2026初中数学9年级上(湘教)专题3.2 平行线分线段成比例(讲义)(解析版)

专题3.2平行线分线段成比例（举一反三讲义）





1

....................................................................................................................................2



2“#”

...................................................................................................................................................4



3“X”

..................................................................................................................................................7



4“A”

..................................................................................................................................................9



5“8”

.................................................................................................................................................12



6!"#$%&'()*(+,

..............................................................................16



7!"#$%&'(-.(+,

..................................................................................20



8/012!"#34

..................................................................................................25



95678!"#

......................................................................................................31



10578!"#

..................................................................................................37

知识点平行线分线段成比例

1.9:;<

=>?@ABCDECF(GH"#I

2.J

KLMN

OPEQR?SSE!TU?mEnVWAEBECEDEEEFEX=E=

퐴퐵퐷퐸퐴퐵

푙1푙2푙3퐵퐶퐸퐹퐴퐶

E=EI

퐷퐸퐵퐶퐸퐹

퐷퐹퐴퐶퐷퐹⋯

3.YZ

V%&'A[(?D\]=[(^=[_`)CF(GH"#I

OPa1bcPa2bCdESEXe=E=EI

퐴퐵퐴퐶퐴퐵퐴퐶

퐵퐶퐷퐸퐵퐷퐶퐸퐴퐷퐴퐸⋯

【题型1判断比例式正误】

1

a24-25fghi·jklmno·p"qrbOPEWDcEcF!Ts([cctE

uvccEUVWEw['xw['EXiyzA{|(}a~~b△퐴퐵퐶퐴퐵퐴퐶퐵퐶

퐷퐸퐸퐹퐴퐹퐴퐹퐷퐸퐺퐵퐹퐸퐷

AI=BI=CI=DI=

퐴퐷퐷퐸퐴퐸퐵퐶퐴퐷퐷퐺퐴퐺퐷퐸

퐵퐷퐴퐶퐴퐸퐹퐺

€

A퐶퐹퐷퐸퐸퐺퐵퐶

!

:

‚ƒ„w['(…†c!"#{‡Eˆw['(…†‰F=E

EEŠ‹Œ!"#{‡Ž!‰F4I퐷퐸퐵퐹

퐸퐹4

4

∥퐴퐵퐷퐸∥w['퐵퐶xw['E

=E∵E퐵퐹퐸퐷E

∴퐷퐸퐵퐹퐸퐹∥퐴퐵퐷퐸∥퐵퐶

====E‘A’Ž|E“,

”•

퐴퐷퐴퐺퐴퐸퐵퐹퐷퐸

퐶퐹

–x∴퐵퐷퐹퐺퐶퐸E‘퐶퐹D’Ž—|E—“,

”•

퐶퐹≠퐵퐶

=E‘B’Ž—|E—“,

”•

퐴퐸퐷퐸

퐴퐶퐵퐶

===E˜—A{™VE‘C’Ž—|E—“,

”•

퐷퐺퐵퐹퐷퐸퐴퐷

퐵퐷

‘’

퐸퐺퐹퐶AI퐹퐶퐵퐷퐴퐸

š1-1

a24-25fght·›œž·Ÿ

bOPE$¡UVWGE¢

===3EXiy£Z)¤(}ab퐴퐵∥퐶퐷∥퐸퐹,퐴퐹퐵퐸

퐴퐺1,퐺퐷2,퐷퐹

1111

IIII

A=2B=5C=2D=5

퐵퐺퐵퐺퐵퐶퐴퐵

퐺퐶퐵퐸퐵퐸퐸퐹

€

B

!

:

¥¦‚ƒ!"#{‡E§q¨©{‡ª«}4:

(¬E%>D=

>?ECF(GH"#I

1

4

4

AIEX==2E|E‘:’Ž—“,

”•

퐵퐺퐴퐺

퐺퐶퐺퐷

퐴퐵∥퐶퐷11

BIEX===123=6E¤E‘:’Ž—“,

”•

퐵퐺퐴퐺퐴퐺

퐵퐸퐴퐹퐴퐺+퐺퐷+퐷퐹++

퐴퐵∥퐸퐹121

CIEX===123=2E|E‘:’Ž“,

”•

퐵퐶퐴퐷퐴퐺+퐺퐷+

퐵퐸퐴퐹퐴퐺+퐺퐷+퐷퐹++

퐴퐵∥퐸퐹11

DIEX===23=5E|E‘:’Ž—“,

”•

퐴퐵퐴퐺퐴퐺

퐸퐹

‘’

