第一章勾股定理(讲义)(解析版)_2026语数英7-9年级上《举一反三系列》多版本_2026《初中数学•举一反三系列》8上多版本_2026《初中数学•举一反三系列》8上(北师大)_第一章勾股定理(讲义)

第一章  勾股定理（举一反三讲义）全章题型归纳
 2024
 ...................................................................................................................................................................3
 1   !"#$% ................................................................................................................3
 2  &'()*+ ........................................................................................................................9
 3  ,-./01 ..........................................................................................................................13
 4  234567898# ......................................................................................................................16
:; .................................................................................................................................................................19
 5  '<=()0>?@-A/ ......................................................................................19
 6  BCDEF ..............................................................................................................25
 7  0GH ..................................................................................................................................29
 8  0I ..................................................................................................................................35
 9  'JKLM ..................................................................................................................38
 10 NOP0QR ..............................................................................................................41
知识点1  勾股定理
1.S
7898#<78T0>?@UVWT0>?XYZ aXb @ c [\]^78
98#0<78T@WTX_`𝒂𝟐+ 𝒃𝟐= 𝒄𝟐.
Y"a^X△𝐴𝐵𝐶b7898#XO4cK078T a deXc*078T b deX
WT c def.
知识点2  勾股定理的验证
0gGhijkl"mnoXpq?mbr,#stuvwxyX"#lzu{)X|}[$%
~0€u‚ƒ„…0.<q?†]^"#$%-‡<ˆ‰XŠ‚$%‹ŒŽU€X‘’
U“tŽ†bGH0”•?m.
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l"mgG0–—˜™
1
	
3 	


2

1
2(𝑎+ 𝑏) × (𝑎+ 𝑏)X
1
2𝑎𝑏+ 1
2𝑐2 + 1
2𝑎𝑏
3
 
1
2 (𝑎+ 𝑏)
2
= 1
2 𝑐2 + 𝑎𝑏
4!"
𝑎2 + 𝑏2 + 2𝑎𝑏= 𝑐2 + 2𝑎𝑏
5#$%&'(
𝑎2 + 𝑏2 = 𝑐2
知识点3  勾股定理的应用
šrBC›7898#œ0FXhiIY
-1)žŸ7898#0 ¡<T*X'¢9T*£
-2)žŸ7898#0 ¡–T*X¤¥¦<T*0€£
-3)BC§¨>?€0©ªF£
-4)«¬?­®‡œ()0*+FXBC¯°¯±40–²„³F.
知识点4  直角三角形的判别条件
1. ')*+,
-
.𝑎/ 𝑏/ 𝑐01𝑎2 + 𝑏2 = 𝑐2234
5	

´6789:;<
=%&'(->'(
2. 7?@
5A=	

-B*
C
DE7?

FG@
5	
/HI
FG
JK
C.LE7?
9MN.-OPQRL.-O
知识点5  勾股数
1. %&S
')
01𝑎2 + 𝑏2 = 𝑐2-
TUS/<=%&S
019:
①
SV5TUS
②MWUS-OPQRXUS-O
YZ
%&S-US[\=%&S/+ 3/4/5 - 2 [ 6/8/10 \
=%&S.
]^
①𝑛2−1/2𝑛/𝑛2 +1𝑛XQ 1 -_`Sa
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②4𝑛/4𝑛2−1/4𝑛2 +1n =TUS
2. 7?%&S-Bbc:
(1)d'
5TUSa
(2)d'RX-SefghMW-S/i8jkRX-S-OfghMW-S-OPa
(3)lmnM/oRXS-OQghMWS-OP/p5%&S,Apq5´
【培优篇】
【题型1  利用勾股定理探究图形面积】
µ 1-24-25 ¶·¸·¹º»¼·½¾/Y"X¿Rt △𝐴𝐵𝐶4X∠𝐴𝐶𝐵= 90∘X[\rT𝐴𝐶À𝐵𝐶Á¦ÂÃ
?#𝐴𝐶𝐷𝐸@Ã?#𝐵𝐹𝐺𝐶XÄÅ𝐸𝐹À𝐴𝐹´ÆžŸ𝐴𝐵2−𝐴𝐶20ÇXÈÉ'Ê098#$%b-   /
A´98#𝐴𝐵𝐹
B´98#𝐴𝐶𝐹
C´98#𝐴𝐸𝐹
D´98#𝐴𝐵𝐶
ËÌA
[ÍÎ*𝐸𝐴X𝐹𝐵ÏVÐ HXŠ‚7898#›Ã?#ÑҍX∠𝐴𝐶𝐵= 90∘X∠𝐶𝐵𝐹= 90°X
∠𝐶𝐴𝐸= 90°XÓT#𝐴𝐶𝐵𝐻bÔ#X∠𝐻= 90°X𝐴𝐻= 𝐵𝐶X𝐴𝐵2−𝐴𝐶2 = 𝐵𝐶20ǞŸXš'𝑆△𝐴𝐵𝐹=
1
2𝐵𝐹⋅𝐴𝐻=
1
2𝐵𝐶2´
ÕBB
Î*𝐸𝐴X𝐹𝐵ÏVÐ HX
∵¿Rt △𝐴𝐵𝐶ÀÃ?#𝐴𝐶𝐷𝐸@Ã?#𝐵𝐹𝐺𝐶4X∠𝐴𝐶𝐵= 90∘X∠𝐶𝐵𝐹= 90°X∠𝐶𝐴𝐸= 90°X
∴∠𝐶𝐴𝐻= ∠𝐶𝐵𝐻= 90°X
∴ÓT#𝐴𝐶𝐵𝐻bÔ#X
∴∠𝐻= 90°X𝐴𝐻= 𝐵𝐶X
∵𝐴𝐵2−𝐴𝐶2 = 𝐵𝐶20ǞŸX
∴šr'Ê𝑆△𝐴𝐵𝐹=
1
2𝐵𝐹⋅𝐴𝐻=
1
2𝐵𝐶2´
Ö×
A´
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ÐØÙhiÚÛ´ÜÝÞßX7898#ÑÒXÃ?#0ÑÒXÔ#05@Ñ
ÒÀ98#$%†XbB0à´
“† 1-1-24-25 ¶·¸·áâãä·½4/Y"b¶åæU07898#ço0“f"”´è"4Ã?
