2020中考數(shù)學(xué) 專題練習(xí)軸對(duì)稱相關(guān)的幾何綜合題型

2020

中考數(shù)學(xué)

專題練習(xí)：軸對(duì)稱相關(guān)的幾何綜合題型（含答案）典例探究

CE

E

F

CE

M

M

MN

N

=4

=7

B

AC

CE

BCGF

AE

E

AC

=

CE

CE

eq

\o\ac(△,在)

M

M

M

M

題型精練

DE

EB

EC

DE

EB

）如圖（），=

=2DE=8

____________EB C eq

\o\ac(△,在)

,

）；

.

eq

\o\ac(△,Rt)

eq

\o\ac(△,是)

DE

E

ECEBC

）點(diǎn)

M

重合），以

DE

N

DE

EF

eq

\o\ac(△,PQ) EDPeq

\o\ac(△,PQ)

EDP

E

F

EF

BEC

EF

，若

o,

o，則

EF BE

o

o

EF

BE

ABCD

AC

BC

BBPE

ACBPE

E

B

BFPE

F

AC

BFPE

=

ABCD

BFPE

E

EF

CE

F

E

BECFD

N

.

F

BE

FDBE

EF

.

ABCD

ABE

B

.

eq

\o\ac(△,在)

=CE

CE

M

N

NA=

N

CE

F

F

FHCE

MN

EF

EF

__________

=EH

EH=4

A

B

EF

E、F

、EF

、M、N

E、F

、EF

、M、N

△

，在△

E、F

、EF

，若EFC

°，連結(jié)，判斷△

eq

\o\ac(△,Rt)

=

=2=3

MN

=_______

_______

=2

=

=

=1

=m

==

m

M

M

CE

E

E

M

M

MN

=

MN

E

=

DE

F

BF

MDF

=

BE

CE

E

.E

=F

EC

AFE

MN

E

BF

NM

F.

F

M

F

F

,

,AASBF

MN

E

.E

F

ANBEBF

=

BF=

ANBE

BCGF

N

M

FB

=

=

=

=

=

=

=

=

=

=

M

CE

=

=

BF

==

=FBP

=

=

=

=

==

FBP

=

,

B

BAD

=+

F

=

CF

EG=ED

=

===eq

\o\ac(△,，) .CF===DCE=GCE==CF++FGC

FG=FC=

==+

.

=

,BE

=

,

,BE

.BEDE

=DE.=CE.

,

.

.

.

=

.=. =,

E

C

=CE.Q

Q

，=.

,

F

.

，

.≌.

,

.

F

,

.

C

,

.

CDE

.CDE

.

FD

FE.

,

.

FD

FE.FC

.

DFE

,

.

.

.

E

,

=

=BE=

,

=

=BE=

.=DE

E=BE.BCE

.

=

=

.

=

DE

E

=

,

.

.

,

,ASA=.==,=

.

= .

eq

\o\ac(△,DG)

EDP

1+1+

E

Peq

\o\ac(△,∴∠)

EDP

F

C

ED

EP

ED

ED

ED

EDP

EDP

EF

=

BE

AF

+

EF=BE+AF.1+2+ 2+3+==1=BEC==,=,BEC.

BFBE=CF

BFEF=EC+CF=BE+AF.

ABCD

BG

，∠，∠

AAS PE

BG

ASA

ACBACB

BPFASA

BF PE

PE

BG

PE

=

=

PEPE

=

ECB

=

PE

EF

EC

AEF

BEC

.AEF

BECBEC

.

BE

.

BE

E

EG

.

BE

.

N

BEC

.

BECN

.

EC

.

BE.BE

.AFEAEF

.EF

EC

AEF

CEB.AFECEB.HFE

.

EF

FHE

BECFHEECB.ECB.BE .

BE

.

.FHE

EG

.

BEC

BCE.

.

EC

EG

.

EG

FHE.FHE.

.

.

,

.

.

BE

.

.

.

CE

CE

.

ABCD

EC

BNE

.

EC

CE

EC

.

E

EFBC

CB

FEBF

BF

EF

eq

\o\ac(△,Rt) EFC

EFFCEC

.

.

EF

EF=

.CE

E

F1=3==

M

C

F

F

M

EC

=

4+5+4=3=

E

N

,

NA=

,

.AN=.

=AM.

.=7=.FHCE

.EH=.=EH.

.

PE=BE

EPB

.EBP

.

.

,

.

.

eq

\o\ac(△,≌)

,

===8

=2

HF

Q

，

HF

HE

Q

F

HF

∥

HF

Q

，

HF

HE

HE∥

，HE

EFC

Q

EFC

°

EFC

eq

\o\ac(△,?)

Q

FD

GF

FD

FGD

FDG

°

△

= eq

\o\ac(△,

)

EF

DEDF

ECB.==

==BECE==CF.=2=2FCE=2

FE

.

DE===

EF=CE+CF=2

==

=

EDF=

=

.

EF

DF

DE.DFE

F

DFE

(

.

.

DE

MANDF

N.

=

=o

.==m

=

=o

.1=o

3=

.

P

M

E

=n

=m

.DE=DF=EF.==1+2+

CE

F

NF

C

AB

AEBC2=3=

ECEAEECF

EC

1=

F1=2=

CBME

=3

.

M

=

=

=

=

E

MN

=

==

==

=

=

==

=

=

=

eq

\o\ac(△,D)eq

\o\ac(△,＇)

=

=

=

==

EP=

PE=

==PE+

DE

F

BF=DF

1=

BF

FD

N

.

N

FD

BF

BF