初中數(shù)學(xué)輔助線（Junior high school mathematics auxiliary line）

初中數(shù)學(xué)輔助線(Juniorhighschoolmathematicsauxiliaryline)

Juniorhighschoolmathematicsauxiliaryline

1.triangleproblemaddingauxiliarylinemethod

Method1:questionsaboutthemiddlelineofatriangle,often

doublingthemiddle1ine.Itiseasytosolvetheproblemwith

themidpointoftheproblem,oftenusingthemiddlelineofthe

triangle.

Method2:thesubjectthatcontainsthebisector,oftentakes

theangularbisectorasthesymmetryaxis,andusestheproperty

ofthebisectorandtheconditioninthequestiontoconstruct

thecongruenttriangle,thussolvestheproblembyusingthe

knowledgeofcongruenttriangle.

Method3:theconclusionisthatthetwolinesegmentsareequal,

theauxiliarylinesareoftendrawnintocongruenttriangles,

orsometheoremsaboutthebisectionlineareused.

Method4:conclusionisalinewithanotherlineandthethird

lineisequaltothiskindofproblems,oftenwithlongorshort

cutmethod,theso-calledtenniniallymethodistoputthethird

linesaredividedintotwoparts,onepartofthecardisequal

tothefirst1ine,andtheotherpartisequaltosecondlines.

Theaddingmethodofauxiliarylinein2.parallelogram

Theparallelogram(includingrectangular,square,diamond)two

groupofdiagonalanddiagonaledges,hassomeofthesame

nature,sotherearealsocommoninaddingauxiliarylinemethod,

theobjectiveistocreatelineparallel,vertical,similar,

congruenttriangles,theparallelogramproblemintoatriangle,

squareandotherissuesofcommonprocessing,hasthefollowing

severalcommonmethods,forexamplesimplesolutionsas

follows:

(1)diagonalortranslationdiagonal:

(2)theverticalstructureoftheedgeoppositethevertex;the

righttriangle

(3)parallellinesconnectingthediagonalintersectionpoint

withthemidpointofonesideorthediagonalintersectionpoint

areconstructed,andtheparallelormiddlelineoftheline

segmentisconstructed

(4)alinesegmentconnectingthevertextoapointonthe

oppositeedgeorextendingtheline,constructingatriangle,

asimilaroranequaltriangle.

(5)makeadiagonallineperpendiculartothevertexandform

alineparallelortrianglecongruent

3.laddercommonlyusedintheauxiliarylineoftim

Atrapeziumisaspecialquadrilateral.Itisasynthesisof

parallelogramandtriangleknowledge.Byaddingproper

auxiliary1ines,thetrapezoidalproblemisreducedtoa

parallelogramoratriangleproblem.Theauxiliarylineis

addedasabridgetosolvetheproblem:

(1)moveonewaistinthetrapezoid.

(2)trapezoidoutsidetranslationonewaist

(3)insidetheladder,movetwowaist

(4)extendtwowaist

(5)theendsofthetopandbottomoftheladderarehigh

(6)translationdiagonal

(7)connecttheacmeofaladderandthemidpointofawaist.

(8)themidpointofonewaististheparallellineoftheother

waist.

(9)makemiddlepositionline;

Ofcourse,intherelevantdemonstrationandcalculationofthe

ladder,theauxiliarylinesaddedarenotnecessarilyfixedand

single.Itisthekeytosolvetheproblemofturningthe

trapeziumproblemintoparallelogramortrianglethroughthe

auxiliarylineofthebridge.

Amethodofmakingauxiliarylines

I.midpoint,middleline,extension1ine,parallelline.

Apointincaseofconditionsinline,themedian1ine,then

amidpoint,extendthemidlineormedianlineastheauxiliary

line,theextensionofasegmentisequaltothemedianline

orline;anotherauxiliarylineisthemidpointforparallel

linesorlinesegmentsknownside,inordertoachievethe

applicationofatheoremthepurposeofmakingorhelp.

Two:vertical,bisector,flip,congruent,even.

