圓的經(jīng)典類型題變形

1.AB00

CD

AB=10.CD=

A.B

CD

I.

AE=1,

ABCDAB=7,ZB=WBC-AD=1.CD

AB±P.PA.D

AP

AB

PBC

2ABOOCDAECDE.0MCDM.BFCDF.CEDF,②

AE-BF2?0M.”CD

⑴

⑶

3

BC

AD

E,F

0G

求證:00

ABCD

G,0G=a.

A.

AD/7BC,AB=2V3a,ZABC=60°,B00AB,BC

0B,0E,0FBF=2a,的形狀,說明理由:⑵若

2AABCZBACBCD,EAABD^AAEC.

AB?AC=AD：+AD?DE;(5).

?EC+AC?BE=AE?BC

BE：=AE?ED;(2).AB?CD=AD?CE;(3).AB?AC=AD?AE;(4).

AB?AC=AD：+BD?DC;(6)AB?ECAE?BD;AC?BE=AE?DC;

AD：=AB?AC-BD?DC?

AB

1.AABCZBACBCD,EEEFABF,EGACG,

BF=CG.

2.AABCZABCZBACAE.BDL

(1)EI=BE;(2)EI=4,AE=8,DEE&AE?DE;

DE=y,yxx

AIZiABCI,BE.

ZkABC3.IE=2.AEx,

IE^DE’+AD<DE;⑹(IE-DE)=BD?C.

3

ME:MN;

CE=ME€E.

1

S.g：S.q

E

AB:AC;

AB,AC

5)BEBD

M,NdkBD

AE10,AM=12?BE5,CE

AB：AC

BE

4?AABCZBACBCD,E.BO0CE,AR

N.(1)BE△BCM(2).BEZMBC;ABEM^>ACBM;(4)XBM:=ME?MC;(5).BX

*DE=BD*EN:(6).CE:BC=BM:CM:(7).CE=：BC：=EM:CM;(8)sAD*EN=BX?DC;(9)AB:BE=BN：EN;(10).AB

*BN=AN?EC?5.AABCZBACBCD,EB00

CE,ARM.N.(l)sEEGBCG,EEFBNF,①EG=EF;②BF=^BC;KBE

EM

N

(KBSoMSnc.

6AABCZABCZBACAE.BDLAIZiABCI,NNF

AMAEF,①NF=AN;②AC?CN=CD?EN

E

7?ABCZBACBCD,EABCG,ZG

AB,ACMN,AMAN;(2)AGMG;GA‘：G^MBINC：(4)XAM:=BM?CN;(5)

1;MAB

86CRBAGR,(1).AG=DG;(2).BG?CG=DG：;BC

ΔMANtaanBAAMG<^ACNG;(4).AANG^ABMG.CAN

9?6CADAQ,

DE?BD￡MQ

AB*AQ=AC*50;(4).ADCQ(5)DQAB;(7)AQCD^ACED;AQ‘=CQ?

10.AABCZBAC

PH,'AB?OC=AE?BC,

BC

P,O0

D,E.AB.CE

ZABE=ZPBC;(2).APBE^AABC;(3).Pf-PB^BE?BC?

11?AABCZBAC

EF:=ED:-BD?DC;(2).F

ZB

BC

D,E

E

EC

AEF

12.ΔABCZBACBCD,EAC000,E

EC(EC<0D)

BE=CE=1,AB

ACF,0BGEEHACH.BCM.

ME=MC;(2).

AC

2EM

13.AABCZBACBCD,E?EECAC

F,AEBF.

MAABCZBACBCD,EA0ACOE

F,(l)xBFAE?(2)ABOEN,AN=AC.

15?AABCZBAC

E.AM.ANB,C.

^N=AE:+ME

BCD,E.A,MN00

AEZMAN;(2).ER:=CN(3(4)AM￡N=MR(5)AM

RECN

16?ABC00OOL00O0BC00D,AB,AC00：ADE

(1)MN=^BC：

AC=3,S.ABCa,

AE=iAD;(3).2DC?BD=AB?AC

■■

S$3?

AB=5,AC=3,MN

AB=5.

