1、初中几何常见辅助线作法50种初中常见辅助线作法任何几何题目都需分析题目条件和结论找到解题思路,本讲从常见的条件和结论出发说明50种辅助线作法,分三角形部分、四边形部分、解直角三角形部分、圆。每种辅助线作法均配备了例题和练习。三角形部分1在利用三角形三边关系证明线段不等关系时,如果直接证不出来,可连结两点或延长某边构造三角形,使结论中出现的线段在一个或几个三角形中,再利用三边关系定理及不等式性质证题.例:如图,已知D、E为ABC内两点,求证:ABACBDDECE. 证法(一):将DE向两边延长,分别交AB、AC于M、N 在AMN中, AM ANMDDENE 在BDM中,MBMDBD 在CEN中,
2、CNNECE 得AMANMBMDCNNEMDDENEBDCEABACBDDECE证法(二)延长BD交AC于F,延长CE交BF于G,在ABF和GFC和GDE中有,ABAFBDDGGFGFFCGECEDGGEDE有ABAFGFFCDGGEBDDGGFGECEDEABACBDDECE注意:利用三角形三边关系定理及推论证题时,常通过引辅助线,把求证的量(或与求证有关的量)移到同一个或几个三角形中去然后再证题.练习:已知:如图P为ABC内任一点, 求证:(ABBCAC)PAPBPCABBCAC2.在利用三角形的外角大于任何和它不相邻的内角证明角的不等关系时,如果直接证不出来,可连结两点或延长某边,构造三
3、角形,使求证的大角在某个三角形外角的位置上,小角处在内角的位置上,再利用外角定理证题.例:已知D为ABC内任一点,求证:BDCBAC证法(一):延长BD交AC于E,BDC是EDC 的外角,BDCDEC同理:DECBACBDCBAC证法(二):连结AD,并延长交BC于FBDF是ABD的外角,BDFBAD同理CDFCADBDFCDFBADCAD即:BDCBAC3.有角平分线时常在角两边截取相等的线段,构造全等三角形. 例:已知,如图,AD为ABC的中线且1 = 2,3 = 4,求证:BECFEF证明:在DA上截取DN = DB,连结NE、NF,则DN = DC 在BDE和NDE中,DN = DB1
4、 = 2ED = EDBDENDEBE = NE同理可证:CF = NF在EFN中,ENFNEFBECFEF4. 有以线段中点为端点的线段时,常加倍延长此线段构造全等三角形.例:已知,如图,AD为ABC的中线,且1 = 2,3 = 4,求证:BECFEF证明:延长ED到M,使DM = DE,连结CM、FMBDE和CDM中, BD = CD1 = 5ED = MDBDECDMCM = BE又1 = 2,3 = 4 123 4 = 180o3 2 = 90o即EDF = 90oFDM = EDF = 90oEDF和MDF中ED = MDFDM = EDFDF = DFEDFMDFEF = MF在C
5、MF中,CFCM MFBECFEF(此题也可加倍FD,证法同上)5. 在三角形中有中线时,常加倍延长中线构造全等三角形.例:已知,如图,AD为ABC的中线,求证:ABAC2AD证明:延长AD至E,使DE = AD,连结BEAD为ABC的中线BD = CD在ACD和EBD中BD = CD 1 = 2AD = EDACDEBDABE中有ABBEAEABAC2AD6.截长补短作辅助线的方法截长法:在较长的线段上截取一条线段等于较短线段;补短法:延长较短线段和较长线段相等.这两种方法统称截长补短法.当已知或求证中涉及到线段a、b、c、d有下列情况之一时用此种方法:abab = cab = cd例:已知
6、,如图,在ABC中,ABAC,1 = 2,P为AD上任一点,求证:ABACPBPC证明:截长法:在AB上截取AN = AC,连结PN在APN和APC中,AN = AC1 = 2AP = APAPNAPCPC = PNBPN中有PBPCBNPBPCABAC补短法:延长AC至M,使AM = AB,连结PM在ABP和AMP中AB = AM 1 = 2AP = APABPAMPPB = PM又在PCM中有CM PMPCABACPBPC练习:1.已知,在ABC中,B = 60o,AD、CE是ABC的角平分线,并且它们交于点O求证:AC = AECD2.已知,如图,ABCD1 = 2 ,3 = 4. 求证