Incaseswherethereisabisectoroftheverticalortheangle,

thegraphcanberotatedby180degreesbythemethodofaxial

symmetryandbyotherconditions,andthecontourofthe

auxiliarylinewillcomeintobeing.Thesymmetryaxisisoften

thebisectoroftheverticalortheangle.

Three:experimentiftheedgesareequalandrotated.

Apolygononbothsidesofthecornersincaseofequalorequal

conditionsinthecorner,sometimeswitheachother,thenthe

angleofgraphrotationtosomeextent,youcangetcongruent,

thenauxiliarylinepracticesstillcomeintobeing.Thecenter

ofsymmetryvariesfromsubjecttocenter,sometimeswithout

center.Itcanbedividedinto"heart"and"nomind“rotation

ofthetwo.

Four:maketheangle,theflat,thesimilar,thedifference,

thedifference,theproductandthecommercialview.

Apolygononbothsidesofthecornersincaseofequalorequal

conditions,tosegmentorangleanddifferenceproductisoften

associatedwithsimilarbusiness.Inmakingtwotrianglesalike,

therearegenerallytwoways:first,tocreateanauxiliary

angleequaltotheknownangle;second,tomovealineinthe

triangle.Averse:"angle,flat,similaranddifferenceproduct

dealerssee.”

Five:areatofindthebottomhigh,multilateralchangethree

sides.

Inthecaseofthearea,thesquareandtheproductoftheline

segmentintheconditionandtheconclusioncanstillbe

regardedastheareaofthesolution,andoftenthebottomor

theheightistheauxiliaryline,whiletheequalbottomor

equalheightofthetwotriangleisthekeytothethinking.

Ifapolygonisencountered,theideaiscutupintoatriangle;

otherwise,itisestablished.

Inaddition,MingandQingDynastiesinChinaareafor

mathematicianstoprovethePythagoreantheorem,theauxiliary

lineapproach,namely"patching“hasmorethan200kinds,most

oftheareaattheendofthree,themultilateralboundary”.

Juniorhighschoolgeometrycommonauxiliarylineformula

Peoplesaygeometryisverydifficult,andthedifficultyis

intheauxiliaryline.Auxiliaryline,howtoadd?Grasp

theoremsandconcepts.

Wemuststudyhard,findoutthelaw,andrelyonexperience.

Triangle

Thegraphhasanangularbisectorthatcanbeperpendicularto

bothsides.Youcanalsolookatthefigureinhalf,symmetrical,

thentherelationshipis.

Anglebisector,parallelline,isoscelestriangletoadd.Angle

bisector,plusverticalline,trythreelinestogether.

A1ineofverticalbisector,oftenjoinedtobothends.Line

segmentsanddifferences,andhalftimes,extended,shortened,

andtestable.

Thelinesegmentsanddifferenceinequalitiesmovetothesame

triangle.Themiddleofthetriangleistwopoints,andthe

connectionisamiddleline.

Thereisamiddlelineinthetriangle,extendingthemiddle

lineandsoon.

Quadrilateral

Theparallelogramappearsandthecenterofsymmetrydivides.

Theproblemhowtoconvertladder,anddelta.

Movethewaist,movediagonally,andlengthenthetwowaistto

makeithigh.Ifthereisamiddlepointofthewaist,carefully

connectthemiddleline.

Theabovemethoddoesnotwork,overwaistmidpointcongruent

creation.Similartotheline,lineparalleltothehabit.

Theplotratioformulaisthekeysegmentforchange.Direct

proofofdifficulty,equalsubstitution,lesstrouble.

Thehypotenuseaboveline,alargeproportionofitems.

Examplesofcommonlyusedmethodsforassistinglinesin

triangles

Onetimesthecenterline

1:ABCADBCisknown,onthesideofthemidline,asanisosceles

righttrianglewithABedge,ACedgeisrectangularshape

anisotropy,asshowninFigure5-2,confirmationofEF=2AD.

Two,takefromauxiliary1inemethod.

InABC,AD/BAC/ACBsplit,/B=2,AB=ACCDconfirmation.

