数姐有话 对于初二的同学来说,三角形与全等三角形,才是同学们正式接触几何,而在这块内容中,辅助线又是必不可少的,所以,希望同学们好好学习这块内容,对于以后学习更难的几何知识打下基础! 例:已知D为△ABC内任一点,求证:∠BDC>∠BAC 证明: (一):延长BD交AC于E, ∵∠BDC是△EDC 的外角, ∴∠BDC>∠DEC 同理:∠DEC>∠BAC ∴∠BDC>∠BAC 证法(二):连结AD,并延长交BC于F ∵∠BDF是△ABD的外角, ∴∠BDF>∠BAD 同理∠CDF>∠CAD ∴∠BDF+∠CDF>∠BAD+∠CAD 即:∠BDC>∠BAC 例:已知,如图,AD为△ABC的中线且∠1 = ∠2,∠3 = ∠4, 求证:BE+CF>EF 证明: 在DA上截取DN = DB,连结NE、NF, 则DN= DC 在△BDE和△NDE中, DN = DB ∠1 = ∠2 ED = ED ∴△BDE≌△NDE ∴BE = NE 同理可证:CF = NF 在△EFN中,EN+FN>EF ∴BE+CF>EF 例:已知,如图,AD为△ABC的中线,且∠1 = ∠2,∠3 = ∠4,求证:BE+CF>EF 证明: 延长ED到M,使DM = DE,连结CM、FM △BDE和△CDM中, BD = CD ∠1 = ∠5 ED = MD ∴△BDE≌△CDM ∴CM = BE 又∵∠1 = ∠2,∠3 = ∠4 ∠1+∠2+∠3 + ∠4 = 180° ∴∠3 +∠2 = 90° 即∠EDF = 90° ∴∠FDM = ∠EDF = 90° △EDF和△MDF中 ED = MD ∠FDM = ∠EDF DF = DF ∴△EDF≌△MDF ∴EF = MF ∵在△CMF中,CF+CM >MF BE+CF>EF (此题也可加倍FD,证法同上) 例:已知,如图,AD为△ABC的中线,求证:AB+AC>2AD 证明: 延长AD至E,使DE = AD,连结BE ∵AD为△ABC的中线 ∴BD = CD 在△ACD和△EBD中 BD = CD ∠1 = ∠2 AD = ED ∴△ACD≌△EBD ∵△ABE中有AB+BE>AE ∴AB+AC>2AD 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: ①a>b ②a±b = c ③a±b = c±d 例:已知,如图,在△ABC中,AB>AC,∠1 = ∠2,P为AD上任一点, 求证:AB-AC>PB-PC 证明: ⑴截长法:在AB上截取AN = AC,连结PN 在△APN和△APC中, AN = AC ∠1 = ∠2 AP = AP ∴△APN≌△APC ∴PC = PN ∵△BPN中有PB-PC<BN ∴PB-PC<AB-AC ⑵补短法:延长AC至M,使AM = AB,连结PM 在△ABP和△AMP中 AB = AM ∠1 = ∠2 AP = AP ∴△ABP≌△AMP ∴PB = PM 又∵在△PCM中有CM >PM-PC ∴AB-AC>PB-PC 练习: 1.已知,在△ABC中,∠B = 60°,AD、CE是△ABC的角平分线,并且它们交于点O 求证:AC = AE+CD 2.已知,如图,AB∥CD,∠1 = ∠2 ,∠3 = ∠4. 求证:BC = AB+CD ①观察要证线段在哪两个可能全等的三角形中,然后证这两个三角形全等。 ②若图中没有全等三角形,可以把求证线段用和它相等的线段代换,再证它们所在的三角形全等. ③如果没有相等的线段代换,可设法作辅助线构造全等三角形. 例:如图,已知,BE、CD相交于F,∠B = ∠C,∠1 = ∠2,求证:DF = EF 证明:∵∠ADF =∠B+∠3 ∠AEF = ∠C+∠4 又∵∠3 = ∠4 ∠B = ∠C ∴∠ADF = ∠AEF 在△ADF和△AEF中 ∠ADF = ∠AEF ∠1 = ∠2 AF = AF ∴△ADF≌△AEF ∴DF = EF 例:已知,如图Rt△ABC中,AB = AC,∠BAC = 90°,过A作任一条直线AN,作BD⊥AN于D,CE⊥AN于E,求证:DE = BD-CE 证明:
