2020年中考数学二轮核心考点讲解第01讲翻折问题专题解析版

【中考数学二轮核心考点讲解】第

01

讲

翻折问题专题

..

.

.

.

】（

=4

E

FG

F

EFG

E

GF

NBE

=4====4==

E

DE=

=2

eq

\o\ac(△,Rt)

DE=2=1= =+=5

eq

\o\ac(△,Rt)

=

=2AN== GF=EF=

E

BE=4EC=2BE=2=BEC

eq

\o\ac(△,Rt)

BEF

EF=BE+BF=12+EFEF=

EFG= = =

=3=4

E

=2EB

EP

PE

PE

EPEC

=3=4=====3

DE=PE==3=2EB=2EB=1

=

DE==4

eq

\o\ac(△,PC)'==4==PEBPE+BEP+BPEBEP=BEPPECeq

\o\ac(△,') ==），=3=+3 =3EC'=

=EC'+=DE

=1=

=4=3

=3

E

DE

DE

_________.

==4==3===3=3=1=

DE

=E=

eq

\o\ac(△,Rt)

eq

\o\ac(△,Rt)

eq

\o\ac(△,Rt)

eq

\o\ac(△,Rt)

），==1=2+=E+==

】（

=4=6

E

F

CF

CF

BF=6

E

BE=3=4

= =5=

BF=FE=BE=ECBFC

CF=

=

=

EF

E

M

N

=

=

==6

=

==6

+== +3+= =

= =

+2 +3

+2

==

=ED=3=6=++= =

= =2=6

=2

=2=4

eq

\o\ac(△,Rt)

=5=MN=65=1=N= =

= =

==

==

==6

=2

= +3=

==6

=2

eq

\o\ac(△,Rt)

= = =2

=6

=6

M

N

MN

E

=2

=

eq

\o\ac(△,和)

），=== =2

），点

），

重合），经过点

．设

=

=m

m

=6

eq

\o\ac(△,Rt)

=

=2=+=6+=2 =

）；eq

\o\ac(△,）) eq

\o\ac(△,、)

eq

\o\ac(△,、)

eq

\o\ac(△,OB)eq

\o\ac(△,，)

+++== = ==m=11=6

=11=6m = m=

+6）；

PE

E

=E+EPCE+EPC eq

\o\ac(△,PC)E eq

\o\ac(△,C) =

eq

\o\ac(△,PC)E

eq

\o\ac(△,和)

eq

\o\ac(△,PC)E eq

\o\ac(△,OC)），'='===PE==6=mEC =m=

+6+36=0= =

）．

=15=

FG

E

CE

EFG

BE

PE=15=

=9=12=3

==15

CE

EC=9==3BE= = =3 FG

E

BF=EFBEFG====tan=

=3= PE=

EF=PE+PFEF=EF=

+EF+

=EFG= =

=

=．（

=5=

E

重合），

FCE

==eq

\o\ac(△,′) ），= = ==tan=

=

=

=

=

=

=

BEEG==4EG=

=

eq

\o\ac(△,Rt)

=

=

=5=3=4=1 = =

=EC=EG=

=

E

BE

=3DE=1

= eq

\o\ac(△,AD)E=DE=E=1=

===BE=BEE=

eq

\o\ac(△,Rt)

=+=9+=5．（

=8=10N

M

MNeq

\o\ac(△,，把)

MN

E

E

E

=

=

==5

E

FG

N

FG

=4==5

eq

\o\ac(△,Rt)

=3FE=53=2

=

==4

eq

\o\ac(△,Rt)

=FE+

=4+=

=

．（

=6

E

=4

F

ABFE

EF

=3BF

eq

\o\ac(△,A)EGE=4=2=4'=

'=CFB

BF=

=

E

F

EF

=4

EF

EF

==6

EF

EF

EF=

=

=eq

\o\ac(△,Rt)

eq

\o\ac(△,Rt)

=

=

=2

EF

=4

EBE=

CE），当

F

==

===EC+BCE=+=FOE

E===BE=

BE=3=5==5==4=4

=BE+=+=

=4=8=

=2E

eq

\o\ac(△,，将)

eq

\o\ac(△,叠至)

DE

F

FG=1

=

=

=

=2

eq

\o\ac(△,A)DE=E=

eq

\o\ac(△,和)

eq

\o\ac(△,′) CF=

===CF+GF=+1

eq

\o\ac(△,Rt)

+=++1=++1

=0=2=2

r=r=2r=4r

eq

\o\ac(△,Rt)

+=+r=r

r=

eq

\o\ac(△,，在)

=6=8=10

BCIJ

E

ME

IC

N

IJ

N

E=10EN=6=8=10+=

== =

=eq

\o\ac(△,Rt)

eq

\o\ac(△,Rt)

N = = EN= N

=eq

\o\ac(△,S)

Neq

\o\ac(△,S)

= ×10×

×6×

=24

EF

F

F

KCF=K=b

KFKFCFKC=bBK= bKC=2KF= ==F+CF b+2=b+ + = b=b = =

==

=

eq

\o\ac(△,，将)

eq

\o\ac(△,到)

eq

\o\ac(△,得到)

==eq

\o\ac(△,和)

eq

\o\ac(△,′) ），

===

=2=2× =

=

×2=1= eq

\o\ac(△,Rt)

=3=4

E

E

DE

eq

\o\ac(△,翻折)

F

=

BE

EG

EGBEG = =

BE==3=4=

=5 =

= EG=

=

==2=2

DF==EDF=

eq

\o\ac(△,和)

），=+EDF=

+DE==+=3+2=13DE=+EG=

+

13+

+

==

BE=

EF

BF

=BF

M

E

F

=5=2

EF

==BCF+FBC+=CBF

eq

\o\ac(△,和)

BCF

BCF）．=BFEFAM

F

FH

AM

=FHE++==EFAM+HFE+FEH=FEHFEH=

==FH=5

eq

\o\ac(△,和)

FHE

FHE），EF=AM= = = E

DE

FBFC

EF

M

==eq

\o\ac(△,将)

DE

DEFDFE=

eq

\o\ac(△,Rt)

eq

\o\ac(△,Rt)

eq

\o\ac(△,Rt)

eq

\o\ac(△,Rt)

==++==

====EF=BE+=+=+=

=

= =3

E

FF

F

=6=10

CE

E

=6=10

==

===10==6==10EF=EDBF= = =8CF=BF=108=2

CE=

EF=ED=6

eq

\o\ac(△,Rt)

CEF

+==

CE

EG

E

DE=CE==10AFE=FE=DEEFG

eq

\o\ac(△,Rt)

CEG

eq

\o\ac(△,和)

FEG

eq

\o\ac(△,Rt)

CEGFEG），=FG=FG=

=+FG=10+==10

eq

\o\ac(△,Rt)

+=10+=

BE，点

，若

=4=9E

CE

MN

MN

MNF

FB

FD

GF

=

K

=3

CK

I

==EBC=

）【画一画】，如图

=

=9=9

=

=BFGBFG==DF== ==4

eq

\