7、:BC = ABCD 7.条件不足时延长已知边构造三角形.例:已知AC = BD,ADAC于A,BCBD于B求证:AD = BC证明:分别延长DA、CB交于点EADAC BCBDCAE = DBE = 90o在DBE和CAE中DBE =CAEBD = ACE =EDBECAEED = EC,EB = EAEDEA = EC EBAD = BC8.连接四边形的对角线,把四边形问题转化成三角形来解决问题.例:已知,如图,ABCD,ADBC 求证:AB = CD 证明:连结AC(或BD)ABCD,ADBC1 = 2 在ABC和CDA中,1 = 2 AC = CA3 = 4 ABCCDAAB = CD
8、练习:已知,如图,AB = DC,AD = BC,DE = BF,求证:BE = DF9.有和角平分线垂直的线段时,通常把这条线段延长。可归结为“垂直加平分出等腰三角形”.例:已知,如图,在RtABC中,AB = AC,BAC = 90o,1 = 2 ,CEBD的延长线于E求证:BD = 2CE证明:分别延长BA、CE交于FBECFBEF =BEC = 90o在BEF和BEC中1 = 2 BE = BEBEF =BECBEFBECCE = FE =CFBAC = 90o , BECFBAC = CAF = 90o 1BDA = 90o1BFC = 90oBDA = BFC在ABD和ACF中BA
9、C = CAFBDA = BFCAB = ACABDACFBD = CFBD = 2CE练习:已知,如图,ACB = 3B,1 =2,CDAD于D,求证:ABAC = 2CD10.当证题有困难时,可结合已知条件,把图形中的某两点连接起来构造全等三角形.例:已知,如图,AC、BD相交于O,且AB = DC,AC = BD,求证:A = D证明:(连结BC,过程略)11.当证题缺少线段相等的条件时,可取某条线段中点,为证题提供条件.例:已知,如图,AB = DC,A = D 求证:ABC = DCB 证明:分别取AD、BC中点N、M,连结NB、NM、NC(过程略)12.有角平分线时,常过角平分线上
10、的点向角两边做垂线,利用角平分线上的点到角两边距离相等证题.例:已知,如图,1 = 2 ,P为BN上一点,且PDBC于D,ABBC = 2BD,求证:BAPBCP = 180o证明:过P作PEBA于EPDBC,1 = 2 PE = PD在RtBPE和RtBPD中BP = BPPE = PDRtBPERtBPDBE = BDABBC = 2BD,BC = CDBD,AB = BEAEAE = CDPEBE,PDBCPEB =PDC = 90o在PEA和PDC中PE = PDPEB =PDCAE =CDPEAPDCPCB = EAPBAPEAP = 180oBAPBCP = 180o练习:1.已知
11、,如图,PA、PC分别是ABC外角MAC与NCA的平分线,它们交于P,PDBM于M,PFBN于F,求证:BP为MBN的平分线2. 已知,如图,在ABC中,ABC =100o,ACB = 20o,CE是ACB的平分线,D是AC上一点,若CBD = 20o,求CED的度数。13.有等腰三角形时常用的辅助线作顶角的平分线,底边中线,底边高线例:已知,如图,AB = AC,BDAC于D,求证:BAC = 2DBC证明:(方法一)作BAC的平分线AE,交BC于E,则1 = 2 = BAC又AB = ACAEBC2ACB = 90oBDACDBCACB = 90o2 = DBCBAC = 2DBC(方法二
12、)过A作AEBC于E(过程略)(方法三)取BC中点E,连结AE(过程略)有底边中点时,常作底边中线例:已知,如图,ABC中,AB = AC,D为BC中点,DEAB于E,DFAC于F,求证:DE = DF证明:连结AD.D为BC中点,BD = CD又AB =ACAD平分BACDEAB,DFACDE = DF将腰延长一倍,构造直角三角形解题例:已知,如图,ABC中,AB = AC,在BA延长线和AC上各取一点E、F,使AE = AF,求证:EFBC证明:延长BE到N,使AN = AB,连结CN,则AB = AN = ACB = ACB, ACN = ANCBACBACNANC = 180o2BCA
13、2ACN = 180oBCAACN = 90o即BCN = 90oNCBCAE = AFAEF = AFE又BAC = AEF AFEBAC = ACN ANCBAC =2AEF = 2ANCAEF = ANCEFNCEFBC常过一腰上的某一已知点做另一腰的平行线例:已知,如图,在ABC中,AB = AC,D在AB上,E在AC延长线上,且BD = CE,连结DE交BC于F求证:DF = EF证明:(证法一)过D作DNAE,交BC于N,则DNB = ACB,NDE = E,AB = AC,B = ACBB =DNBBD = DN又BD = CE DN = EC在DNF和ECF中1 = 2NDF