17AEZBAC,ABO0EO0BFEEFF,AGE

EFfG,

(1).FG

O0

(2).

0A=5.BE=2>5

BF,AG

ZBAC=60\

ABCEAC=AB?BF

18.MN00C.ABO0.AEMN.BF

ACZBAE;(2).AB=AE+BF;(3).EF:=1AE?BF.

1V1NO0G.E.F.AC(1).

19BC00ZBA0=ZCA0,0BOFABE,EM0AM,E

ABEM0FAM-AC=2BE?EM;ZBA0=ZCA03

(2).MA=0M,OMEF

20.ABO0ZBAD=ZCADDDFAEF,B,D,E

CF=EF;DFO0DF=EF?AF;(4hCD:=EF?AE;(5).BF?GF=EF?AF?

A

21?ZBAD=ZCADBD=DE,ABG)0AB=BE,BD=CE?

A

22?ZBAD=ZCAD,AE00AEBCD,AG,CBF,CG,BG(1)ZACG=Z

F;(2)AC^AG?AF;(3)ZAGC=ZBGF?

23?ZBAE=ZCAEAEBFD,BDDH;⑵.AGEF;③.GH

GC;△ADHS^FDES^FGH.

24?ZBAE=ZCAEAEBFD,ACEFG,BD=DH;(2)GHGC

25?00PABAD,PCM,BCAD,PD

E,N,BD,MX.d)NBE008,AB90°,MN

26?

CF=6,EF=6^2,

ZD=90%ZBAE=ZCAE,ACO0AECEE,

CEDE=3.EF=5AEGEAC

DF00

G,BE:=CG?

27?

(2)DE=CE.

ZB=90°.ZBAF=ZCAF,ABG>0

FE0

DEO0

AC

28.(1)

00

AC€>0

PA00

ZBAD=ZCAD,PA

PE:=PB?PC,

00ZP=45°.PA=10,00

ZBAD=ZCADa

(2)

GA

29

GE

G

GD

BA

ZXAEG

0

A

ZEFG=ZEDG;（2）.

30.ABCDAC,BD0,00zAABDACW=4AE?AF.

31?ABAABC0CP?0C,ADP,ZBAD=ZBAC

0E=^AC；（2）X

AC=6,AB=10

PC

B

32?Z\ABFO0.CDABE,AF,CDP,CG,CF00C,F,CPG,AB

H,PHADM,GPGD?GCAB00(2).MHMP;

33.000AB,Z\ABDDB00C,AME,BDF,BE.(l).pp2Mp

AM?EF=DM?CE;BC=5,BD=7,CF=2DF,AM=4MF,MFCE

31.AB00CAO0A,BC00D,DEO0E,ZABC=ZEBD,AD,O,DE,BE

x:-(2+m)x+2m:-m+3=0(DE<BE)AC

35ZXABCZABE=ZCBE,BC00CF00C,ACO0E.CAO0

BEF,BECDG?(l)AB=AC;(2).DG:GE=3：BG=3k,kAC;

FCpBF+FCb,=c,a：xHx^c=0

36?AABCAB=AC,ZABF=ZCBF.BGG)0ACO0F,ADBCD,FMBG

37AC=AG,ABZiAGC0ABCGE,ADFΔAFC"

ΔACD;CD=2,AD=3,AC=4,CE

3&

A3

00BC

AD.過點

作

ZW、N?(1)

(3)BC=6,cosC=|,DV

39BFAADF0BAB?0D,C?0±BCADf

E,BFADG,AABC^AEAB;(2).AB:-EB:=AE?ED;(3).AB=3BF=15,AE:ED=1：3.BC

C

4000O0：MX0?BE0?E,BE=DE,BD=m,BE=n,AC,DEx

*+q=0AC=BD;m,npq：xqX:-(jn:mp)x+l=0

ZDBE=30\O0：

41BEO0BE=6,BEPEPEB,AAEA＜30

AD,DDDFABF,DADBHADH,ED,FH.AE=2.

