2.证明:
连接BF和EFoVBC=ED,CF=DF,BCF玄EDF二△BCF^AEDF边角
边)。
二BF=EF,/CBF=/DEF连接BE心BEF中,BF=EFoa/EBF=/BEFO
又•••/ABC/AEDa/ABE/AEBaAB=AE在^ABFftAAEF中,
AB=AE,BF=EF/ABF/ABE/EBF=ZAEB/BEF=/AEFa△ABF^AAEF
a/BAF/EAF(/1=/2)o
3.证明:
过E点,作EG//AC,交AD延长线于G则/DEG/DCA/DGE/2
又•••CD=DE.△ADC^AGDE(AASaEG=AC-EF//AB••/DFE/1v/仁/2
a/DFE/DGE.EF=EG.EF=AC
4.证明:
在AC上截取AE=AB连接ED/AD平分/BAC••/EAD/BAD
又vAE=ABAD=AD./AED^/ABD(SASa/AED/B,DE=DVAC=AB+BD
AC=AE+CECE=DE./C=/EDC//AED/C+/EDC=/C;/B=2/C
5.证明:
在AE上取F,使EF=EB连接CFvCEIAB••/CEB=/CEF=90°
vE吐EF,CE=CE•••△CEB^ACEFa./B=/CFEv/B+/D=180°,/
CFE^/CFA=180°•••/D=/CFAvAC平分/BADa/DAC=/FAC
又vAOAC•••△ADC^AAFC(SAS••AD=AF••AE=AF+FE=AD+BE
6.证明:
在BC上截取BF=BA连接EF.
v/ABE/FBE,BE=BEa/ABE^AFBE(SAS),/EFB/A;
AB平行于CD,a/A+/D=18O;又v/EFB+ZEFC=180,a/EFC=/D;
又v/FCE/DCE,CE=CE,a/FCE^ADCE(AAS),FC=CD:
.BC=BF+FC=AB+CD.
7.vAB//ED,AE//BDaAE=BD又vAF=CD,EF=BC.AAEF^ADCB
a/C=/F
8.延长AD至H交BC于H;BD=DCA/DBC/DCB/仁/2;/DBC/仁/DCB/2;/ABC/ACBA.AB=ACAABD^AACD/BAD/CAD;AD是等腰三角形的顶角
平分线•••ADLBC
9.VAOM与MOB都为直角三角形、共用OM且/MOAMMOB.MA=MB•••/MABhMBA•/OAMhOBM=9度a/OAB=9O/MAB/OBA=9O/MBA•••/OAB/OBA
10.证明:
做BE的延长线,与AP相交于F点,VPA//BC
a/PAB+/CBA=18O,又V,AEBE均为/PAB和/CBA的角平分线
a/EAB/EBA=9Oa/AEB=9O,EAB为直角三角形
在^ABF中,AE!
BF,且AE为/FAB的角平分线•••△FAB为等腰三角形,
AB=AF,BE=EFS^DEF与△BEC中,/EBC/DFE,且BE=EF/DEF/CEB•••△DEF^ABECaDF二baAB=AF=AD+DF=AD+BC
11.证明:
在AB上找点E,使AE=AC-AE=AC/EAD/CADAD=AD•••△ADE^AADCDE=CD/AED/CVAB=AC+C,DaDE=CD=AB-AC=AB-AE=BE
/B=/ED玄C=/B+/EDB=/B
12.分析:
通过证明两个直角三角形全等,即Rt△DEC^Rt△BFA以及垂线的性质得出四边形BEDF是平行四边形.再根据平行四边形的性质得出结论.
解:
(1)连接BE,DF.VDEIAC于E,BF丄AC于F,,a/DEC/BFA=9O,DE
//BF,在Rt△DEC和Rt△BFA中,VAF=CEAB=CDaRt△DEC^Rt△BFA
aDE=BFa四边形BEDF是平行四边形.aMB=MpME=M;
(2)连接BE,DF.VDEIAC于E,BFIAC于F,,a/DEC/BFA=9O,DE//
BF,在Rt△DEC和Rt△BFA中,vAF=CEAB二CDaRt△DEC^Rt△BFA
aDE=BFa四边形BEDF是平行四边形.aMB=MpME=M.F
13.
(1)VDC//AE,且DC=AEa四边形AECD是平行四边形。
于是知AD二EC且
/EAD/BEC由AE二BEaAAED^AEBC
(2)^AEC△ACD△ECD都面积相等。
14.证明:
延长BACE两线相交于点FVBEICEa/BEF/BEC=9O
心BE^yBEC中/FBE^CBE,BE=BE,/BEF/BEC/•△BEF^ABEC(ASA)
aEF=ECaCF=2CEv/ABD/ADB=9O,/ACF/CDE=9O又v/ADB/CDE
a/ABD/ACF在^ABD^n^ACF中/ABD/ACF,AB=AC,/BAD/CAF=9O•••△ABD^AACF(ASA)aBD=CFaBD=2CE
15.证明:
VBE//CF/./E=/CFM/EBM/FCIVBE二CF.ABEM^ACFM
•••BM=CMAM>^ABC的中线.
