHADCHDBAC=BHH=BFH BF=BHAA證明：延長(zhǎng)AD到H,使得DH=AD,連接BHD為BC中點(diǎn)/.BD=DC在△ADC和△HDB中。AADC=￡BDHBD=CDADCHDB(SAS)AC=BHH=HACEA=EFHAE=AFEBFH=AFEBH=BFBF=ACBBHACAD HADCHDBBFD BFACBFFDHDH=FDHCCDHBDFBUJ//CmJ證明：延長(zhǎng)FD至H,使得