#𝐴𝐵𝐶𝐷XÃ?#𝐸𝐹𝐺𝐻XÃ?#𝑀𝑁𝐾𝑇0$%[\u𝑆1X𝑆2X𝑆3XÆÃ?#𝐸𝐹𝐺𝐻0T*u 6XÈ𝑆1 + 𝑆2 +
𝑆30Ç-    /
A´16
B´17
C´18
D´20
ËÌC
[ÍéÚÛ0IXŽ0ŸêÐb@Ã?#ÀæU98#0ÑÒXrNnæ>
?†UŸê´Š‚¶å7898#æUXÓT#𝐴𝐵𝐶𝐷X𝐸𝐹𝐺𝐻X𝑀𝑁𝐾𝑇ëbÃ?#XÊ
𝐶𝐺= 𝐹𝑀= 𝑁𝐺,𝐶𝐹= 𝐹𝑁= 𝐷𝐺X슂𝑆1 = (𝐶𝐺+ 𝐷𝐺)2X𝑆2 = 𝐺𝐹2X𝑆3 = (𝐹𝑀−𝐹𝑁)2íš'B´
ÕBB
¿Rt △𝐶𝐹𝐺4X&
𝐶𝐺2 +𝐶𝐹2 = 𝐺𝐹2
∵¶å7898#æUXÓT#𝐴𝐵𝐶𝐷X𝐸𝐹𝐺𝐻X𝑀𝑁𝐾𝑇ëbÃ?#X
∴𝐶𝐺= 𝐹𝑀= 𝑁𝐺X𝐶𝐹= 𝐹𝑁= 𝐷𝐺X
∴𝑆1 = (𝐶𝐺+ 𝐷𝐺)2
= 𝐶𝐺2 + 𝐷𝐺2 + 2𝐶𝐺⋅𝐷𝐺
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= 𝐶𝐺2 + 𝐶𝐹2 + 2𝐶𝐺⋅𝐷𝐺
= 𝐺𝐹2 +2𝐶𝐺⋅𝐷𝐺£
𝑆2 = 𝐺𝐹2£
𝑆3 = (𝐹𝑀−𝐹𝑁)2
= 𝐹𝑀2 + 𝐹𝑁2−2𝐹𝑀⋅𝐹𝑁
= 𝐶𝐺2 + 𝐶𝐹2−2𝐶𝐺⋅𝐷𝐺
= 𝐺𝐹2−2𝐶𝐺⋅𝐷𝐺
∵Ã?#𝐸𝐹𝐺𝐻0T*u 6X
∴𝐺𝐹2 = 6X
∴𝑆1 + 𝑆2 + 𝑆3 = 𝐺𝐹2 +2𝐶𝐺⋅𝐷𝐺+ 𝐺𝐹2 +𝐺𝐹2−2𝐶𝐺⋅𝐷𝐺= 3𝐺𝐹2 = 18
Ö×
C´
“† 1-2-24-25 ¶·¸·îïðñ·½¾/¿Rt △𝐴𝐵𝐶4X∠𝐶= 90°X𝐴𝐶≥𝐵𝐶Xr𝐴𝐵uT¿98#¦
}ÂÃ?#𝐴𝐵𝐷𝐸´¿Ã?#𝐴𝐵𝐷𝐸ò}ÂÃ?#𝐴𝐿𝐾𝐽ÀÃ?#𝐷𝑃𝑄𝑅X𝐴𝐿= 𝐴𝐶X𝐷𝑃= 𝐵𝐶X𝑆1À𝑆2À𝑆3[
\]^ÓT#𝐸𝑃𝑀𝐽ÀÓT#𝐵𝐿𝑁𝑅ÀÓT#𝐾𝑀𝑄𝑁0$%X𝑆1À𝑆2À𝑆3~0,€        
´
ËÌ𝑆3 = 𝑆1 + 𝑆2
[Íó𝐴𝐵= 𝑐X𝐴𝐶= 𝑏X𝐵𝐶= 𝑎X'𝑆1 + 𝑆20ÇXìGHÓT#𝐾𝑀𝑄𝑁uÃ?#X‘'𝑆30ÇX
¿Rt △𝐴𝐵𝐶4X&š𝑎2 + 𝑏2 = 𝑐2Xíš'B´
ÕBó𝐴𝐵= 𝑐X𝐴𝐶= 𝑏X𝐵𝐶= 𝑎X
È𝐴𝐵= 𝐵𝐷= 𝐷𝐸= 𝐴𝐸= 𝑐X𝐴𝐿= 𝐿𝐾= 𝐽𝐾= 𝐴𝐽= 𝑏X𝑃𝐷= 𝐵𝐶= 𝑃𝑄= 𝑄𝑅= 𝐷𝑅= 𝑎X
∴𝐵𝐿= 𝐴𝐵−𝐴𝐿= 𝑐−𝑏X𝐿𝑁= 𝐵𝑅= 𝐵𝐷−𝐷𝑅= 𝑐−𝑎X𝐽𝐸= 𝐴𝐸−𝐴𝐽= 𝑐−𝑏X𝐸𝑃= 𝐷𝐸−𝐷𝑃= 𝑐−𝑎X
𝑁𝐾= 𝐿𝐾−𝐿𝑁= 𝑏−(𝑐−𝑎) = 𝑏−𝑐+ 𝑎X
∴𝑆1 + 𝑆2 = 2(𝑐−𝑏)(𝑐−𝑎) = 2𝑐2 +2𝑎𝑏−2𝑎𝑐−2𝑏𝑐X
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∵𝐽𝐸= 𝐵𝐿= 𝑐−𝑏X𝑄𝑅= 𝑃𝑄X
∴𝑄𝑀= 𝑄𝑁X
ô∵∠𝑄= ∠𝐾= 90°X∠𝐴𝐿𝐾= ∠𝐴𝐵𝐷= 90°X
∴𝐿𝐾∥𝐵𝐷X
∴∠𝑄𝑁𝐾= ∠𝑄𝑅𝐷= 90°X
∴ÓT#𝐾𝑀𝑄𝑁uÃ?#X
∴ÓT#𝐾𝑀𝑄𝑁0$%u𝑆3 = 𝑁𝐾2 = (𝑏−𝑐+ 𝑎)2 = 𝑎2 + 𝑏2 + 𝑐2 +2𝑎𝑏−2𝑎𝑐−2𝑏𝑐X
¿Rt △𝐴𝐵𝐶4X
∵∠𝐶= 90°X
∴𝑎2 + 𝑏2 = 𝑐2X
∴𝑆3 = 2𝑐2 +2𝑎𝑏−2𝑎𝑐−2𝑏𝑐= 𝑆1 + 𝑆2´
ÖËÌu
𝑆3 = 𝑆1 + 𝑆2´
