中考数学三角形四边形综合复习

几何综合（三角形四边形）

C

ABDACDBE

DABeq

\o\ac(△,≌) DCE

DAEC

ABCDACBD

OACABAB

ACBD

AC

AOD

MON

ON

=

OM

C M

ABCDDBADAB

PNABAC

CD

E

eq

\o\ac(△,AE) ADE

AED

ACFCDDE

F

C

E

ABCABACD

BCEABBADAE

CB

EAD

F

AFAE

CFBECF

ABCDECBDE

ABFAEDCED

G

DF

CEDDAG

BEAG

AB∥DE∠ACB=∠CDE，AC=CDAB=CD

E

C

FDABCFFACBEAAC

CD

D

C

A

B

E

ABCADABAD

ABAEACAE

ACBECD

BADADCAEDFBC

EFAE

DFG

AEDF

ADABAE

DF

F

E

ABAC

DEABACAEADBCCEA D B

ABCDABBC

DCEDE

AB

eq

\o\ac(△,F) BEF

AF

DEB C

AACDBABBDACBEBCDE

E

CABCDAADCABADE

BCDEDC

ABCD

E

A DABCADBCAD

BECF

EF

B

A

CBECF

EEC

D

BF

AABCDAB

ADBADCBDBCD

EB

C

AB2

2

A

DDB

C

EF

BCBF=CEAD

BC

A

DABCD

FAEABCD

F

B

CE

F

E

C

ABCDAD

BE

BCBAEABE

AEBF

FCABCF

□

E

EF

BCAD

AE=CF

ABCDF

BCDFAC

M

M

MECD

EAM=2

D E

F

已知：如图，点

C

AB

ABDE

CFO，且

ACBDD E

FDECF

E

F

ABCDABCDE

ABDE

．若CE

，且

C

E

，C，E

∥

DECE

△≌△CDE

DAB

C

E

A．如图，四边形

ABCD

CD=

,

AB

DB

C三角形四边形答案

eq

\o\ac(△,∵) DACDBEDADCDBDEADCBDE

EADCCDBBDECDB

CDE,

CDE,

DE,OA

ACOB

BD

DABDCE

,

DABDCE.eq

\o\ac(△,∵) DABeq

\o\ac(△,≌) DCE

ADCE=

ADC=

DCE

ADCDAEC

ABCDACBD

O

x

x

OAOB

OAx

x

OBxACABBAC

eq

\o\ac(△,Rt) OABOAABOBABxx

ACOA

eq

\o\ac(△,S)

AOD

ACOA

eq

\o\ac(△,S)

AOD

eq

\o\ac(△,S)

AOB

AOAB

BAB

DCE,

CE,ABCD

ACBDOOBOD

OPMONCOADOBCAPDBNOAOBACBD

ABACBAC ACDADCDAFED=AF=AD

EF=DEEFCFC

DEx

EF=xCE=-x

EFC

)

x

DE

EC

DCE.

CDE

EDC,CDE

DE.

F

=

BF

AFE

AFEEF

DE

BF

EF

DE

(

,

......................

E

.......................................................................

ED...............................................................................

E

BC

EF

BC

,

BC

AECF ABCAFEC

AECF

=8

=8

,

,

ABAC

D

BC

CADBAD

EABBAD

CADEABACFABE

,

ACFABE

BECF

ABCD

ADBC

CED

ADE

G

DF

AGDG

DAG

ADE

CED

DAG

AEDCEDAGEDAG

AEDAGE

AEAG

AG

AE

AEBAB

∵AB∥DE∴∠ABC=∠E∴△ABC≌△CED∴AB=CD

B

BMFD

MACBACB

AACABC

BCAC

tan

ABCFBCM

EFDF

EEDF

证明：∵CE⊥AF，FB⊥AF∴∠DEC

＝∠DFB＝90°又∵AD

为

BC

边上的中线∴BD＝CD又∵∠EDC

＝∠FDB∴△BFD≌△CED∴BF=CE

解：（1）在

Rt△ABC

中，AB＝2

，∠BAC＝30°∴BC＝

，AC＝3．如图（1），作

DF⊥AC∵Rt△ACD

中，AD＝CD∴DF＝AF＝CF＝

∵BP

平分∠ABC∴∠PBC＝30°∴CP＝BC·tan30°＝1∴PF＝

∴DP＝

PF

DF

＝

．

（2）当

P

点位置如图（2）所示时，根据（1）中结论，DF＝又

PD＝BC＝

，∠ADF＝45°∴cos∠PDF＝

DF

＝

∴∠PDF＝30°∴∠PDA＝∠ADF－∠PDF＝15°当

P

点位置如图（3）所示时，同（2）可得∠PDF＝30°．∴∠PDA＝∠ADF＋∠PDF＝75°

ADABAEACDAB=EACDAB+DAC=EAB

EDACEAB

CD

BEABCD

BADADC=BADADC

,

F E AEDF

BC=

AD

DC

=AB

DFCAEBDFCBE=ABBFEFADBCEFGADGEG EF

EGEG

EG= EG= AG=

FG=FGE=AGD=DG=

DF=

ABE eq

\o\ac(△,AC)

，

，

.ABEACDC

eq

\o\ac(△,4)

ECB

o

ED

EC

o

E

第题图

B

E

E

F

B

B

FB

FD

EF

EDDF

EEG

CBG

ECG

EGC

o

EC

ECG

o

EG

EC

ECG

EC

o

EB

EG

EB

BEFEF

FB

EB

ACDBBAC

DBABACDBE

,

,

BE,

E

CBACDBEBCDE

A

BADABADABDADBADC

BDCADCADB

BDCBDCDEE

BCDCBC=DE

ADABAFBDDF

ADABAFBDDF

AFDAFDADB

F

E

DF

DF

ABCDABADDCBC

ABCADBDCD

CFADBEADCFDBED

BEDCFDBEDCFD

BDECDF

BDCDBEDCFDBECF

C

CEDB

ABEACECODDCAB DCEBBDCEBE

DC

AEABBEDCABADDBAC

ACEACAE

CAB

C

CHAE

H

ACH

CHACCAB

ABCD

A

D

B

C

EA

E

B

C

CE

D

F

AEDF AEB DFC

EBFCE

ïïïï

BAEABCBFïïïï

BAEABC

BECF

ABEDCF

DF? ?DFCïïî

BE

CFABEDCF

BC.

ABCD.

BGAEGBGABGEABCDAD

E

BCBE

EC

FBEGABEAFEABAFBEFEBEF

BGE

B

E

CBGGE

ABGABAF

ECFFC

EF

EC

ABCF

eq

\o\ac(□,∴)

ABCDBC//AD

△∽△

CF AM

F

BCCF

CF AM

B//DC

1=1=

2=

MECDCE

ABCD

3=F

BCCF

CF

CEeq

\o\ac(△,在)

CMFeq

\o\ac(△,和)

CME3=CMFCME

=

2=2=CE

CE

CE

CE

AC