∴∠1+∠2 = 90o ∠1+∠3 = 90° ∴∠2 = ∠3 ∵BD⊥AN CE⊥AN ∴∠BDA =∠AEC = 90° 在△ABD和△CAE中, ∠BDA =∠AEC ∠2 = ∠3 AB = AC ∴△ABD≌△CAE ∴BD = AE且AD = CE ∴AE-AD = BD-CE ∴DE = BD-CE 例:AD为△ABC的中线,且CF⊥AD于F,BE⊥AD的延长线于E 求证:BE = CF 证明:(略) 例:已知AC = BD,AD⊥AC于A,BCBD于B 求证:AD = BC 证明:分别延长DA、CB交于点E ∵AD⊥AC BC⊥BD ∴∠CAE = ∠DBE = 90° 在△DBE和△CAE中 ∠DBE =∠CAE BD = AC ∠E =∠E ∴△DBE≌△CAE ∴ED = EC,EB = EA ∴ED-EA = EC- EB ∴AD = BC 例:已知,如图,AB∥CD,AD∥BC 求证:AB = CD 证明:
∵AB∥CD,AD∥BC ∴∠1 = ∠2 在△ABC和△CDA中, ∠1 = ∠2 AC = CA ∠3 = ∠4 ∴△ABC≌△CDA ∴AB = CD 练习: 已知,如图,AB = DC,AD = BC,DE = BF, 求证:BE = DF 例:已知,如图,在Rt△ABC中,AB = AC,∠BAC = 90°,∠1 = ∠2 ,CE⊥BD的延长线于E 求证:BD = 2CE 证明: 分别 延长BA、CE交于F ∵BE⊥CF ∴∠BEF =∠BEC = 90° 在△BEF和△BEC中 ∠1 = ∠2 BE = BE ∠BEF =∠BEC ∴△BEF≌△BEC ∴CE = FE =1/2CF ∵∠BAC = 90° , BE⊥CF ∴∠BAC = ∠CAF = 90° ∠1+∠BDA = 90° ∠1+∠BFC = 90° ∠BDA = ∠BFC 在△ABD和△ACF中 ∠BAC = ∠CAF ∠BDA = ∠BFC AB = AC ∴△ABD≌△ACF ∴BD = CF ∴BD = 2CE 练习: 已知,如图,∠ACB = 3∠B,∠1 =∠2,CD⊥AD于D, 求证:AB-AC = 2CD 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) 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) 例:已知,如图,∠1 = ∠2 ,P为BN上一点,且PD⊥BC于D,AB+BC = 2BD, 求证:∠BAP+∠BCP = 180° 证明:过P作PE⊥BA于E ∵PD⊥BC,∠1 = ∠2 ∴PE = PD 在Rt△BPE和Rt△BPD中 BP = BP PE = PD ∴Rt△BPE≌Rt△BPD ∴BE = BD ∵AB+BC = 2BD,BC = CD+BD,AB = BE-AE ∴AE = CD ∵PE⊥BE,PD⊥BC ∠PEB =∠PDC = 90° 在△PEA和△PDC中 PE = PD ∠PEB =∠PDC AE =CD ∴△PEA≌△PDC ∴∠PCB = ∠EAP ∵∠BAP+∠EAP = 180° ∴∠BAP+∠BCP = 180° 练习: 1.已知,如图,PA、PC分别是△ABC外角∠MAC与∠NCA的平分线,它们交于P,PD⊥BM于M,PF⊥BN于F,求证:BP为∠MBN的平分线 2. 已知,如图,在△ABC中,∠ABC =100o,∠ACB = 20°,CE是∠ACB的平分线,D是AC上一点,若∠CBD = 20°,求∠CED的度数。 ⑴作顶角的平分线,底边中线,底边高线 例:已知,如图,AB = AC,BD⊥AC于D, 求证:∠BAC = 2∠DBC 证明: (方法一)作∠BAC的平分线AE,交BC于E,则∠1 = ∠2 = 1/2∠BAC 又∵AB = AC ∴AE⊥BC ∴∠2+∠ACB = 90° ∵BD⊥AC ∴∠DBC+∠ACB = 90° ∴∠2 = ∠DBC ∴∠BAC = 2∠DBC (方法二)过A作AE⊥BC于E(过程略) (方法三)取BC中点E,连结AE(过程略) ⑵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,DE⊥AB于E,DF⊥AC于F, 求证:DE = DF 证明:连结AD. ∵D为BC中点, ∴BD = CD 又∵AB =AC ∴AD平分∠BAC ∵DE⊥AB,DF⊥AC ∴DE = DF ⑶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:EF⊥BC 证明:延长BE到N,使AN = AB,连结CN,则AB = AN = AC ∴∠B = ∠ACB, ∠ACN = ∠ANC ∵∠B+∠ACB+∠ACN+∠ANC = 180° ∴2∠BCA+2∠ACN = 180° ∴∠BCA+∠ACN = 90° 即∠BCN = 90° ∴NC⊥BC ∵AE = AF ∴∠AEF = ∠AFE 又∵∠BAC = ∠AEF +∠AFE ∠BAC = ∠ACN +∠ANC ∴∠BAC =2∠AEF = 2∠ANC ∴∠AEF = ∠ANC ∴EF∥NC ∴EF⊥BC ⑷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DN∥AE,交BC于N,则∠DNB = ∠ACB,∠NDE = ∠E, ∵AB = AC, ∴∠B = ∠ACB ∴∠B =∠DNB ∴BD = DN 又∵BD = CE ∴DN = EC 在△DNF和△ECF中 ∠1 = ∠2 ∠NDF =∠E DN = EC ∴△DNF≌△ECF ∴DF = EF (证法二)
⑸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:DE⊥BC 证明:(证法一)过点E作EF∥BC交AB于F,则 ∠AFE =∠B ∠AEF =∠C ∵AB = AC ∴∠B =∠C ∴∠AFE =∠AEF ∵AD = AE ∴∠AED =∠ADE 又∵∠AFE+∠AEF+∠AED+∠ADE = 180o ∴2∠AEF+2∠AED = 90o 即∠FED = 90o ∴DE⊥FE 又∵EF∥BC ∴DE⊥BC (证法二)过点D作DN∥BC交CA的延长线于N,(过程略) (证法三)过点A作AM∥BC交DE于M,(过程略) ⑹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 = 60o AE = AB = BE ∵AB = AC ∴AE = AC ∠ABC =∠ACB ∴∠AEC =∠ACE ∵∠EAC =∠BAC-∠BAE = 80°-60° = 20° ∴∠ACE = 1/2(180°-∠EAC)= 80° ∵∠ACB= 1/2(180°-∠BAC)= 50° ∴∠BCE =∠ACE-∠ACB = 80°-50° = 30° ∵∠PCB = 30° ∴∠PCB = ∠BCE ∵∠ABC =∠ACB = 50°, ∠ABE = 60° ∴∠EBC =∠ABE-∠ABC = 60°-50° =10° ∵∠PBC = 10° ∴∠PBC = ∠EBC 在△PBC和△EBC中 ∠PBC = ∠EBC BC = BC ∠PCB = ∠BCE ∴△PBC≌△EBC ∴BP = BE ∵AB = BE ∴AB = BP ∴∠BAP =∠BPA ∵∠ABP =∠ABC-∠PBC = 50°-10° = 40° ∴∠PAB = 1/2(180°-∠ABP)= 70° 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 = 60o ∵EB = EC ∴E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 ∴EA⊥BC ∠AEB = 1/2∠BEC = 30° =∠PCB 由解法一知:∠ABC = 50° ∴∠ABE = ∠EBC-∠ABC = 10°=∠PBC ∵∠ABE =∠PBC,BE = BC,∠AEB =∠PCB ∴△ABE≌△PBC ∴AB = BP ∴∠BAP =∠BPA ∵∠ABP =∠ABC-∠PBC = 50°-10°= 40° ∴∠PAB = 1/2(180o-∠ABP) = 