ÐØéhiÚÛõ0IÀÃ?#05›ÑÒÀö†÷R1‡UŸêXø+cXù±
1bBà´
“† 1-3-24-25 ¶·¸·úûüä·½4/ýþ9þÿ½0,ĀāĂÓåæU07898#lo
Y" 1 a^0"#XO4ÓT#𝐴𝐵𝐸𝐷@ÓT#𝐶𝐹𝐺𝐻ëbÃ?#X㥹$%mÊõ7898#9T
* aXbXc ~0–åĆićĈ
𝑎2 + 𝑏2 = 𝑐2´
(1)ĉĊċ,ĀāĂ0Čk­zčnö
žŸ
¿Rt △𝐴𝐵𝐶4X∠𝐴𝐶𝐵= 90°X𝐵𝐶= 𝑎X𝐴𝐶= 𝑏X𝐴𝐵= 𝑐´'G
𝑎2 + 𝑏2 = 𝑐2´
GH
&" 1 šŸ𝑆正方形𝐴𝐵𝐸𝐷= 4𝑆△𝐴𝐵𝐶+ 𝑆正方形𝐶𝐹𝐺𝐻X
∵𝑆正方形𝐴𝐵𝐸𝐷= 𝑐2X𝑆△𝐴𝐵𝐶= ______X
Ã?#𝐶𝐹𝐺𝐻T*u______X
∴𝑐2 = 4 ×
1
2𝑎𝑏+ (𝑎−𝑏)2 = 2𝑎𝑏+ 𝑎2−2𝑎𝑏+ 𝑏2X
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í𝑎2 + 𝑏2 = 𝑐2´
(2)Y" 2X¿ △𝐴𝐵𝐶4X∠𝐶= 90°X𝐵𝐶= 𝑎X𝐴𝐶= 𝑏X𝐴𝐵= 𝑐Xr𝐴𝐵u78T¿𝐴𝐵0ĎďÂUĐ78
△𝐴𝐵𝐷XO4𝐴𝐵= 𝐵𝐷X∠𝐴𝐵𝐷= 90°XkÐ D Â𝐷𝐸⊥𝐶𝐵XđĒuÐ E´Ċ<qēŒ0?m]^Ĕ#𝐴𝐶𝐸𝐷
0$%XĕGH𝑎2 + 𝑏2 = 𝑐2£
(3)ċ" 1 40Óå7898#4cK078T[\Á¦Î*‹Œ0*+XŽY" 3 a^0“,Ėė”´ Æ
𝑎= 12X𝑏= 9X“,Ėė”¦ĘęĚ-"4„(}[/0ě*+u 108X'påĖė"Ì0$%´
ËÌ(1)
1
2𝑎𝑏X𝑎−𝑏
(2)•BÍ
(3)Ėė0$%u 393
[ÍéÚÛõ0gG@1XBBCF0à´
-1/‚¡X 𝑆△𝐴𝐵𝐶=
1
2𝑎𝑏,ì&"#b&ÓåæU07898#loY"1a^"#XĜĝ<q?
m]^Ã?#𝐴𝐵𝐸𝐷0$%XíšB£
-2/‚¡XjkGH △𝐴𝐵𝐶≌△𝐵𝐷𝐸,íš56Ê𝐵𝐶= 𝐷𝐸= 𝑎,𝐴𝐶= 𝐵𝐸= 𝑏X<q?m[\]^
ÊĔ#𝐴𝐶𝐸𝐷= 𝑆△𝐴𝐵𝐶+ 𝑆△𝐴𝐵𝐷+ 𝑆△𝐵𝐸𝐷= 𝑎𝑏+
1
2𝑐2@𝑆梯形𝐴𝐶𝐸𝐷=
1
2(𝐴𝐶+ 𝐷𝐸)(𝐶𝐵+ 𝐵𝐸) =
1
2(𝑎+ 𝑏)2XìĞ
†“#획B£
-3/‚¡XćR"#X“,Ėė”¦ĘęĚ ("4„(}[)0ě*+u108Xš
𝐴𝐷+ 𝐵𝐷= 108 ÷ 4 = 27.ôó 𝐴𝐷= 𝑥,Ö𝐵𝐷= 27−𝑥.ô¿ △𝐵𝐶𝐷4X𝐵𝐶² + 𝐶𝐷² = 𝐵𝐷²,ÈX'Ê𝑥ĝšĞ
†®‡B´
ÕB-1/GH
&"šŸ𝑆正方形𝐴𝐵𝐸𝐷= 4𝑆△𝐴𝐵𝐶+ 𝑆正方形𝐶𝐹𝐺𝐻X
∵𝑆正方形𝐴𝐵𝐸𝐷= 𝑐2X𝑆△𝐴𝐵𝐶=
1
2𝑎𝑏X
Ã?#𝐶𝐹𝐺𝐻T*u𝑎−𝑏X
∴𝑐2 = 4 ×
1
2𝑎𝑏+ (𝑎−𝑏)2 = 2𝑎𝑏+ 𝑎2−2𝑎𝑏+ 𝑏2X
í𝑎2 + 𝑏2 = 𝑐2´
ÖËÌu
1
2𝑎𝑏X𝑎−𝑏£
-2/B
 ∵𝐷𝐸⊥𝐵𝐶X
∴∠𝐷𝐵𝐸+ ∠𝐵𝐷𝐸= 90°X
∵∠𝐴𝐵𝐷= 90°X
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∴∠𝐴𝐵𝐶+ ∠𝐷𝐵𝐸= 90°X
∴∠𝐴𝐵𝐶= ∠𝐵𝐷𝐸X
ô∠𝐶= ∠𝐵𝐸𝐷= 90°X 𝐴𝐵= 𝐵𝐷X
∴△𝐴𝐵𝐶≌△𝐵𝐷𝐸(AAS).