14、=EDN = EC DNFECFDF = EF(证法二)过E作EMAB交BC延长线于M,则EMB =B(过程略)常过一腰上的某一已知点做底的平行线例:已知,如图,ABC中,AB =AC,E在AC上,D在BA延长线上,且AD = AE,连结DE求证:DEBC证明:(证法一)过点E作EFBC交AB于F,则AFE =BAEF =CAB = ACB =CAFE =AEFAD = AEAED =ADE又AFEAEFAEDADE = 180o2AEF2AED = 90o 即FED = 90o DEFE又EFBCDEBC(证法二)过点D作DNBC交CA的延长线于N,(过程略)(证法三)过点A作AMBC交DE
15、于M,(过程略)常将等腰三角形转化成特殊的等腰三角形-等边三角形例:已知,如图,ABC中,AB = AC,BAC = 80o ,P为形内一点,若PBC = 10o PCB = 30o 求PAB的度数.解法一:以AB为一边作等边三角形,连结CE则BAE =ABE = 60oAE = AB = BEAB = ACAE = AC ABC =ACBAEC =ACEEAC =BACBAE = 80o 60o = 20oACE = (180oEAC)= 80oACB= (180oBAC)= 50oBCE =ACEACB = 80o50o = 30oPCB = 30oPCB = BCEABC =ACB =
16、50o, ABE = 60oEBC =ABEABC = 60o50o =10oPBC = 10oPBC = EBC在PBC和EBC中PBC = EBCBC = BCPCB = BCEPBCEBCBP = BEAB = BEAB = BPBAP =BPAABP =ABCPBC = 50o10o = 40oPAB = (180oABP)= 70o解法二:以AC为一边作等边三角形,证法同一。解法三:以BC为一边作等边三角形BCE,连结AE,则EB = EC = BC,BEC =EBC = 60oEB = ECE在BC的中垂线上同理A在BC的中垂线上EA所在的直线是BC的中垂线EABCAEB = BE
17、C = 30o =PCB由解法一知:ABC = 50oABE = EBCABC = 10o =PBCABE =PBC,BE = BC,AEB =PCBABEPBCAB = BPBAP =BPAABP =ABCPBC = 50o10o = 40oPAB = (180oABP) = (180o40o)= 70o14.有二倍角时常用的辅助线构造等腰三角形使二倍角是等腰三角形的顶角的外角例:已知,如图,在ABC中,1 = 2,ABC = 2C,求证:ABBD = AC证明:延长AB到E,使BE = BD,连结DE则BED = BDEABD =EBDEABC =2EABC = 2CE = C 在AED和
18、ACD中E = C1 = 2AD = ADAEDACDAC = AEAE = ABBEAC = ABBE即ABBD = AC平分二倍角例:已知,如图,在ABC中,BDAC于D,BAC = 2DBC求证:ABC = ACB证明:作BAC的平分线AE交BC于E,则BAE = CAE = DBCBDACCBD C = 90oCAEC= 90o AEC= 180oCAEC= 90oAEBCABCBAE = 90oCAEC= 90oBAE = CAEABC = ACB加倍小角例:已知,如图,在ABC中,BDAC于D,BAC = 2DBC求证:ABC = ACB证明:作FBD =DBC,BF交AC于F(过
19、程略)15.有垂直平分线时常把垂直平分线上的点与线段两端点连结起来.例:已知,如图,ABC中,AB = AC,BAC = 120o,EF为AB的垂直平分线,EF交BC于F,交AB于E求证:BF =FC证明:连结AF,则AF = BFB =FABAB = ACB =CBAC = 120oB =CBAC =(180oBAC) = 30oFAB = 30oFAC =BACFAB = 120o30o =90o又C = 30oAF = FCBF =FC练习:已知,如图,在ABC中,CAB的平分线AD与BC的垂直平分线DE交于点D,DMAB于M,DNAC延长线于N求证:BM = CN16. 有垂直时常构造