퐴퐵∥B퐸퐹I퐺퐹퐺퐷+퐷퐹+

š1-2

a24-25fght·l­œ®·Ÿ

bOPEs)EWDstEWEEFstE¯

EEXiy£Z)E¤(}ab△퐴퐵퐶퐴퐵퐴퐶

퐷퐸∥퐵퐶퐷퐹∥퐵퐸

AI=BI=CI=DI=

퐴퐹퐴퐸퐷퐸퐷퐹퐴퐷퐹퐸퐴퐹퐷퐸

퐴퐸퐵퐶퐴퐵퐴퐷

€

D퐴퐶퐵퐸퐸퐶퐵퐸

!

:

‚ƒ„¡°%&'({±…†E!"#E²³%&'¡°}4

(¬I

ˆ!"#‰F=E=E´µ²³E‰F==E‰34I

퐴퐹퐴퐸퐷퐸퐷퐹퐸퐹퐷퐸퐴퐷

4

4

E퐴퐸퐴퐶E퐵퐶퐵퐸△퐷퐹퐸∽△퐵퐸퐶퐸퐶퐵퐶퐴퐵

∵퐷퐸∥퐵퐶퐷퐹∥퐵퐸

==E==E=E=E

퐷퐸퐴퐷퐴퐸퐷퐹퐷퐴퐴퐹

∴퐵퐶퐴퐵퐴퐶퐵퐸퐴퐵퐴퐸∠퐴퐸퐷∠퐶∠퐵퐸퐶∠퐷퐹퐸

=E=EE

퐴퐹퐴퐸퐷퐸퐷퐹

∴퐴퐸퐴퐶퐵퐶퐵퐸△퐷퐹퐸∽△퐵퐸퐶

=E

퐸퐹퐷퐸

∴퐸퐶퐵퐶

=E

퐴퐷퐸퐹

퐸퐶

‘’

∴퐴퐵D.

š1-3

OPEs)Ec!Txc[t(WEWx[tAWEuvU

VWIXiy£Z)A{|(}a△퐴퐵퐶퐷퐸b퐴퐵퐴퐶퐷퐸∥퐵퐶퐹퐵퐶퐴퐹퐷퐸

퐺

AI=BI=CI=DI=

퐴퐷퐴퐸퐴퐺퐴퐸퐵퐷퐶퐸퐴퐺퐴퐶

퐴퐵퐺퐹퐴퐷퐺퐹

€

C퐸퐶퐵퐷퐴퐸퐸퐶

!

:

‚ƒ!"#{‡(YZEYZ1

V%&'(A[(?D\¶=[a^=

[(_`bECF(GH"#•YZ2

V%&'(A[E·¯±\¶=[¡U(?ECD

F(%&'(%[$¸%&'(%[GH#Iˆ¹Ž‰Fº€I

4

4

s)EE

∵△퐴퐵퐶퐷퐸∥퐵퐶

=E

퐴퐷퐴퐸

퐴퐶

‘’Ž∴퐴퐵A£Z¤E—,

”•

s)EE

∵△퐴퐹퐶퐸퐺∥퐶퐹

=E

퐴퐺퐴퐸

퐸퐶

∴퐺퐹—A{™VE

∵퐸퐶퐵퐷

=—A{|E

퐴퐺퐴퐸

퐵퐷

‘’Ž∴퐺퐹B£Z¤E—,

”•

s)EE

∵△퐴퐵퐶퐷퐸∥퐵퐶

=E

퐵퐷퐶퐸

퐴퐸

‘’Ž∴퐴퐷C£Z|E“,

”•

s)EE

∵△퐴퐹퐶퐸퐺∥퐶퐹

=E

퐴퐺퐴퐸

퐸퐶

‘’Ž∴퐺퐹D£Z¤E—,

”•

‘’CI

【题型2“#”字型】

2

a2025·»l¼½·¾¿ÀÁbOPEsw[')EEWEstEUVW

FE¢=1:3E=3EX(`xab퐴퐵퐶퐷퐴퐷∥퐵퐶퐴퐵퐸퐹∥퐴퐷퐶퐷

퐴퐸:퐴퐵퐷퐹퐹퐶

AI6BI4CI5DI4.5

€

A

!