∴𝐵𝐶= 𝐷𝐸= 𝑎,𝐴𝐶= 𝐵𝐸= 𝑏£
&¡X¢–q?m
𝑆梯形𝐴𝐶𝐸𝐷= 𝑆△𝐴𝐵𝐶+ 𝑆△𝐴𝐵𝐷+ 𝑆△𝐵𝐸𝐷 =
1
2𝑎𝑏+
1
2𝑐2 +
1
2𝑎𝑏
= 𝑎𝑏+
1
2𝑐2£
¢ğq?m
𝑆梯形𝐴𝐶𝐸𝐷= 1
2 (𝐴𝐶+ 𝐷𝐸) (𝐶𝐵+ 𝐵𝐸)
= 1
2 (𝑎+ 𝑏) (𝑎+ 𝑏)
=
1
2(𝑎+ 𝑏)2X
∴
1
2(𝑎+ 𝑏)2 = 𝑎𝑏+ +
1
2𝑐2X
∴𝑎² + 2𝑎𝑏+ 𝑏2 = 2𝑎𝑏+ 𝑐2X
∴𝑎² + 𝑏² = 𝑐²£
-3/&¡XY"X
∵“,Ėė”¦ĘęĚ ("4„(}[)0ě*+u108X
∴𝐴𝐷+ 𝐵𝐷= 108 ÷ 4 = 27X
ó𝐴𝐷= 𝑥,È𝐵𝐷= 27−𝑥X
¿ △𝐵𝐶𝐷4X 𝐵𝐶² + 𝐶𝐷² = 𝐵𝐷²
∴𝑎2 + (𝑏+ 𝑥)2 = (27−𝑥)2X
ċ𝑎= 12,𝑏= 9ĠġšX
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(9 + 𝑥)2 +144 = (27−𝑥)2X
 ∴𝑥= 7X
∴ĢÃ?#0T*UV𝑎−𝑏= 12−9 = 3X
∴Ėė0$%u
1
2𝐵𝐶× 𝐶𝐷× 4 + 3 × 3 =
1
2 × 12 × 16 × 4 + 3 × 3 = 393´
【题型2  由勾股定理求线段长度】
µ 2-24-25 ģ·¸·ĤĥĦħ·Ĩ)Ýĩ/Y"X¿T*u 8 0Ã?#𝐴𝐵𝐶𝐷4X𝐸b𝐵𝐶Ī–ÐX𝐹b𝐶𝐷
0Î*(Ī–ÐXÄÅ𝐴𝐸X𝐴𝐹X𝐴𝑀>[∠𝐸𝐴𝐹Ï𝐶𝐷VÐ𝑀´Æ𝐵𝐸= 𝐷𝐹= 2XÈ𝐹𝑀0*+u-īī/
A´
24
5
B´
34
5
C´
48
5
D´
12
5
ËÌB
[ÍéÚÛÃ?#0ÑÒÀ98#æU05NÑÒUXUŸêXŠ‚Ã?#0ÑÒN9
8#æU05NÑÒXGH𝐴𝐸= 𝐴𝐹X8>[(0ÑÒN98#æU05NÑÒXGH𝐸𝑀= 𝐹𝑀X
ó𝐸𝑀= 𝑥XÈ𝐹𝑀= 𝑥X𝑀𝐶= 10−𝑥X𝐶𝐸= 6X¿Rt △𝑀𝐶𝐸4Š‚'BíšXÞßÃ?#0ÑÒÀ
98#æU05NÑÒÀbB0à´
ÕBB
∵ÓT#𝐴𝐵𝐶𝐷bÃ?#X
∴𝐴𝐵= 𝐴𝐷X∠𝐴𝐵𝐸= ∠𝐴𝐷𝐹= 90° X
∴¿Rt △𝐴𝐵𝐸@Rt △𝐴𝐷𝐹4X
𝐴𝐵= 𝐴𝐷
∠𝐴𝐵𝐸= ∠𝐴𝐷𝐹
𝐵𝐸= 𝐷𝐹
 
X
∴Rt △𝐴𝐵𝐸≌Rt △𝐴𝐷𝐹(SAS)X
∴𝐴𝐸= 𝐴𝐹£
∵𝐴𝑀>[∠𝐸𝐴𝐹X
∴∠𝐸𝐴𝑀= ∠𝐹𝐴𝑀X
∴¿ △𝐴𝐸𝑀@ △𝐴𝐹𝑀4X
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𝐴𝐸= 𝐴𝐹
∠𝐸𝐴𝑀= ∠𝐹𝐴𝑀
𝐴𝑀= 𝐴𝑀
 
X
∴△𝐴𝐸𝑀≌△𝐴𝐹𝑀(SAS)X
∴𝐸𝑀= 𝐹𝑀X
∵ÓT#𝐴𝐵𝐶𝐷bÃ?#X
∴𝐵𝐶= 𝐶𝐷= 8X∠𝐵𝐶𝐷= 90°X
ó𝐸𝑀= 𝑥XÈ𝐹𝑀= 𝑥X𝑀𝐶= 𝐶𝐷−𝐷𝑀= 8−(𝑥−2) = 10−𝑥X
𝐶𝐸= 𝐵𝐶−𝐵𝐸= 8−2 = 6X
¿Rt △𝑀𝐶𝐸4XŠ‚X𝐸𝑀2 = 𝑀𝐶2 +𝐶𝐸2X
í𝑥2 = (10−𝑥)2 + 62X
B
𝑥=
34
5 X
∴𝐹𝑀=
34
5 X
Ö×
B´
“† 2-1-24-25 Ĭ·¸·îâĭĮ·½¾/Y"X¿ △𝐴𝐵𝐶4X𝐴𝐵T0đ7>[(𝐷𝐸[\Ï𝐴𝐵À𝐴𝐶
TVÐ𝐷@Ð𝐸Xį𝐵𝐶2 +𝐶𝐸2 = 𝐴𝐸2´
(1)ÄÅ𝐵𝐸X'G
∠𝐶= 90°£
(2)Æ𝐴𝐶= 4,𝐵𝐶= 2X'𝐶𝐸0*´
ËÌ(1)•BÍ
(2)𝐶𝐸=
3
2
[ÍİÚÛõđ7>[(0ÑÒ@XB0àbÞßrĪŸêд
-1/Š‚đ7>[(0ÑҍÊ𝐴𝐸= 𝐵𝐸XćR𝐵𝐶2 +𝐶𝐸2 = 𝐴𝐸2Ê𝐵𝐶2 +𝐶𝐸2 = 𝐵𝐸2íšGH£
-2/ó𝐶𝐸= 𝑥XÈ𝐴𝐸= 𝐵𝐸= 4−𝑥X¿Rt △𝐵𝐶𝐸4XŠ‚Ğ?­'B횴
ÕB-1/GH
 ∵𝐴𝐵T0đ7>[(u𝐷𝐸X
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∴𝐴𝐸= 𝐵𝐸X
∵𝐵𝐶2 +𝐶𝐸2 = 𝐴𝐸2X
∴¿ △𝐵𝐶𝐸4X𝐵𝐶2 +𝐶𝐸2 = 𝐵𝐸2X
