截长补短辅助线模型

1、截长补短辅助线模型模型：截长补短L1 1AB CD如图，若证明线段AB、CD、EF之间存在EF = AB11+ CD,可以考虑截长补短法EF截长法：如图，在EF上截取EG=AB ,再证明GF=CD即可.111补短法：如图，延长明AH = EF即可.AB至H点，使BH = CD ,再证EIG1F1ABH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已 知的线段；补短，指将一条短线端延长，延长部分等于已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，已知在 ABC 中，/ C= 2/ B，

2、/ 1 = 7 2 .求证：AB = AC + CD .图证法一，截长法：如图，在 AB上取一点E,使AE= AC,连接DE./ AE = AC ,7 1 = 7 2, AD = AD , ACD AED , CD = DE , 7 C=7 3 .7 C= 2 7 B , 7 3= 27 B = 7 4 +7 B , 7 4=7 B , DE = BE , CD = BE./ AB = AE + BE , AB = AC + CD .证法二，补短法: 如图,延长 AC到点E ,使CE = CD ,连接DE ./ CE = CD, / 4 =7 E . 7 3=7 4+7 E, 7 3 = 27

3、 E ./ 3= 2/B,./ E =Z B ./ 1 = Z 2, AD = AD , EAD BAD , AE = AB. 又: AE = AC + CE ,AB = AC + CD .例2:如图，已知OD平分/ AOB , DC丄OA于点C, / A = Z GBD .求证：AO + BO = 2CO ./ CD = CD , DC 丄 OA , ACD ECD ,/ A = Z CED ./ A = Z GBD , / CED = Z GBD , 180°/ CED = 180°/ GBD , / OED = / OBD ./ OD 平分/ AOB , / AOD

4、=/ BOD ./ OD = OD , OED OBD , OB = OE, AO + BO = AO + OE= OE + 2CE + OE = OE + CE+ OE + CE= 2 ( CE + OE)= 2CO .1. 如图，在厶ABC中，/ BAC = 60°, AD是/ BAC的平分线，且AC = AB + BD .求/ ABC 的度数【答案】证法一：补短延长AB到点E,使BE = BD .在厶BDE中，/ BE = BD ,/ E =Z BDE ,/ ABC = Z BDE +Z E = 2/ E .又 AC = AB + BD , AC = AB + BE , AC

5、= AE ./AD 是/ BAC 的平分线，/ BAC = 60°,/ EAD = Z CAD = 60°弋=30° ./ AD = AD , AED ACD，/ E =Z C ./ ABC = 2 / E，/ ABC = 2 / C ./ BAC = 60°,/ ABC + Z C = 1800 600= 1200, 3 / ABC = 1200,./ ABC = 800 .2证法二：在 AC上取一点F，使AF = AB，连接DF. AD是/ BAC的平分线，/ BAD =Z FAD ./ AD = AD , BAD FAD ,/ B =Z AFD ,

6、 BD = FD ./ AC = AB + BD , AC = AF + FC FD = FC，/ FDC = Z C ./ AFD =Z FDC + Z C,/ B =Z FDC + Z C = 2/ C ./ BAC + Z B + Z C= 1800 , 3 / ABC = 1200 ,ABC = 800 .22. 如图，在 ABC 中，/ ABC = 600 , AD、CE 分别平分/ BAC、/ ACB .求证：AC = AE + CD .【答案】如图，在 AC边上取点F，使AE = AF，连接OF ./ ABC =BAC +Z ACB = 180°/ ABC = 120&

7、#176; ./ AD、CE 分别平分/ BAC、/ ACB ,/ OAC = / OAB = DBAC , / OCA = / OCB = DACB ,2 2/ AOE = / COD = / OAC + / OCA = ? BAC ? ACB = 60°,2/ AOC = 180°/ AOE = 120° ./ AE = AF，/ EAO = / FAO, AO = AO , AOE AOF (SAS), / AOF = / AOE = 60°, / COF = / AOC / AOF = 60°, / COF = / COD ./ CO

8、= CO , CE 平分/ ACB , COD COF (ASA ), CD = CF ./ AC = AF + CF, AC = AE + CD ,3. 如图，/ ABC +/ BCD = 180°, BE、CE 分别平分/ ABC、/ DCB .求证：AB + CD = BC .图FC【答案】证法一：截长如图，在 BC上取一点F，使BF = AB，连接EF ./ 1 = / ABE , BE = BE, ABE FBE， / 3=/ 4 ./ ABC + / BCD = 180°,BE、CE 分别平分/ ABC、/ DCB ,./ 1 + / 2= 1 / ABC +

9、1 / DCB2 2=-X180°= 90° ,2/ BEC = 90° , / 4+/ 5= 90°,/ 3 +/ 6= 90° ./ 3=/ 4 , / 5=/ 6 ./ CE = CE,/ 2=/ DCE , CEF也厶 CED , CF= CD ./ BC = BF + CF , AB = BF , AB + CD = BC证法二：补短如图，延长BA到点F ,使BF = BC ,连接EF ./ 1 = / ABE , BE = BE , BEF 也厶 BEC ,圈 EF = EC, / BEC = Z BEF ./ ABC + / BC

10、D = 180°,BE、CE 分别平分/ ABC、/ DCB ,11/ 1 + Z 2=/ ABC + 丄/ DCB221=-X180°= 90° ,2/ BEC = 90°/ BEF = / BEC = 900, / BEF + / BEC = 1800, C、E、F三点共线./AB / CD , / F =/ FCD ./ EF = EC，/ FEA = / DEC , AEF DEC , AF = CD ./ BF = AB + AF , BC = AB + CD .4. 如图，在厶 ABC 中，/ ABC = 900 , AD 平分/ BAC 交

11、 BC 于 D, / C= 300 , BE 丄 AD 于 点 E .求证：AC AB = 2BE .【答案】延长BE交AC于点M ./ BE 丄 AD ,AEB =/ AEM = 900 ./ 3= 900/ 1 , / 4 = 900/ 2, / 1 = / 2 ,/ 3=/ 4, AB = AM ./ BE 丄 AE , BM = 2BE ./ ABC = 900 , / C= 300 , / BAC = 600./AB = AM , / 3=/ 4 = 600 , / 5= 900/ 3 = 300 , / 5=/ C , CM = BM , AC AB = CM = BM = 2BE

12、 .5. 如图，Rt ACB中，A = BC , AD平分/ BAC交BC于点D , CE丄AD交AD于点F , 交AB于点E .求证：AD = 2DF + CE .【答案】在AD上取一点G,使AG = CE,连接CG . / CE 丄 AD ,/ AFC = 90°，/ 1 + Z ACF = 90° ./ 2+Z ACF = 90°,二/ 1 = 7 2 ./ AC = BC , AG = CE, ACG CBE，/ 3 =7 B = 45°, 7 2+7 4= 90°-7 3 = 45°.7 2=7 1= - 7 BAC = 22.5°,2.7 4= 450-7 2 = 22.5°,.7 4=7 2= 22.5°.又 CF= CF, DG 丄 CF, CDF CGF , DF = GF ./ AD = AG + DG , AD = CE+ 2DF .6. 如图，五边形 ABCDE 中，AB = AE , BC + DE = CD, 7 B+7 E= 180° .求证：AD 平分 7 CDE .【答案】如图，延长 CB到点F,使BF = DE，连接AF、AC . 7 1 + 7 2= 180&