:

‚ƒ„!"#I‹Œ!"#4I

4

4

=1:3E

=1:2∵E퐴퐸:퐴퐵

∴sw['퐴퐸:퐸퐵)EEE

∵퐴퐵퐶퐷E퐴퐷∥퐵퐶퐸퐹∥퐴퐷

∴퐴퐷∥퐵퐶∥1퐸퐹3

=2=E

퐴퐸퐷퐹

퐹퐶퐹퐶

4F∴퐸퐵=6I

‘’

퐹퐶AI

š2-1

a24-25Âghi·ÃÄ`Å·Ÿ

bOPEE¢=2EXxab

퐴퐵퐷퐸

퐴퐷∥퐵퐸∥퐶퐹퐵퐶퐷퐹

23

AI3BI2CI2DI3

€

A

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:

‚ƒ!"#EƋŒ!"#FÇ==2EȘ‰34I

퐷퐸퐴퐵

퐸퐹퐵퐶

4

4

E=2E

퐴퐵

∵퐴퐷∥퐵퐸∥퐶퐹퐵퐶

==2E

퐷퐸퐴퐵

∴퐸퐹2퐵퐶

=3E

퐷퐸

∴‘’

퐷퐹AI

š2-2

OPE?abcEWAEBs?atEWCEDs?ctE"ACEBD!TU?b

VWEEFEXiy"($∥∥A{¡™(}ab

퐴퐸

퐴퐶

AIBICIDI

퐶퐸퐵퐹퐵퐹퐴퐵

€

퐴퐶B퐵퐷퐹퐷퐶퐷

!

‹Œ!"#E‰FÇ=.

퐵퐹퐴퐸

퐵퐷

4

4

abcE퐴퐶

=E∵∥∥

퐵퐹퐴퐸

∴퐹퐷퐸퐶

=•

퐵퐹퐴퐸

퐴퐶

‘’É

∴퐵퐷B.

WÊ

:

‚ƒ„!"#{‡E¨©%>D=>?ECF(GH"#}4

(¬I

š2-3

a24-25fght·l­Ë½·p"qrbOPE?E¢=3E=4EX(Ì

퐵퐷

}ab퐴퐵∥퐶퐷∥퐸퐹퐴퐶퐶퐸퐵퐹

4433

AI7BI3CI7DI4

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C

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‚ƒ„!"#{‡(H2E|ÍTGH"}4

(¬I12!

"#{‡Eκ"c$c(ÏEŠ´µQR"`ÐÑÒÌI

4

4

퐵퐷E퐵퐹퐴퐶퐴퐸

∵퐴퐵∥퐶퐷∥퐸퐹

=E

퐵퐷퐴퐶

퐵퐹퐴퐸

∴=3E=4E=+=7E

∵퐴퐶퐶퐸퐴퐸퐴퐶퐶퐸

3

=7E

퐵퐷

‘’

∴퐵퐹CI

【题型3“X”字型】

3

a24-25Âghi·ÃÄ`Å·Ÿ

bOPEE!TUccVWccE!TUcc

VWccE¢=6E=4E=3EX푎"∥푏∥푐(`xa푚푎b푏푐퐴퐵퐶푛푎푏푐

퐷퐸퐹퐴퐵퐵퐶퐸퐹퐷퐸

AI1.5BI4.5CI7.5DI10.5

€

B

!

:

¥¦‚ƒ!"#E¨©(ÓÒE!"#(RÍ}4

(

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4

4

E=6E=4E=3E

6∵푎∥푏∥푐퐴퐵퐵퐶퐸퐹

=E4=3E

퐴퐵퐷퐸퐷퐸

퐸퐹

∴퐵퐶189

=4=2=4.5E

‘’

∴퐷퐸BI

š3-1

a24-25Âghi·-Ô·Ÿ

bOPEE$¡UVWE¢==3E

퐴퐵∥퐶퐷∥퐸퐹퐴퐹퐵퐸퐺퐴퐺퐷퐹

=1EX(Ìxab

퐵퐶

퐺퐷퐵퐸

3234

AI4BI3CI7DI7

€

D

!