∴∠𝐶= 90°£
-2/B
ó𝐶𝐸= 𝑥XÈ𝐴𝐸= 𝐵𝐸= 4−𝑥X
∴¿Rt △𝐵𝐶𝐸4X𝐸𝐶2 +𝐵𝐶2 = 𝐵𝐸2X
í𝑥2 + 22 = (4−𝑥)2X
B
𝑥=
3
2X
í𝐶𝐸=
3
2´
“† 2-2-25-26 ¶·¸Ī·æþ·ıIJÝĩ/Y"X¿ △𝐴𝐵𝐶4X𝐴𝐵= 5,𝐴𝐶= 4,𝐵𝐶= 3X𝐷𝐸b𝐴𝐵0đ
7>[(X𝐷𝐸[\Ï𝐴𝐶À𝐴𝐵VÐ EÀD´
(1)'G
 △𝐴𝐵𝐶b7898#£
(2)'𝐴𝐸0*´
ËÌ(1)•BÍ
(2)
25
8
[ÍéÚÛXPXđ7>[(ÑÒXB0àbijGH78X슂đ7
>[(ÑÒst()XŠ‚Ğ?­'B´
-1/Š‚PíšGH£
-2/ÄÅ𝐵𝐸XŠ‚𝐷𝐸b𝐴𝐵0đ7>[(XŽ𝐴𝐸= 𝐵𝐸Xó𝐴𝐸= 𝐵𝐸= 𝑥XÈ𝐸𝐶= 4−𝑥X¿Rt △𝐴𝐵𝐶4X
Š‚Ğ?­'B획ŽËÌ´
ÕB-1/GH
∵𝐴𝐵= 5X𝐴𝐶= 4X𝐵𝐶= 3X
∴𝐴𝐶2 +𝐵𝐶2 = 𝐴𝐵2
∴∠𝐴𝐶𝐵= 90°
∴△𝐴𝐵𝐶b7898#£
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-2/B
ÄÅ𝐵𝐸X
∵𝐷𝐸b𝐴𝐵0đ7>[(X
∴𝐴𝐸= 𝐵𝐸X
∴ó𝐴𝐸= 𝐵𝐸= 𝑥XÈ𝐸𝐶= 4−𝑥X
∵¿Rt △𝐴𝐵𝐶4X𝐸𝐶2 +𝐵𝐶2 = 𝐵𝐸2X
∴(4−𝑥)2 + 32 = 𝑥2X
∴𝑥=
25
8 X
∴𝐴𝐸=
25
8 ´
“† 2-3-24-25 ģ·¸·ºïĴĵ·Ĩ)Ýĩ/Y"X¿UT98#𝐴𝐵𝐶4XÐ P ¿Oò}Xį
𝐵𝑃= 12X𝐶𝑃= 5X∠𝐶𝑃𝐵= 150°Xċ △𝐴𝐵𝑃ĶÐ B ķPÿĸ?ÁĹs60°Ž △𝐶𝐵𝐷´
(1)'Ð P ›Ð D ~0LM£
(2)'()𝐴𝑃0*´
ËÌ(1)12
(2)13
[ÍĺhiÚÛĹs0ÑÒXUT98#05@ÑÒXB98#UXB¡XÂÊĻ
ļ(bBà´
-1/ÄÅ𝑃𝐷´Š‚UT98#05@ÑÒXĹs0ÑÒÊ △𝑃𝐵𝐷bUT98#Xíš'B£
-2/UT98#0ÑÒ¤ △𝑃𝐶𝐷b7898#Xì&'B횴
ÕB-1/B
Y"XÄÅ𝑃𝐷´
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∵△𝐴𝐵𝐶bUT98#X
∴∠𝐴𝐵𝐶= 60°,𝐵𝐴= 𝐵𝐶´
∵△𝐶𝐵𝐷b △𝐴𝐵𝑃ĶÐ B PÿĸĹsŽ0X
∴△𝐶𝐵𝐷≌△𝐴𝐵𝑃X
∴𝐵𝑃= 𝐵𝐷´
∵∠𝑃𝐵𝐷= ∠𝐴𝐵𝐶= 60°X
∴△𝑃𝐵𝐷bUT98#´
∴𝑃𝐷= 𝑃𝐵= 12XíÐ P ›Ð D ~0LMb 12´
-2/ ∵∠𝐶𝑃𝐵= 150°X △𝑃𝐵𝐷bUT98#´
∴∠𝐷𝑃𝐶= 90°X
∴△𝑃𝐶𝐷b7898#X
∴𝑃𝐶2 +𝑃𝐷2 = 𝐶𝐷2X
∴𝐶𝐷2 = 52 + 122X
∴𝐶𝐷= 13´
&-1/Ÿ △𝐶𝐵𝐷≌△𝐴𝐵𝑃´
∴𝐴𝑃= 𝐶𝐷= 13´
【题型3  勾股数（树）的运用】
µ 3-24-25 ¶·¸·îâĽä·½4/œ–åT*u 1 0Ã?#Xľk 1 ˆ“¯*”ĝX¿Ŀ0ŀĎŁ
Ī¯Ê<åĢÃ?#XO49åÃ?#Ęo098#b7898#Xìľk 1 ˆ“¯*”ĝX#o0"#Y" 1
a^´YZłŃ“¯*”ńXĿċ““ŅņŇň”Y" 2 a^XÆ“¯*”õ 2025 ˆĝX#o0"#4aœ0
Ã?#0$%@b-    /
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A´2026
B´2025
C´22025
D´22022−1
ËÌA
[ÍéÚÛõXÉʼnŠ‚Ŋ…ŋ–ˆŽ0Ō0Ã?#0$%@›ōÃ?#0$
%~0€bBËé0ഊ‚'Ê“¯*”õ 1 ˆĝ#o0"#4aœ0Ã?#0$%@X
ćR"#ěćŎŏXŠ‚ŎŏBË횴
ÕBB
Y"X&¡XÃ?# A 0$%u 1X
&XÃ?# B 0$% + Ã?# C 0$% = Ã?# A 0$% = 1X
∴“¯*”õ 1 ˆĝ#o0"#4aœ0Ã?#0$%@u 2X