16.证明:
在^ABC^AACD中AB=ACBD=DCAD=ADABD^^ACD
•••/ADB2ADC./BDF=/FDWABDFgFDC中BD=DC
/BDF/FDCDF=DF」FBD^AFCD.BF=FC
17.VAB=DCAE=DFCE=FBCE+EF=EF+.RBABE^ACDF
•••/DCB/abfab=dcbf=ceaabf^acde.af=de
18.证:
•••AB平行CD(已知)•••/B=/C(两直线平行,内错角相等)
•••M在BC的中点(已知)•••EM=F(中点定义)在^BME^n^CMF中
BE=CF(已知)/B=/C(已证)EM=FM(已证)△BME全等与△CMF(SAS
•/EMB/FMC(全等三角形的对应角相等)•••/EMF/EMB/BMF/FMC/
BMF/BMC=180(等式的性质)•••E,MF在同一直线上
19.
证明:
•••AF=CE.AF+EF=CE+EFAE=CRBE//DFBEA=/DFC
20.证明:
VAB=AC•/EBC/DCB/BDlACCE!
AB•/BEC/CDB
BC=CB公共边)•••△EBC^ADCB・.BE=CD
21./C=/E=90度/B=/EAD=9Q度-/BACBC=AE\ABC^ADAEAD=AB=5
22.证明•••AB=AC.AABC是等腰三角形•••/B=/C又tME=M,△BEMfHACEM^
直角三角形•••△BEM^等于△CEM.MB=MC
23.
(1)证明:
•••/ACB=90,•/ACD/BCE=90,而ACLMN于D,BEXMN
于E,./ADC/CEB=90,/BCE/CBE=90,•/ACD/CBE
在Rt△ADC和Rt△CEB中,{/ADC/CE氐ACD/CBEAC=CB•Rt△ADC^Rt
△CEB(AAS,•AD=CEDC=BE•DE=DC+CE=BE+;D
(2)不成立,证明:
在^ADCffiACEB中,{/ADC/CEB=90/ACD/CBEAC=CB
•••△ADC^ACEB(AAS,•AD=CEDC=BE.DE=CE-CD=AD-BE
24.
(1)证明•••AElABA/EAB/EAC/CAB=90度vAFXAC••/CAF/CAB/
BAF=90度•/EAC/BAF/AE=ABAF=AC.AEAC^AFABiEC=B/ECA/F
(2)
(2)延长FB与EC的延长线交于点GV/ECA/F(已证)•/G=ZCAF
v/CAF=90度.EClBF
25.证明:
(1)vBE!
AC,CFlAB••/ABM/BAC=90,/ACN/BAC=90
•/ABM/ACN/BM=ACCN=AB.△ABM^ANAUAM=AN
(2)v^NAC./BAMhNv/N+/BAN=90/•/BAM#BAN=90
即/MAN=90•••amian
26.连接BF、CE证明△ABF^ADEC(SAS然后通过四边形BCEF寸边相等的证
得平行四边形BCEF从而求得BC平行于EF
27.在AB上取点N,使得AN=AC/CAE/EAN,AE为公共边,二△CAE^AEAN•••/ANE/ace又vAC平行BD/./ace/BDE=18(而/ANE/ENB=18O•••/ENB/BDEZnbe/ebnbe为公共边△EBN^Aebd•••bd=bn.ab=an+bn=ac+bd
28.证明:
vAD是中线•••BD=CD-DF=DE/BDE/CDF
•••△BDE^ACDF••/BED/CFD.BE//CF
29.证明:
vdelACBF丄AC,•••/DEC/AFB=9O,在Rt△dec和Rt△BFA中,DE=BFAB=CD.Rt△DEC^Rt△BFA•••/C=/a,/.AB//cd
30.结论:
CE>DE当/AEB越小,贝UDE越小。
证明:
过D作AE平行线与AC交于F,连接FB由已知条件知AFDE为平行四边形,
ABEC为矩形,且△DFB为等腰三角形。
RTABAE中,/AEB为锐角,即/AEB<9OvDF//AEa/FDB/AEB<9O
△DFB中/DFB/DBF=(18O-/FDB)/2>45°RTAAFB中,/FBA=9O-/DBF
<45°/AFB=9O-/FBA>45aAB>AVAB=CEAF=DE.CE>DE
31.先证明△ABC^ABDC的出角ABC角DCB在证明△ABE^ADCE
得出ae=de
32.证明:
作CG平分/ACB交AD于G••/ACB=9O./ACG=/DCG=45
v/ACB=9OAC=BCa/B=/BAC=45a/B=/DCG/ACG
vCF丄ad/./ACF/DCF=9Ov/ACF+ZCAF=9Oa/CAF=/DCF
/ACG/baaACG^ACBE・.cg=be
CD=BD•••△CDG尢BDE••/ADC/BDE