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‚ƒ„!"#I‹ŒEFÇ=EŠÕÖJŒÑҏ‰34I

퐵퐶퐴퐷

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4

E퐴퐵∥퐶퐷∥퐸퐹퐵퐸퐴퐹

3∵퐴퐵1∥퐶퐷4∥퐸퐹

==313=7E

퐵퐶퐴퐷+

‘’

∴퐵퐸퐴퐹DI++

š3-2

a24-25fght·›œž·p"qrbOPEQR?E?!T$?cc

13123

UVAcBcC%WE?!T$?ccUVDcEcF%WE푙∥푙2∥$푙UVW퐴퐶OE¢=푙푙E푙

=3EX(`}퐷퐹I푙1푙2푙3퐴퐶퐷퐹퐵퐶퐴퐵

퐷퐸퐷퐹

€

6

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(¬I‹Œ

!"#{‡FºE‰Fº£ÚI

4

4

?E

∵푙1∥푙2∥푙3

=E

퐴퐵퐷퐸

퐸퐹

∴퐵퐶=E=3E

∵퐵퐶=3퐴퐵E퐷퐸

∴퐸퐹=+=3+3=6E

∴‘€x

퐷퐹퐸퐷6I퐸퐹

š3-3

QR%>Û¡(?EE!TD?l4VWEEED?VWEEE?

$¡UVWOE¯=32E=푙150푙2E푙3=8E=2I퐴퐵퐶푙5퐷퐸퐹푙4

푙5퐴퐵퐵퐶퐸퐹퐸푂

(1)3(`•

퐷퐸

(2)3(`I

24

€

푂퐵(1)

5

(2)52

4

!

a1bˆEYº=E‰34•

퐷퐸퐴퐵

푙1∥푙2∥푙3퐸퐹퐶퐵

a2bˆEYº=E‰34I

푂퐵푂퐸

4

a퐵퐸1∥b4

퐴퐷퐴퐵퐷퐸E

123

=E∵푙∥푙∥푙

퐷퐸퐴퐵

퐸퐹퐶퐵

∴32

=E

850

퐷퐸

∴24

•

=5

a∴퐷퐸2b4

E

=E∵퐵퐸∥퐴퐷

푂퐵푂퐸

∴퐴퐵퐷퐸2

=24E

32

푂퐵5

∴

52I

=4

WÊ

:

‚ƒ∴푂퐵!"#{‡E4

(¬}¨©!"#{‡EÜV)‚Ý‚

I

【题型4“A”字型】

4

a2025·›œœÞ·¾¿ÀÁbOPEß[àáâFÇE¢==6E

=4EX(`xab△퐴퐵퐶퐵퐶△퐷퐸퐹퐸퐶2퐵퐸

퐶퐺퐷퐹

AI6BI8CI9DI10

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¥¦‚ƒ„â(…†E!"#E

ƋŒâ(…†FE=EŠ‹Œ!"#3ºE¹

‰4I

퐴퐵∥퐸퐺퐴퐶퐷퐹퐴퐺

4

4

ãßä[àáâFÇE

E∵=△퐴퐵퐶E퐵퐶△퐷퐸퐹

∴퐴퐵∥퐸퐺퐴퐶퐷퐹

==2.

퐶퐸퐶퐺

퐴퐺

∴퐵퐸=4E

∵퐶퐺=2E

∴퐴퐺==+=6I

∴‘’

퐷퐹A퐴퐶I퐴퐺퐶퐺

š4-1

a24-25Âghi·åæç跟

bOPEs)EE!T}[Et(WE¯

I¢=2E=4E=1EX(`}ab△퐴퐵퐶퐸퐹퐴퐵퐴퐶퐸퐹∥

퐵퐶퐴퐸퐵퐸퐴퐹퐹퐶

1

AI2BI1CI2DI3

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¬}ˆFÇGH[#34I

‹ŒGH[#yº™EÕÖQR[(`ÐE3ºI

4

4

E퐹퐶

∵2퐸퐹∥1퐵퐶

=E4=E

퐴퐸퐴퐹

퐹퐶퐹퐶

4F

∴퐸퐵=2E

‘’