ŒšX“¯*”õ 2 ˆĝ#o0"#4aœ0Ã?#0$%@u 3X
∴“¯*”õ 3 ˆĝ#o0"#4aœ0Ã?#0$%@u 4X
……
∴“¯*”õ 2025 ˆĝ#o0"#4aœ0Ã?#0$%@u 2026´
Ö×
A´
“† 3-1-24-25 ¶·¸Ī·ºŐőä·½¾/ĞÓŒ,4b,0–Œb-īī/
A´
1
3X
1
4X
1
5
B´0.3X0.4X0.5
C´5X12X13
D´32X42X52
ËÌC
[ÍÙÚÛõ,0ŸêXœĒ𝑎2 + 𝑏2 = 𝑐209åÃö,XŔu,XÞßrĪŸêbBËé
0à£
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éŕ'GbŖu,XpŗŘÊ9T0*XřigG<ĢT0>?@UVJ*T0>?Xíš'B´
ÕBB
AÀŚu 
1
3X
1
4X
1
5ëēbö,XarĿśēb,XÖé×ŜēR¡£
BÀŚu0.3X0.4X0.5ëēbö,XarĿśēb,XÖé×ŜēR¡£
CÀŚu132 = 52 + 122XarĿśb,XÖé×ŜŝR¡£
DÀŚu32 = 9X42 = 16X52 = 25X92 + 162 ≠252XarĿśēb,XÖé×ŜēR¡´
Ö×
C´
“† 3-2-24-25 ¶·¸·ŞşŠ·½4/𝑎2 + 𝑏2 = 𝑐2éšŢb–åV𝑎X𝑏X𝑐0?­Xœ
Ēpå?­0Ãö,B(𝑎,𝑏,𝑐)j”de,Œ´ţŤťŦŧŨũÊõ–嫬,Œ0†XŠ‚İ
†šr«¬ÊY,Œ
(3,4,5),(5,12,13),(7,24,25),⋯´[ÍĪ$,ŒšrŊ…X4 = 1 × (3 + 1)
,12 = 2 × (5 + 1),24 = 3 × (7 + 1),⋯X[ÍĪ$ŎŏX¢ 9 å,Œu           
´
ËÌ-19X180X181/
[ÍéhiÚÛõ,XB0àbŪÊ,‚~0€XÞß´
&,Œ
-3X4X5/X-5X12X13/X-7X24X25/…4X4 = 1 × [(1 × 2 + 1) + 1]X12 = 2 × [(2 × 2 + 1) + 1]
X24 = 3 × [(3 × 2 + 1) + 1]X…š¢ 9 Œ,40,u
9 × [(9 × 2 + 1) + 1] = 180X‘Ê
-19X180X181/´
ÕBB
&,Œ
-3X4X5/X-5X12X13/X-7X24X25/…4X4 = 1 × [(1 × 2 + 1) + 1]X
12 = 2 × [(2 × 2 + 1) + 1]X24 = 3 × [(3 × 2 + 1) + 1]X…š¢ 9 Œ,40,u
9 ×
[(9 × 2 + 1) + 1] = 180X‘Ê-19X180X181/´
ÖËÌu-19X180X181/´
“† 3-3-24-25 Ĭ·¸Ī·îâūú·½4/Y"b–ŬŭŮ0.XO4aœ0ÓT#ëbÃ?#X
aœ098ëb7898#´Æ𝐴X𝐵X𝐶X𝐷0T[\b5X4X3X2XÈJ0Ã?#𝐹0$%u      
´
ËÌ54
[ÍéÚÛõ0IXŠ‚Ã?#$%†ij'Ê𝑀𝑁2À𝐻𝐼20ÇX슂Ã?#0Ñҍ
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Ž𝑄𝑃2À𝑃𝐺20ÇXJĝ𝑄𝐺2 = 𝑄𝑃2 +𝑃𝐺2X퍎Ã?#𝐹0$%XÞß0Ib
B0à´
ÕBB
∵𝐴X𝐵X𝐶X𝐷0T[\b5X4X3X2Xaœ098ëb7898#X
∴𝑀𝑁2 = 52 + 42 = 41X𝐻𝐼2 = 32 + 22 = 13X
∵aœ0ÓT#ëbÃ?# X
∴𝑄𝑃2 = 𝑀𝑁2 = 41X𝑃𝐺2 = 𝐻𝐼2 = 13X
∴X𝑄𝐺2 = 𝑄𝑃2 +𝑃𝐺2 = 41 + 13 = 54X
∴J0Ã?#𝐹0$%u54X
ÖËÌu
54´
【题型4  网格中判断直角三角形】
µ 4-24-25 ¶·¸·ŞşŠ·½4/Y"X¿4 × 4234XŋåĢÃ?#0T*ë‹UX23(0
ÏÐŔu3Ð.3Ð𝐶›"403Ð𝐴X𝐵š«o7898#XÈpů03Ð𝐶œ-   /
A´2 å
B´3 å