Three,extendtheknownsidestructuretriangle:

Forexample:Figure7-1:givenAC=BD,ADACinAgroup,BC

groupofBDinB:AD=BC,confirmation

Analysis:toAD=BC,ofwhichcontainAD,BCthereareseveral

congruenttriangles,DeltaADCanddeltaplan:BCD,DeltaAOD

anddeltaBOC,DeltaABDanddeltaBAC,butaccordingtothe

existingconditionsarenotequaltothedifferenceofangle

congruentcard,therefore,cantrytomakeanewangle.Letthe

angleastwotriangularpublicangle.

Proof:extendedDA,CB,respectively,andtheirelongationat

Epoint,

ADACBCgroupBDgroupofdreams(known)

L/CAE/DBE==90DEG(verticaldefinition)

InthedeltaDBEanddeltaCAE

Dreams

StardeltaDBE=CAE(AAS)

ED=EB=EA*EC(congruenttrianglescorrespondingequal

sides)

L=EDEAECEB

Namely:AD=BC.

Whenconditionsareinsufficient,newconditionscanbecreated

byaddingauxiliarylinestocreateconditionsforthequestion

Four,connectthequadrilateraldiagonal,thequadrilateral

problemintoatriangletosolve.

Forexample:asshowninfigure8-1:AB,CD,ADandBCprovethat

AB=CD.

Analysis:thegraphisquadrilateral,weonlylearnthe

knowledgeabouttriangle,andmusttransformitintotriangle

tosolve.

Proof:connectAC(orBD)

ABCDADBC"Dreams”(known)

L/l=2/3=4,//(twolineparallel,alternateanglesequal)

InthedeltaABCanddeltaCDA

Dreams

StardeltaABC=CDA(ASA)

ABCD(r=congruenttrianglescorrespondingequalsides)

Five.Thelineisusuallyextendedwhenthereisalinesegment

perpendiculartotheanglebisector.

Forexample:asshowninFigure9-1:inRtDeltaABC,AB=AC

/BAC=90/1=2DEGangle,theextensionofBDinECEgroup.

Confirmation:BD=2CE

Analysis:toBD=2CE,thoughttoconstructthesegments2CE,

CEandABCanglebisectorandvertical,thoughttoextendthe.

Proof:extendedBAandCEtopointF,respectively.

BECF(known)groupofdreams

L/BEF/BEC==90DEG(verticaldefinition)

InthedeltaBEFanddeltaBEC,

Dreams

StardeltaBEF=BEC(ASA)*CE=FE=CF(congruenttriangles

correspondingequalsides)

AnCF/BAC=90?BEdreams(known)

L/BAC/CAF==90deg/1+/BDA=90deg/1+/BFC=90

DEG

L/BDA=/BFC

InthedeltaABDanddeltaACF

StardeltaABD=ACF(AAS)*BD=CF(congruenttriangles

correspondingequalsides)*BD=2CE

Six.Connecttheknownpointsandconstructcongruent

triangles.

Forexample:known:figure10-1;ACandBDintersectat0,and

AB=DC,AC=BD/A/D=confirmation.

Analysis:Card/A=/D,permittheirtriangleDeltaABOand

deltaDCOcongruent,andonlyAB=DCandtheverticalangle

ofthetwoconditions,acondition,topermitthecongruent,

onlytofindanotherothercongruenttriangles,AB=DC,AC=

BD,ifconnectBC,DeltaABCandDeltaDCBiscongruent,soget

/A=/D.

Proof:connectBC,DeltaABCanddeltaDCBin

Dreams

StardeltaABC=DCB(SSS)

L/A/D=(congruenttrianglescorrespondingequalsides)

Seven,takelinesegments,midpointstructure,congruent,

threetangible.

Forexample:figure11-1:AB=DC/A=/D/ABC/DCB=

confirmation.