1/2(180°-40°)= 70° ⑴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 = 2∠C, 求证:AB+BD = AC 证明:延长AB到E,使BE = BD,连结DE 则∠BED = ∠BDE ∵∠ABD =∠E+∠BDE ∴∠ABC =2∠E ∵∠ABC = 2∠C ∴∠E = ∠C 在△AED和△ACD中 ∠E = ∠C ∠1 = ∠2 AD = AD ∴△AED≌△ACD ∴AC = AE ∵AE = AB+BE ∴AC = AB+BE 即AB+BD = AC ⑵平分二倍角 例:已知,如图,在△ABC中,BD⊥AC于D,∠BAC = 2∠DBC 求证:∠ABC = ∠ACB 证明:作∠BAC的平分线AE交BC于E,则∠BAE = ∠CAE = ∠DBC ∵BD⊥AC ∴∠CBD +∠C = 90o ∴∠CAE+∠C= 90o ∵∠AEC= 180o-∠CAE-∠C= 90o ∴AE⊥BC ∴∠ABC+∠BAE = 90o ∵∠CAE+∠C= 90o ∠BAE = ∠CAE ∴∠ABC = ∠ACB ⑶加倍小角 例:已知,如图,在△ABC中,BD⊥AC于D,∠BAC = 2∠DBC 求证:∠ABC = ∠ACB 证明:作∠FBD =∠DBC,BF交AC于F(过程略) 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 =1/2FC 证明:连结AF,则AF = BF ∴∠B =∠FAB ∵AB = AC ∴∠B =∠C ∵∠BAC = 120o ∴∠B =∠C∠BAC =1/2(180°-∠BAC) = 30° ∴∠FAB = 30° ∴∠FAC =∠BAC-∠FAB = 120°-30° =90° 又∵∠C = 30° ∴AF = 1/2FC ∴BF =1/2FC 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,DM⊥AB于M,DN⊥AC延长线于N 求证:BM = CN 例:已知,如图,在△ABC中,∠B =2∠C,AD⊥BC于D 求证:CD = AB+BD 证明: (一)在CD上截取DE = DB,连结AE,则AB = AE ∴∠B =∠AEB ∵∠B = 2∠C ∴∠AEB = 2∠C 又∵∠AEB = ∠C+∠EAC ∴∠C =∠EAC ∴AE = CE 又∵CD = DE+CE ∴CD = BD+AB (二)延长CB到F,使DF = DC,连结AF则AF =AC(过程略) 例:已知,如图,在△ABC中,BC = 2AB, ∠ABC = 2∠C,BD = CD 求证:△ABC为直角三角形 证明:过D作DE⊥BC,交AC于E,连结BE,则BE = CE, ∴∠C =∠EBC ∵∠ABC = 2∠C ∴∠ABE =∠EBC ∵BC = 2AB,BD = CD ∴BD = AB 在△ABE和△DBE中 AB = BD ∠ABE =∠EBC BE = BE ∴△ABE≌△DBE ∴∠BAE = ∠BDE ∵∠BDE = 90° ∴∠BAE = 90° 即△ABC为直角三角形 例:已知,如图,在△ABC中,∠A = 90°,DE为BC的垂直平分线 求证:BE2-AE2 = AC2 证明:连结CE,则BE = CE ∵∠A = 90° ∴AE2+AC2 = EC2 ∴AE2+AC2= BE2 ∴BE2-AE2 = AC2 练习: 已知,如图,在△ABC中,∠BAC = 90°,AB = AC,P为BC上一点 求证:PB2+PC2= 2PA2 例:已知,如图,在△ABC中,∠B = 45°,∠C = 30°,AB =根号2,求AC的长. 解:过A作AD⊥BC于D ∴∠B+∠BAD = 90°, ∵∠B = 45o,∠B = ∠BAD = 45°, ∴AD = BD ∵AB2 = AD2+BD2,AB =根号2 ∴AD = 1 ∵∠C = 30°,AD⊥BC ∴AC = 2AD = 2 |
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