C´4 å
D´5 å
ËÌD
[ÍŠ‚230ŰÐrNUĐ7898#0ÑÒX[űŲĈXŪÊŝR¡0Ð𝐶Xíš'B´
ÕBB
Y"X
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3Ð𝐶›"403Ð𝐴X𝐵š«o7898#XÈpů03Ð𝐶œ5å
Ö×
D´
“† 4-1-24-25 ¶·¸Ī·Ĥĥéų·½4/Y"X △𝐴𝐵𝐷@ △𝐴𝐵𝐶0ŴÐŵ¿T*u 1 0ĢÃ?#2
33ÐĪXÈ∠𝐵𝐴𝐶0+,u-   /
A´120°
B´135°
C´150°
D´165°
ËÌB
[ÍÎ*𝐵𝐴Ž EXÄÅ𝐶𝐸Xij0PGH △𝐴𝐸𝐶b7898#XŶ‘š
∠𝐴𝐸𝐶= 90°XĜĝUTŷU8N98#0ò8@š∠𝐸𝐴𝐶= ∠𝐸𝐶𝐴= 45°XJĝŸz8Źz
획ÊËÌ´
ÕBB
Y"XÎ*𝐵𝐴Ž EXÄÅ𝐶𝐸X
&¡š
𝐶𝐸2 = 12 + 22 = 5X𝐴𝐸2 = 22 + 12 = 5X𝐴𝐶2 = 12 + 32 = 10X
∴𝐴𝐸2 +𝐶𝐸2 = 𝐴𝐶2X
∴△𝐴𝐸𝐶b7898#X
∴∠𝐴𝐸𝐶= 90°X
∵𝐸𝐴= 𝐸𝐶X
∴∠𝐸𝐴𝐶= ∠𝐸𝐶𝐴=
1
2(180°−∠𝐴𝐸𝐶) = 45°X
∴∠𝐵𝐴𝐶= 180°−∠𝐸𝐴𝐶= 135°X
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Ö×
B´
ÐØéhiÚÛõ›23FX0PXUTŷU8X98#0ò8@X
Ÿz8Źz'8+UŸêÐXÜÝÞßNOPbB0à´
“† 4-2-24-25 ¶·¸·źĮŻż·½¾/Y"XÐ AXBXCXD ŵ¿Ã?#233ÐĪXÈ
∠𝐷𝐴𝐶−∠𝐵𝐴𝐶=        
°´
ËÌ45°/45 +
[ÍİÚÛõXŽŷŔ@UĐ7898#0ÑÒ@5XÂÐ𝐵V()𝐴𝐶0ŷŔÐ𝐵′XÄ
Å𝐴𝐵′,𝐷𝐵′X&ŷŔš∠𝐵′𝐴𝐶= ∠𝐵𝐴𝐶Xí∠𝐷𝐴𝐶−∠𝐵𝐴𝐶= ∠𝐷𝐴𝐶−∠𝐵′𝐴𝐶= ∠𝐵′𝐴𝐷XČH △𝐴𝐵′𝐷bUĐ
7898#Xíš'B´
ÕBB
Y"XÂÐ𝐵V()𝐴𝐶0ŷŔÐ𝐵′XÄÅ𝐴𝐵′,𝐷𝐵′X
&ŷŔš∠𝐵′𝐴𝐶= ∠𝐵𝐴𝐶X
í∠𝐷𝐴𝐶−∠𝐵𝐴𝐶= ∠𝐷𝐴𝐶−∠𝐵′𝐴𝐶= ∠𝐵′𝐴𝐷X
óĢÃ?#0T*u 1 X
&X𝐴𝐵′2 = 𝐷𝐵′2 = 12 + 22 = 5,𝐴𝐷2 = 12 + 32 = 10X
∴𝐴𝐵′2 +𝐷𝐵′2 = 𝐴𝐷2X
∴△𝐴𝐵′𝐷bUĐ7898#X
∴∠𝐵′𝐴𝐷= 45°Xí∠𝐷𝐴𝐶−∠𝐵𝐴𝐶= 45°´
ÖËÌu
45°´
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“† 4-3-24-25 ¶·¸·žſžä·½4/Y"X∠1X∠2¿23ĪƀƁXÈ∠1 + ∠2 =     °´
ËÌ45
[ÍéÚÛõ2Ƃ98#´ÜÝÞßX0PXUĐ7898#5@ÑÒX
>ƃÑÒXbB0à´
>ƃ𝑂𝐷Ž𝐴𝐵XÄÅ𝐵𝐶XÈ∠2 = ∠𝐴𝐵𝐹XGH △𝐴𝐵𝐶bUĐ7898#X∠𝐴𝐵𝐶= 45°X∠𝐴𝐶𝐵= 90°X¿
△𝐴𝐸𝐹@ △𝐵𝐸𝐶4XŠ‚∠1 = ∠𝐶𝐵𝐸X∠1 + ∠2 = 45°´
ÕBB
Y"X>ƃ𝑂𝐷Ž𝐴𝐵XÄÅ𝐵𝐶X
È∠2 = ∠𝐴𝐵𝐹X
∵𝐵𝐶2 = 12 + 42 = 17X𝐴𝐶2 = 12 + 42 = 17X𝐴𝐵2 = 32 + 52 = 34X
∴𝐴𝐶2 +𝐵𝐶2 = 𝐴𝐵2X𝐴𝐶= 𝐵𝐶X
∴△𝐴𝐵𝐶bUĐ7898#X
∴∠𝐴𝐵𝐶= 45°X∠𝐴𝐶𝐵= 90°X
¿ △𝐴𝐸𝐹@ △𝐵𝐸𝐶4X
∵∠𝐴𝐹𝐸= ∠𝐵𝐶𝐸= 90°X∠𝐴𝐸𝐹= ∠𝐵𝐸𝐶X
∴∠1 = ∠𝐶𝐵𝐸X
∴∠1 + ∠2 = ∠𝐶𝐵𝐸+ ∠𝐴𝐵𝐹= 45°´
ÖËÌu
45´
【拔尖篇】
【题型5  利用勾股定理求两条线段的平方和（差）】
µ 5Y"X△𝐴𝐵𝐶@ △𝐸𝐶𝐷ëbUĐ7898#X△𝐴𝐵𝐶0ŴÐ A ¿ △𝐸𝐶𝐷0WT DE Ī´ĞćĈ
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O4ä0œ-    /