Analysis:AB=DC/A=D/AD,thoughtofasthemidpointof

N,NC,NBconnection,thentheSASaxiomsofdeltaABNDelta

DCN=,soBN=CN,ABN=//DCN.Theonlycard/NBC=M/NCB,

andBCMN,themidpointconnection,theSSSaxiomsofdeltaNBM

=NCM/NBC=NCB/so.Proofofproblem.

Proof:takethemidpointofAD,BC,N,M,connectNB,NM,NC.

ThenAN=DN,BM=CM,

InthedeltaABNanddeltaDCN

Dreams

StardeltaABN=DCN(SAS)

L/ABN=NB/DCN=NC(congruenttrianglescorrespondingedges,

equalangles)

InthedeltaNBManddeltaNCM

Dreams

StardeltaNMB=NCM,r=(SSS)/NBC/NCB(congruenttriangles

correspondingequalangles)/NBC/ABN/r=NCB/DCN/ABC

/DCB=namely.

Twotheauxiliarylinethatisinducedbythebisectorofangles

Formula:abisectorgraph,tomakebothsidesofthevertical.

Youcanalsolookatthefigureinhalf,symmetrical,thenthe

relationshipis.Anglebisector,parallel1ine,isosceles

triangletoadd.Anglebisector,plusverticalline,trythree

linestogether.

Theangularbisectorhastwoproperties:Aandsymmetry;Band

thebisectorofthebisectoroftheangularbisectorareequal

tothedistanceofthetwosidesofthecorner.Thereare

generallytwokindsofauxiliarylineswithangularbisector.

Averticallinefromonepointtotheotherfromthebisector;

Theuseofangularbisectortoconstructsymmetricalfigures

(e.g.,cuttingshortedgesonthelongsideofaside).

Normally,whenarightorverticalconditionoccurs,avertical

lineisusuallyconsidered;inothercases,asymmetric

structureisconsidered.Asforwhichmethodtochoose,combine

thetitle,graphandknownconditions.

Anauxiliarylinewithanangle

(I)interception,construction,congruence

Theproofofgeometryliesinthesuppositionandtheattempt,

butthiskindofattemptandthesuppositionareincertainlaw

basicabove,hopedschoolmatescangraspthecorrelation

geometrylaw,

Insolvinggeometryproblems,trytoguessandtryaccording

tocertainrules.Thefollowingisabriefintroductiontothe

auxiliarylinesinvolvedinthetheoremscommontogeometry.

Figure1-1/AOC=/BOC,suchas0E=0F,DE,DFandconnectwith

deltaOED,DeltatownofOFD,thuscreatingconditionsforour

proofline,angle.

Figure1-2,AB//CD,BE/BCD/CEsplit,splitBCD,EinAD,

toverify:BC=AB+CD.

Analysis:thisproblemisrelatedtotheanglebisectorofangle

bisector,canbeusedtoconstructthesolutionofcongruent

triangles,dividethelinetoconstructtheaxisofsymmetry,

andtheproblemisthatlineandthebadtimesproblem.The

methodintheproofoflinesegmentandthebadtimestothe

commonproblemistoextendthemethodorinterceptmethodto

prove,extendtheshortlineorinalonglinetointercepta

portionofthelengthisequaltotheshortline.Butwhether

orextendedtoprovetheequallineintercept,extendedtoprove

thattheextensionofthelineandalinesegmenttoproveequal

interceptioninterceptionaftertherestofthelinewithan

equal,andachievethepurposeoftheproof.

Simpleproof:inthisquestion,wecaninterceptBF=ABonlong

segmentBC,andthenproveCF=CD,soastoachievethepurpose

ofproof.Thisusesangularbisectortoconstructcongruent

triangles.Anothercongruentselfproof.Theproofofthis

problemcanalsobeextendedtotheextensionlineofBEand

CDtoproveitatonepoint.Tryityourself.

(two)theverticallineonbothsidesoftheanglelineisequal

totheverticalline

Theangleofthebisectorisperpendiculartobothsidesofthe

bisector,andtheproblemisprovedbythedistancebetweenthe

pointsonthebisectorofthebisectorandthedistancebetween

thetwosides.

Figure2-1,knownasAB>AD,CD=BC/BAC=/FAC.