C퐹퐶I

š4-2

a2025·›œ·)‚é

bOPCd(êë)Eìíîï'([`ðx1E(%íñWð

sêë(UWtEWDcE!T}[c$êë(UWEuvEX(`xa△퐴퐵퐶b

퐵퐴퐶퐴퐷퐸퐷퐸

1

AI2BI1CI2DI3

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‚ƒ„!"#{‡E%&')*{‡E²³º}()*}4



1퐷퐸1△퐴퐵퐶

¬IòëWcEˆêë(…†‰REEFÇ==2E==2EȘ²³}(

퐴퐷퐴퐺퐴퐸퐴퐺

)*E‰34I퐺퐻퐸퐺∥퐶퐻퐴퐵퐴퐻퐴퐶퐴퐻퐷퐸△퐴퐵퐶

4

4

OPEòëWcE

퐺퐻

ˆêë(…†‰REE

21퐸퐺∥퐶퐻21

==4=2E==4=2E

퐴퐷퐴퐺퐴퐸퐴퐺

∴퐴퐵c!T}퐴퐻c퐴퐶()WE퐴퐻

∴퐷}퐸퐴퐵()*퐴퐶E

∴퐷퐸△1퐴퐵퐶

=2=1E

‘’

∴퐷퐸BI퐵퐶

š4-3

a24-25Âghi·åæç跟

bOPEs3×4(ï'êë)EccxëWEuvE

UµW(óëVWI¢îï'[`x1EX=I퐴퐵퐶퐶퐷

퐴퐸퐶퐸

61

€

3

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‚ƒ„ôõ{‡c!"#{‡E‹Œôõ{‡3º(`EŠ‹Œ!

"#Fº™3º(`‰I퐶퐷

4

4

OPCdE퐶퐸

11

ˆsêë)(*öE‰R==2=2E

퐴퐵5푀퐷푁퐷푀푁

=+=2E

∴퐷퐹퐷푁퐹푁561

sRt)E=+=3+=E

222

222

△퐶퐷퐹E퐶퐷퐷퐹퐶퐹

∵퐸퐺∥퐷퐹2

==3E

퐶퐸퐶퐺

퐶퐷퐶퐹2

∴E

61=3

퐶퐸2

∴

61

4F

=3I

퐶퐸61

‘€x

3I

【题型5“8”字型】

5

a2025·÷øù½·%¾bOPE)EExG&tAWEµWE(?!TU[E

VWFEGEUúEVWME▱퐴퐵퐶퐷NI¢=3E=2EX퐵퐷(Ìxab푀푁퐴퐵

퐵퐶퐷퐴퐷퐶푀퐹퐸퐹퐸퐺⋅퐸푁

AI6BI8CI10DI15

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¥¦ƒ„w['(…†E!"#I‹Œw['(…†E‰F

EŠˆ!"#‰F=E=Eû˜FÇ=E‰34I

퐸퐺퐵퐸퐸퐹퐵퐸퐸퐺퐸퐹

4

4

퐴퐷∥퐵퐶,퐴퐵∥퐶퐷w['}w['E푀퐸퐷퐸퐸푁퐷퐸푀퐸퐸푁

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=E=E

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∴푀퐸퐷퐸퐸푁퐷퐸

=E

퐸퐺퐸퐹

퐸푁

∴푀퐸=3E=2E

2

∵푀퐹E퐸퐹

32=

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퐸푁

∴+=10I

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š5-1

OPEsw[')E(!BF!T$ACcADUVWEcFE=3E

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∴∠AF=AB=3∠EÿFD=2E

∴BC=AD=AF+FD=5E

∴ADBCE

3

∵∥E

==5

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3

E

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//E퐴퐶퐵퐶퐵퐷퐵퐶

=E∵퐴퐶퐸퐹

퐸퐹퐵퐹

퐵퐶

∴퐴퐶//E

∵퐸퐹=퐷퐵E

퐸퐹퐶퐹

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+=+===1E+=1E

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∴11

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=45°

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∴∠퐶퐺퐸=90°

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∴퐶퐹∥퐸퐺

==1

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∴퐵퐹}퐸퐹()*E

∴퐶퐹1△퐵퐺퐸

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【题型7平行线分线段成比例与三角形的重心的综合】

7

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