ADAEPAEAD=x,

BH=y.yxx

4200的邊柑切于點分別相交于〃是00

0E,CCGOE00G,DG,EG,DGECF.⑴AB00

0

(2)AE?ED=AC?EF;EF=3,tanZACE

M,N(MANAN

A

AN/7CE

2

0

M

N

43

FB,FC.

AC;AB

ADAABCBCD,DAAABC

FB=FC;AD?DF=DB?DC;FB:=FA?FD;(4AD?AF=AB?AC;

ZEAC=120°,BC=6,AD

F,

AD:=DB?DC-AB?

44

1,ZEAABC

ABCD00,ADBD,

EZBECZkABCZBAC

ZAEDAB=&CD=5,

A

ZA=a,

00

ABCD

DF

3.

ADEF

2

a

F,

ZE2,

BFCD

AE,AF,AC00

3

3AB0C,DEABOACDEE

6DEABAF=DE(1)

(2).

O

1?ABO0

BC=3?sinZP=^＞00

5

200

CDABE,P0zl=zC?(1)CBPD：(2)

(2)3,

M

E

00Z1=Z56D?(1

O0

O0

4

4

BC=3,sinP

3.F9

(2)FG

(3)

、

3?(1)

ABO0CDABE,PO0ZCBP=ZC?⑴.CBPD;(2)

3

00

5

ABDC,AB00

0DBCE,(l).

BE=4,AC=6,DE?

600AABCAB0D00±0DACE,BD.(1).BD

ZABC;(2).ZODB=30°BC=OD?

00

ZUBC

7

DBC

A￡=

AB00BC(DOBDO0

0CBDCDOO

丄于

ADOC,0CBDE.(I).

AD

9PA.PBO0AB.AC?0，①0PCB(2).

2ZBAC=ZAPB.

A

B

10AC00BC00P?0ZPBA=ZCc(1)PB?0

(2)0P/7BC.0P=8,BC=2,00

11PA.PBOOAB?O.ZAPB=60°PO,PA,PB,0C

12

AABC

00,AB

CEABF,C

CEG,AD,CE,BCP,Q.(1).PZkAQC

ADBD

3

⑵.tanZABC=-,CF=8,

4

CQ(FP-PQ):=FP?FG.

13

BC

DOO

ABC00

DEACG.ZADG=ZAGDo(1)AD00(2)

(DO

14

<30

CD

(1)

BC,(30

AD.CD

154300BC

BC

AC.(1)(2)

EBC0EBCF.

AB=2,AD=4,EG=2?

BFAD

4.cos

AAPP.

7

2

16.AD00A,AB00BBCAD,<?0C,AC,CCD//AB,

ADD,A0BCM,CC,ZBCPZACD.

(I)PC00AB=9,BC=6,PC

17＜3025,IABCDO0,ACBDH,PCA

ZPDAZABDP00tanZADB=-,PA=

4

AH,BD

ABCD

B

18?

BC00

AABC

(2)o

AB

BN

00ACMN/7BCBCE,ME=1,AM=2,AE=J5?(l)

19ABO0CO00D1BCD,C00D

E,BE.(1).

BE00

(2).AD

BEF,0B=9,sinZABC=

BF

20O0AC00MA,MB00A,B?①hZBAC=25°,ZAMB

2.BBDACE,0D,BDMA,ZAMB

17＜3025,IABCDO0,ACBDH,PCA

21.00AABCAB=AC.AAP//BC.B0PAP00

22.A,B,C,D00ACBDE,

AD=2F0.

23?0AABCAB00

E,CFABF.CE=CF?（1）DEOO

AB=6?BD=3,AEBC

0OFBCF.①AAEB^A0FC;（2）.

DABAECDDC

24AB0AFOOCDAB

AFF.CD=BE=2?（1）FADC

（2）FC00

E,

CDA

D

25O0AABCAB0.DAB.AE

E.ACZEAB.①.DEOO(2)AB=6,AE=—

DCDC

24

BDfIIBC

5

26..PA0C.P0OOA.B.BEPEE.BE00D.FPC.PF=AF?

FAOOG..ZFGX2ZPBC;(2).PC?AB=0P?AG?

27PD

CDF?(1)OO

(3)PF=13,sinA-^-?EF

13

?0

CD

0AB.PD?0

(2)PE=PF：

CD,PE?0

E

AE.