20、垂直平分线.例:已知,如图,在ABC中,B =2C,ADBC于D求证:CD = ABBD证明:(一)在CD上截取DE = DB,连结AE,则AB = AEB =AEBB = 2CAEB = 2C又AEB = CEACC =EACAE = CE又CD = DECECD = BDAB(二)延长CB到F,使DF = DC,连结AF则AF =AC(过程略)17.有中点时常构造垂直平分线.例:已知,如图,在ABC中,BC = 2AB, ABC = 2C,BD = CD求证:ABC为直角三角形证明:过D作DEBC,交AC于E,连结BE,则BE = CE,C =EBCABC = 2CABE =EBCBC =
21、 2AB,BD = CDBD = AB在ABE和DBE中AB = BDABE =EBCBE = BEABEDBEBAE = BDEBDE = 90oBAE = 90o即ABC为直角三角形18.当涉及到线段平方的关系式时常构造直角三角形,利用勾股定理证题.例:已知,如图,在ABC中,A = 90o,DE为BC的垂直平分线求证:BE2AE2 = AC2证明:连结CE,则BE = CEA = 90o AE2AC2 = EC2AE2AC2= BE2BE2AE2 = AC2练习:已知,如图,在ABC中,BAC = 90o,AB = AC,P为BC上一点求证:PB2PC2= 2PA219.条件中出现特殊角
22、时常作高把特殊角放在直角三角形中.例:已知,如图,在ABC中,B = 45o,C = 30o,AB =,求AC的长. 解:过A作ADBC于DBBAD = 90o,B = 45o,B = BAD = 45o,AD = BDAB2 = AD2BD2,AB =AD = 1C = 30o,ADBCAC = 2AD = 2四边形部分20.有平行线时常作平行线构造平行四边形例:已知,如图,RtABC,ACB = 90o,CDAB于D,AE平分CAB交CD于F,过F作FHAB交BC于H求证:CE = BH证明:过F作FPBC交AB于P,则四边形FPBH为平行四边形B =FPA,BH = FPACB = 90
23、o,CDAB5CAB = 45o,BCAB = 90o5 =B5 =FPA又1 =2,AF = AFCAFPAFCF = FP4 =15,3 =2B3 =4CF = CECE = BH练习:已知,如图,ABEFGH,BE = GC求证:AB = EFGH21.有以平行四边形一边中点为端点的线段时常延长此线段. 例:已知,如图,在ABCD中,AB = 2BC,M为AB中点求证:CMDM证明:延长DM、CB交于N四边形ABCD为平行四边形AD = BC,ADBCA = NBA ADN =N又AM = BMAMDBMNAD = BNBN = BCAB = 2BC,AM = BMBM = BC = B
24、N1 =2,3 =N123N = 180o,13 = 90oCMDM22.有垂直时可作垂线构造矩形或平行线.例:已知,如图,E为矩形ABCD的边AD上一点,且BE = ED,P为对角线BD上一点,PFBE于F,PGAD于G求证:PFPG = AB证明:证法一:过P作PHAB于H,则四边形AHPG为矩形AH = GP PHADADB =HPBBE = DEEBD = ADBHPB =EBD又PFB =BHP = 90oPFBBHPHB = FPAHHB = PGPF即AB = PGPF证法二:延长GP交BC于N,则四边形ABNG为矩形,(证明略)23.直角三角形常用辅助线方法:作斜边上的高例:已
25、知,如图,若从矩形ABCD的顶点C作对角线BD的垂线与BAD的平分线交于点E求证:AC = CE证明:过A作AFBD,垂足为F,则AFEGFAE = AEG四边形ABCD为矩形BAD = 90o OA = ODBDA =CADAFBDABDADB = ABDBAF = 90oBAF =ADB =CADAE为BAD的平分线BAE =DAEBAEBAF =DAEDAC即FAE =CAECAE =AEGAC = EC作斜边中线,当有下列情况时常作斜边中线:有斜边中点时例:已知,如图,AD、BE是ABC的高, F是DE的中点,G是AB的中点求证:GFDE证明:连结GE、GDAD、BE是ABC的高,G是AB的中点GE = AB,GD = ABGE = GDF是DE的中点GFDE有和斜边倍分关系的线段时例:已知,如图,在ABC中,D是BC延长线上一点,且DABA于A,AC = BD求证:ACB = 2B证明:取BD中点E,连结AE,则AE = BE = BD1 =BAC = BDAC = AEACB =2 2 =