第三章截长补短模型：截长补短a b cI |E①FI I IE G F②D 如图①，若证明线段AB、CD、EF之间存在EF=AB +CD,可以考虑截长补短法.截长法：如图②，在EF上截取EG = AB,再证明GF =CD即可.补短法：如图③，延长AB至H点，使BH=CD,再证 明AH = EF即可.模型分析截长补短的方法适用于求证线段的和差倍分关系.截长，指在长线端中截取一段等于巳 知的线段；补短，指将一条短线端延长，延长部分等于己知线段.该类题目中常出现等腰三 角形、角平分线等关键词句，可以采用截长补短法构造全等三角形来完成证明过程.模型实例例 1：如图，己知在Z\ABC 中，ZC = 2ZB, Z1 = Z2 .求证：AB=AC+CD.图②证法一，截长法：如图①，在AB上取一点E,使AE = AC,连接DE.•「AE=AC, Z1 = Z2, AD=AD,A AACD^AAED ,A CD = DE, ZC=Z3.VZC = 2ZB,.\Z3=2ZB = Z44-ZB ,AZ4=ZB ,・・・DE=BE ,A CD = BE.•.・AB = AE+BE,・.・AB = AC+CD・证法二，补短法：如图②，延长AC到点E,使CE=CD,连接DE.•.・CE=CD, AZ4=ZE.VZ3 = Z4+ZE, AZ3 = 2ZE. VZ3=2ZB, AZE=ZB .VZ1 = Z2, AD=AD,A AEAD^ABAD, /.AE=AB.又・.・AE=AC+CE,A A AB = AC+CD.例 2：如图，已知 OD 平分/AOB, DCOA 于点 C, ZA=ZGBD.求证：AO + BO = 2CO.证明：段AO上取一点E,使CE=AC,连接DE.VCD=CD, DC1OA, A AACD^AECD, AZA=ZCED.VZA=ZGBD , AZCED=ZGBD , :.180- ZCED= 180- ZGBD , AZOED=ZOBD.•「OD 平分ZAOB, .•.ZAOD=ZBOD,VOD = OD,A AOED^AOBD ,・.・OB = OE, ・.・AO+BO=AO+OE=OE+2CE+OE=OE+CE+OE+CE=2 (CE+OE) =2CO .1. 如图，在AABC 中，ZBAC=60, AD 是ZBAC 的平分线，且 AC=AB + BD.求ZABC 的度数. 【答案】证法一：补短延长AB到点E,使BE=BD.在Z\BDE中，VBE=BD, A ZE=ZBDE,・・・ ZABC= ZBDE+ ZE=2ZE .又・..AC = AB + BD,・.・AC = AB + BE, ..・AC = AE.VAD 是 ZB AC 的平分线，ZBAC = 60(), 「・ ZEAD= ZCAD = 60%2=30 ・•.・AD = AD,AAAED^AACD, AZE=ZC.VZABC = 2ZE, AZABC=2ZC.・.NBAC=60,Z ABC + ZC = 180-60 = 120,・.・ 一 ZABC= 120,... ZABC=80 .2证法二：在AC上取一点F,使AF=AB,连接DF. •「AD是ZB AC的平分线，A ZBAD=ZFAD.VAD = AD,AABAD^AFAD,・.・NB=NAFD, BD=FD.・「AC = AB + BD, AC=AF+FC・・・FD=FC , A ZFDC=ZC.VZAFD=ZFDC+ZC,・.・ ZB=ZFDC+ ZC=2ZC ・VZBAC+ZB + ZC=180(),・・・-ZABC= 120,.・・ZABC = 80・22. 如图，在Z\ABC 中，ZABC=60, AD、CE 分别平分ZBAC、ZACB .求证：AC=AE +CD.【答案】如图，在AC边上取点F,使AE=AF,连接OF. VZABC = 60, /.ZBAC+ZACB = 180-ZABC=l20. •「AD、CE 分别平分匕BAC、ZACB, ・・./OAC=NOAB=^^, ZOCA=ZOCB=^d , 2 2・.