28CD?0CDAB.F.A0BC,E,AO=L(1)ZC(2)

29000CAB0,L00A,D0ACDL

E,F0BCF00G,AC,AG?003,CE5BF-5AD=4,（1）AE

cosZCAGCG

30.

^2^3,

0

-1.0）

180°?

Z

x

CM

50

（1）

5C

（3）

Q

y

（2）

25"

F

GM?

31

E.AC

Z\ABCZABC=93?0AB0

D?AD=2,AE=1?以。。

0B

AB

32AB0）0DBMB.0CADB1VIC

（1）CD（2）AB4DADX,ACD

y.yXX

C

33.AB0）0DBCB.CE=EB.（1）.DE

O0

AD=4?DC=4

4

DE（3）os2C

?

S

33AB00DBCB,DE00

2DE2=OE-CD.

34.AB00DBCB.AC0CBC

0

E.

DC

35ACG>0

B,ZBCE=ZCDE,CDAE

AD-DF?DC.

EO0DAEAE

F.BCO0CDZACE,

$6

DDE

（2）2,AB

PB=2PA.

E,

O0

D

AD,

AP

PEDF）

（1）1,

AC,

0

AP，

CF

C

P

y

30

ZACB=90°ZACB交于點

E,1CEII

P

/WC,CCEAP,丄

（3）3,

O0

PE

CP

O0

為

AD丄

E，

x(?

AP

x

IM.

012@3

37

BC

00

F

D

CD

AB

0

CE

C,E

CF;

00

BD=1,CD

CE=CB,ZBCD=ZCAE

AC

AE

38A4BCAC=BGA300ACO

ZCDE=ZABD(1)DE00(2)若BD=24,sinZCDE=—,

E

0

13

BODE,

AC

39AB0C00

ADO0E,BE.(l).Z\ABD

AB=4,CF=1.DE

A,BBCD,CD=BC,

ACBF,

40ZkABC

BECE;⑵EF00

BAM

E,BE,CE,EEFBC,CMFD.

41ABCAB=AC,O0△ABCB0ACDZBAC=2Z

BA0;M1ABCDZBCDAD=2,CD=3,BC

42AABCACBC,DAB00A,C,D,BCE,DDFBC,O0F?

DBCF(2)AF=EF.

43①.

ABE.

AB

DE

O

ZMEM

ODBCB.CDBA

ZPAE=ZPOA.OE:EA=1:2,PA=6.PD

P.PNM

G)O

ODACBNDNE=x,PM=y,yx

I

DE

x

44.

ZC.EO=OP

AB

0

DC

0D

＜30

BCB?

;(2).BC+BE=CP:

CDBA

PD=2.DC=3?

P.OC

tanZPOA?

45

E.PD=DO.

AB00D

(1).2AE?EB=PA?PB;(2).

BC

0D=2PA

B.CDBA

uE：AB=3PA;AB=BC.

P,DEAB

46.AB0

M,N.PD=4,CD=6.PA=2

0DBCB.PC00D.B.CBMPC.ANAC

ABtanZPDABM.ANx2-ABx+DE2=0

4700AB00CABD,CDCA,ZD=30°,.

①CD00A,BCDE,FCAB

G.CG:=AE-BF?

CG

48

AB

RlAABCZACB=90\DABADO0BCE,ACF,C

G,AEH,EEPABQ（EPQEPBP,BP00

BC

O0

EFED;sinZABC=?AC=5,

CHQE

00

AB

49AABCBC

E,CE

BD+CE

AB

AF

E,0

F.①

AC

0C

00

00.?0BA

AB=10.tanB=p

D,AC=AD,

O0

0A

F

50ZM0N00A,B,O0r1,PA,B

ZMON=80°,ZACB2,C00XPCAPBCZAPB

PC＜50D,r

51OOi

OOiB.D

LAC:CE

52OOi

B.D

00

00

E

E

(DO

C.OQ

00EFAD

sinA.tanZDCE

C.OQ

EFADF.