Proof:/ADC+/B=180?

Analysis:fromC/BADtobothsidesofthevertical.InC/

ADCandangleBandangle.

(three)theverticalstructureofangularbisector;isosceles

triangle

Theverticalangularbisectorfromacornerontheside,sothat

bothsidesandcorneroftheintersection,thencutanisosceles

triangle,pedalforthemidpointonthebottom,theangle

bisectorandbeonthebottomlineandhigh,onetothreelines

bynatureandnatureinisoscelestrianglethebitline,(if

thereisalinesegmentperpendiculartotheangularbisector

inthesubject,thelineisextendedtointersecttheotherside

ofthecorner).

Knownas:3-1,AB>AC/BAD=/DAC,CD,ADinDgroup,Histhe

midpointofBC.

Confirmation:DH=(AB-AC)

Analysis:theextensionofCDandtheintersectionofABat

pointEarecongruenttriangles.Problemprovable.

(four)parallellinesontheothersideofthecornermadefrom

apointontheangleline

Inthecaseofanangularbisector,theparallellineofone

sideofthebisectoronthebisectorofthebisectoris

constructedtoformanisoscelestriangle.Theparallel1ines

oftheangularbisectorareintersectedwiththereverse

extensionlineontheotherside,soanisoscelestriangleis

alsoformed.Asshowninfigures4-1and4-2.

Threebythelineandthedifferencetothinkoftheauxiliary

line

Formula:

Linesegmentsanddifferences,andhalftimes,extended,

shortened,andtestable.Thelinesegmentsanddifference

inequalitiesmovetothesametriangle.

Whenprovingalineequaltotheothertwolinesandwhenthe

generalmethodisonlythenmethod:

1.Length:acutinalonglineisequaltooneoftheother

two,andthenprovesthattheremainderisequaltotheother;

2,improve:ashort1ineextension,theextensionpartisequal

toanothershortsegment,andthenprovethatthenewlineis

equaltothelengthofline.

Toprovetheinequalityoflineanddifference,itisusually

associatedwiththesumoftwolinesinatriangleisgreater

thanthirdsides,andthedifferenceislessthanthirdsides,

soitcanbeprovedinatriangle.

Inthethreesidesofatrianglerelationshipthatunequal

relationshipbetweenthesegment,suchasdirectevidence,can

connecttwooralongTingedgetoformatriangle,theline

appearsintheconclusioninoneorseveraltriangles,and

provedwiththeunequalrelationshipbetweenthethreesides

ofatriangle,suchas:

Inthetrianglecornerandcorneritisnotgreaterthanany

adjacentifdirectcertificatedoesnotcomeout,canconnect

twoorasideextension,

Toverifythestructureofthetriangle,atriangleinthe

Bighornangleposition,smallangleintheinteriorposition

ofthetriangle,theexteriorangletheorem:

Fourtheauxiliarylinethatcomesfromthemiddlepoint

Formula:

Themiddleofthetriangleistwopoints,andtheconnection

isamiddleline.Thereisamiddlelineinthetriangle,

extendingthemiddlelineandsoon.

Inatriangle,ifalittleisknownonthesideofatriangle

midpoint,itshouldfirstthinkoftriangle1ine,middle1ine,

doublecenterlineextendinganditsrelatedproperties

(hypotenuselineproperties,isoscelestrianglebottommidline,

thenthroughtheexplorationofnature),findawaytosolve

theproblem.

(1)themidlinedividestheoriginaltriangleintotwosmall

trianglesofequalsize

AsFigure1,ADisaABCline,SABD=SACD=SABC(deltadelta

deltadeltaABDanddeltaACDisbecausethebottomlevel).

Example1..InFigure2,inDeltaABC,ADisthemedian1ine,

extendingADtoE,andmakingDE=ADandDFthemiddleofdelta

DCE.TheareaoftheknownDeltaABCis2,andtheareaofthe

deltaCDFis.