・ ZAOE= ZCOD= ZOAC+ ZOCA=项』"《B =6Oo,2・・・ ZAOC=180-ZAOE= 120.VAE=AF, ZEAO=ZFAO, AO=AO,.•.△AOE丝△AOF (SAS),AZAOF=ZAOE=600,・•・ ZCOF= ZAOC- N AOF=60,・・./COF=NCOD・VCO=CO, CE 平分匕ACB,「•△COD丝△COF (ASA),・.・CD = CF・•.・AC = AF+CF,・.・AC = AE+CD,3. 如图，ZABC+ZBCD=180, BE、CE 分别平分ZABC> ZDCB .求证：AB+CD = BC.B C【答案】证法一：截长如图①，在BC上取一点F,使BF=AB,连接EF.VZ1 = ZABE, BE=BE,A AABE^AFBE, A Z3=Z4.VZABC+ZBCD=180,BE、CE 分另lj平分匕ABC、ZDCB,AZ1 + Z2=- ZABC+- ZDCB 2 2= -xl8O=9O0 , 2A ZBEC=90 , ・.・N4+Z5=90, Z3 + Z6=90(). VZ3=Z4 , /.Z5 = Z6. VCE=CE, Z2=ZDCE , A ACEF^ACED, ACF=CD.•.・BC = BF+CF, AB = BF, AAB+CD=BC图①证法二：补短如图②，延长BA到点F,使BF=BC,连接EF.VZ1 = ZABE, BE=BE,FAABEF^ABEC, ・.・EF=EC, ZBEC=ZBEF. VZABC+ZBCD=180,BE、CE 分别平分ZABC、ZDCB,AZ1 + Z2=- ZABC+ - ZDCB 2 2= -xl80=90 , 2A ZB EC=90 ,AZBEF=ZBEC=90,AZBEF+ZBEC=180, ・.・C、E、F三点共线. ・「AB〃CD, AZF=ZFCD. VEF=EC, ZFEA=ZDEC, A AAEF^ADEC, ・.・AF=CD・ VBF=AB + AF, ・.・BC = AB + CD・4. 如图，在Z\ABC 中，ZABC=90, AD 平分ZBAC 交 BC 于 D, ZC=30, BEAD 于【答案】延长BE交AC于点M・VBE1AD, AZAEB = ZAEM=90.VZ3=9O-Z1, Z4=90()-Z2, Z1 = Z2,AZ3=Z4, A AB = AM.VBE1AE, 「・BM=2BE・VZABC=90, ZC=30, A ZB AC=60.VAB = AM, .・・N3 = Z4=60,・・・N5=90—N3 = 30,A Z5=ZC, .・・CM = BM,・・・ AC-AB=CM = BM=2BE ・5. 如图，RtAACB A=BC, AD 平分ZBAC 交 BC 于点 D, CE_LAD 交 AD 于点 F, 交AB于点E・求证：AD=2DF+CE.AA【答案】在AD上取一点G,使AG=CE,连接CG. VCE1AD,A ZAFC=90, Zl + ZACF=90.VZ24-ZACF=90, AZ1 = Z2.・「AC = BC, AG=CE,A AACG^ACBE, A Z3= ZB=45,・・・ Z24- Z4=90-Z3=45.VZ2=Z1 = - ZBAC = 22.5,2・.・N4=450—N2=22.5,・・・ Z4=Z2=22.5.又・.・CF=CF, DG1CF,AACDF^ACGF, ADF=GF.•.・AD=AG+DG, ..・AD=CE+2DF・6. 如图，五边形 ABCDE 中，AB = AE, BC+DE=CD, ZB+ZE=180.求证：AD 平分 ZCDE.【答案】如图，延长CB到点F,使BF=DE,连接AF、AC.VZ1 + Z2=18O, ZE+Z 1 = 180, AZ2=ZE.•.・AB = AE, Z2=ZE, BF=DE,A AABF^AAED, AZF=Z4, AF=AD .V BC + DE=CD, .・・BC + BF=CD,即 FC = CD .又・.・AC = AC, A AACF^AACD,AZF=Z3.VZF=Z4,AZ3=Z4,「•AD平分NCDE.。