OOK0A.EAOAD

F.BC.CDQE.CF?①AD:AC=2：

AC:CEADEF

H

0O00AEA00ADOOi

BC.CD.DE.CF.(1).ZBCD=ZBDC+ZA:(2)

ZBCD+ZACD=180°:(3)BD2=BC?DE.

53.001OCh

BA.CA

A.BC

M.N.

B.CBC0QD.DBC

DM=DA=DN:2MDL1VIANB:BC=CD.ABBM.

3

M

4

PC00

PA.PB

PAPB

PCZAPB.

PA=PB

P

A

1.PC＜50PA.PBPCZAPB.0MPAM,ONPB0M=0

PA=PB.PPAPB

20ZPOQOR01OP.OQA.B.CJD.ZOAOiZOCOi

ZOBOiZODChZOAOiZODO

3AB.CD00AB=CDMNAB.CDM,N0

2

4.P.QOOAB.CDZAPQ=ZCQP.

5.O0AB

CD

P'

AB

CD

CD.?P

C,D

CDA

C,D

ZCP'DZCOB

ZCPDZCOB

1.ZkABCZABCZBACAE.BEEAEAABCD,

BD,CD,CE,ZBAC=60(1).ABDE(2).ZBDC=120°,BDCE

2IZXABCBIZkABC0D.ACE,CD,BAF.

ZADFAFG.DGCA;⑵AD=ID;DE=4.BE=5?BI

3ABCZBACZABCAE,BEE,ABCD.BD.CD.CE.

ZBAC=60r\(l).BDEZBDE=6gBDCE

4AABCAB=AC.ABCBCD.E.(l)AABDAAEB

(2)qlAAB=l,AD=x,DE=y.yx(3)AABCZDABa,ZAEB=

P,sin2a+sin：P=1,y

5AABC00PBCBPD.BD=AP.

CQ.(l).AP0,AP0,ΔPDC

6AABC

AB2=AD-AE.(DM1D

AB=AC,D

BC

BC

E

ADAABC

2BC3

1

ABADAE

E

2

7.AP

ZkBDE

AABC00AABCS,CO0AB

（2）.CPABD,BBECDE,BE00

MN

00

E.EA

OO

ZkABC

OO

ADZBACBC

（2）.AB?AC=AD?AE:③.

D.AB?AC=AD?AE

E,00

AE

D.

ZBAC

DMN/7BC.

ZFAC.BC

7

L

O0

AB00

（CC2

LO0CI,C2.AD

ACi.ACz.AB.AD

C2）

LD.（1）.

ACAB.AD

AC：

;（3）?

-AC^AB-AD;

L

(3)

10AB

00（1）FP

CAB

3）,C

O0

C

CD

P

AB

2）,（1）

APD.

AB

1）

CD

,

00

A.B）.AAOPP0

P0BC

（3）

AB=4PD?

A

（2）

P.CAB

C

P

AB

11.^EAABCZABC90°,ZkABCAB/XABD,CDAB

M,ECHBE.FABDEADEF.

ΔBEF△BEFS^BCA;(3)AB6,BCmCM

E,EFABm

6OAO0

R

OA

OB.ilD

DE

0C

0B

(DO

E,

AB

D.

ADDB

1AEOOOAOCOOABD.OB.DEF.AE=4?

ZBAE=30°,BF.DF

2??GXhOO2P,P?0D.?ChE.DAG?ChC.(l).PC

ZAPD:(2)PC2=PA?PE.

D

C

C

7

ZAPCZCPB=60°,

AABC

PC.PA

PBxAB

PC.PAPB

1

AABC

PABPCABE,

2?ABCD

BBECDDA

G）0?ZABC=60\BDZADC

E,AD=2,DC=3,ABDE

AABC

3II

ZDBC=45%

1,

ABCD

ABCD

2

AB

24

DC,AC

BD,

ABCD

DBDBCE

00ZBCD-60\00

BC

4.

ABCD

ZBCD,BC+CDAC?

5.

QO

AABC

C

1.00

2,CA

2品,BD

2,DABA,BDA,DB,DC.

DCZADB

t,

ADBCS

CA,CB

Dt

DCx

D

t

6.AB