Solution:becauseADisthemiddlelineofABC,soS,ACD=S,

ACD=1,ABC=,2=1,andbecauseCDisthemiddlelineofdelta

ACE,soS,Delta,CDE=S,delta,

SinceDFisthemiddlelineofdeltaCDE,soS,Delta,CDF=S,

Delta,CDE=*1=.

StardeltaCDFarea.

(two)themiddle1ineofthetriangleshouldbeconsideredfrom

themiddlepoint

Example2..AsshowninFigure3,inthequadrilateralABCD,

AB=CD,E,FarethemidpointofBCandAD,andtheextended1ines

ofBAandCDarerespectivelyEFextension1inesGandH.Proof:

/BGE=/CHE.

Proof:1inkBDandtakethemidpointofBDasM,andconnect

ME,MF,

DreamsMEisthemedianlineofthedeltaBCD,

RMECD,R/MEF=/CHE,

DreamsMFisthemedianlineofthedeltaABD,

RMFAB,R/MFE=/BGE,

AB=CDME=MFdreams,peryleneperylene/MEF=/MFE,

WhichangleBGE=angleCHE.

(three)themiddlelineshouldthinkofextendingthemiddle

line

Example3.,figure4,knowninDeltaABC,AB=5,AC=3,evenBC

onthecentrallineAD=2,thelengthoftheBC.

Solution:extendADtoE,makeDE=AD,thenAE=2AD=2*2=4.

InaACDandaEBD/ADC=/AD=ED,dreams,EDB,CD=BD,

StardeltaACD=EBD,RAC=BE,

ThusBE=AC=3.

InABE,forAE2+BE2=42+32=25=AB2,theangleE=90degrees,

RBD===,soBC=2BD=2.

4.casesofFigure5,knownasdeltaABC,ADisthebisector

ofangleBAC,ADandBConthesideofthemidline.Confirmation:

DeltaABCisisoscelestriangle.

Proof:extendADtoE,makeDE=AD.

Examp1e3provable:

DeltaDeltaBED=CAD,

TheEB=AC/E=/2,

And/1=/2,

L/1=/E,

AB=EBandAB=AC,namelyR,DeltaABCisanisoscelestriangle.

(four)thenatureofthecenterlineoftherightangled

triangle

5.casesofFigure6,knownasAB//DCABCD,ACladder,BCgroup,

ADgroupBD,verify:AC=BD.

Proof:ABDE,CEEpoint,link,DE,CErespectivelyRtDelta

ABD,DeltaABCRtABonthehypotenuseline,theDE=CE=AB/CDE=

/DCE,therefore.

AB//DCdreams,

L/CDE=/DCE=/2/1,

L/1=/2,

InDeltaADEanddeltaBCE,

DE=CE/1=/imprisonment,2,AE=BE,

StardeltaADEDeltaBCE=AD=BC,R,thusABCDistrapezoid

isoscelestrapezoid,soAC=BD.

(five)thebisectoroftheangleandthelineofaverticalline

shouldbethoughtofthemiddlelineoftheisoscelestriangle

(six)midlineextension

Auxiliarylineofcongruenttriangles

Themethodoffindingcongruenttriangles:

(1)wecanproceedfromtheconclusiontoseewhichtwopossible

congruenttrianglesaretobeprovedinthetwoequalsegments

(orangles);

(2)fromtheknownconditions,wecanseewhichtwotriangles

areequalunderthegivenconditions;

(3)consideringtheconditionsandconclusions,theycan

determinewhichtwotrianglesarecongruentwitheachother;

(4)iftheabovemethodsarenotpossible,wemayconsider

addingauxiliarylinestoformcongruenttriangles.

Themethodofcommonauxiliarylineintriangle:

Extendingthemiddle1ineandconstructingcongruent

triangles;

Usingfoldingtoconstructcongruenttriangles;

Third,drawparallellinesandconstructcongruenttriangles;

Makeisoscelestriangleofconnectionstructure.

Commonauxiliarylinemethodhasthefollowingseveralkinds:

Theisoscelestrianglecanbeusedastheheightonthebottom

edge,andtheproblemissolvedbythenatureof"three1i