①△𝐴𝐶𝐸≌△𝐵𝐶𝐷        ②𝐵𝐷+ 𝐴𝐷= 𝐷𝐸
③∠𝐷𝐴𝐵= ∠𝐵𝐶𝐷        ④𝐴𝐸2 +𝐴𝐷2 = 2𝐵𝐶2
A´1 å
B´2 å
C´3 å
D´4 å
ËÌD
[ÍéÚÛUĐ98#0ÑÒX98#æU05NÑÒX´
& △𝐴𝐵𝐶@ △𝐸𝐶𝐷ëbUĐ7898#XšG∠𝐵𝐶𝐷= ∠𝐴𝐶𝐸X‘Š‚“SAS”šG △𝐴𝐶𝐸≌△𝐵𝐶𝐷XÖć
Ĉ①ä£& △𝐴𝐶𝐸≌△𝐵𝐶𝐷š𝐵𝐷= 𝐴𝐸X‘šGćĈ②ä£& △𝐴𝐵𝐶@ △𝐸𝐶𝐷ëbUĐ789
8#š∠𝐶𝐴𝐵= ∠𝐶𝐵𝐴= 45° ∠𝐸= ∠𝐶𝐷𝐸= 45°XŶ‘G∠𝐷𝐵𝐴= ∠𝐴𝐶𝐷X∠𝐴𝐷𝐵= 90°‘Ž
∠𝐷𝐵𝐴+ ∠𝐷𝐴𝐵= 90°X∠𝐵𝐶𝐷+ ∠𝐴𝐶𝐷= 90°XŚÙ∠𝐷𝐴𝐵= ∠𝐷𝐶𝐵XÖćĈ③䣿Rt △𝐴𝐵𝐷4X𝐵𝐷2 +𝐴
𝐷2 = 𝐴𝐵2X¿Rt △𝐴𝐵𝐶4X𝐴𝐶2 +𝐵𝐶2 = 𝐴𝐵2XŚÙ𝐵𝐷2 +𝐴𝐷2 = 𝐴𝐶2 +𝐵𝐶2XUĠt획Ž𝐴𝐸2 +𝐴𝐷2
= 2𝐵𝐶2XÖćĈ④ä´
ÕB∵△𝐴𝐵𝐶@ △𝐸𝐶𝐷ëbUĐ7898#X
∴𝐶𝐴= 𝐶𝐵X𝐶𝐸= 𝐶𝐷X∠𝐵𝐶𝐴= ∠𝐷𝐶𝐸= 90°X
∴∠𝐵𝐶𝐴−∠𝐷𝐶𝐴= ∠𝐷𝐶𝐸−∠𝐷𝐶𝐴X
í∠𝐵𝐶𝐷= ∠𝐴𝐶𝐸X
∴△𝐴𝐶𝐸≌△𝐵𝐶𝐷(SAS)XÖćĈ①ä£
∵△𝐴𝐶𝐸≌△𝐵𝐶𝐷X
∴𝐵𝐷= 𝐴𝐸X
∴𝐵𝐷+ 𝐴𝐷= 𝐴𝐸+ 𝐴𝐷= 𝐷𝐸XÖćĈ②ä£
∵∠𝐵𝐶𝐴= ∠𝐷𝐶𝐸= 90°X
∴∠𝐶𝐴𝐵+ ∠𝐶𝐵𝐴= 90°X∠𝐸+ ∠𝐶𝐷𝐸= 90°X
∵𝐶𝐴= 𝐶𝐵X𝐶𝐸= 𝐶𝐷X
∴∠𝐶𝐴𝐵= ∠𝐶𝐵𝐴= 45°X∠𝐸= ∠𝐶𝐷𝐸= 45°X
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∵△𝐴𝐶𝐸≌△𝐵𝐶𝐷X
∴∠𝐸𝐴𝐶= ∠𝐷𝐵𝐶X
∵∠𝐸𝐴𝐶= ∠𝐴𝐷𝐶+ ∠𝐴𝐶𝐷= 45° + ∠𝐴𝐶𝐷X
∠𝐷𝐵𝐶= ∠𝐴𝐵𝐶+ ∠𝐷𝐵𝐴= 45° + ∠𝐷𝐵𝐴X
∴∠𝐷𝐵𝐴= ∠𝐴𝐶𝐷X
∵△𝐴𝐶𝐸≌△𝐵𝐶𝐷X
∴∠𝐵𝐷𝐶= ∠𝐴𝐸𝐶= 45°X
∴∠𝐴𝐷𝐵= ∠𝐴𝐷𝐶+ ∠𝐵𝐷𝐶= 45° + 45° = 90°X
∴∠𝐷𝐵𝐴+ ∠𝐷𝐴𝐵= 90°X
∵∠𝐵𝐶𝐷+ ∠𝐴𝐶𝐷= ∠𝐴𝐶𝐵= 90°X
∴∠𝐷𝐴𝐵= ∠𝐷𝐶𝐵XÖćĈ③ä£
∵¿Rt △𝐴𝐵𝐷4X𝐵𝐷2 +𝐴𝐷2 = 𝐴𝐵2X
¿Rt △𝐴𝐵𝐶4X𝐴𝐶2 +𝐵𝐶2 = 𝐴𝐵2X
∴𝐵𝐷2 +𝐴𝐷2 = 𝐴𝐶2 +𝐵𝐶2X
∵𝐵𝐷= 𝐴𝐸X𝐴𝐶= 𝐵𝐶X
∴𝐴𝐸2 +𝐴𝐷2 = 2𝐵𝐶2XÖćĈ④ä´
QĪXä0ćĈœ 4 å´
Ö×
D
“† 5-1ŷ8(Ź‹đ70ÓT#de“đŭ”ÓT#X…œY"a^0“đŭ”ÓT#𝐴𝐵𝐶𝐷Xŷ8(𝐴𝐶,𝐵𝐷
ÏVÐ𝑂´Æ𝐴𝐷= 1,𝐵𝐶= 4XÈ𝐴𝐵2 +𝐶𝐷2 =                
´
  
ËÌ17
[ÍŠ‚đ70S@BË횣
ÕBB
∵𝐴𝐶⊥𝐵𝐷X
∴∠𝐴𝑂𝐷= ∠𝐴𝑂𝐵= ∠𝐵𝑂𝐶= ∠𝐶𝑂𝐷= 90°,
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&X𝐴𝐵2 +𝐶𝐷2 = 𝐴𝑂2 +𝐵𝑂2 +𝐶𝑂2 +𝐷𝑂2,
𝐴𝐷2 + 𝐵𝐶2 = 𝐴𝑂2 + 𝐷𝑂2 + 𝐵𝑂2 + 𝐶𝑂2,
∴𝐴𝐵2 + 𝐶𝐷2 = 𝐴𝐷2 + 𝐵𝐶2,
∵𝐴𝐷= 1,𝐵𝐶= 4,
∴𝐴𝐵2 + 𝐶𝐷2 = 12 + 42 = 17.
ÖËÌu
17´
ÐØéÚÛ0bđ70SX0IXäB“đŭ”ÓT#0SÀù±1b
B0à´
“† 5-2-24-25 Ĭ·¸·îâūú·½¾/Y" 1X𝐴𝐵∥𝐶𝐷XÐ𝐸X𝐹[\¿7(𝐴𝐵X𝐶𝐷ĪXÄÅ𝐸𝐹X
∠𝑀𝐸𝑁= 90°XÐ𝑁¿𝐶𝐷Ī´
(1)Æ∠𝐵𝐸𝑁= 20°X'∠𝐴𝐸𝑀0+,£
(2)Æ𝐸𝑁>[∠𝐵𝐸𝐹Ï𝐶𝐷VÐ𝑁X'G
Ð𝐹b𝑀𝑁04У
(3)Y" 2XkÐ𝐹Â𝐹𝐻⊥𝐶𝐷Ï𝐸𝑁VÐ𝐻XƄy()𝐸𝑀X𝐸𝐻X𝐻𝑁œª,€XĕČH&´
ËÌ(1)∠𝐴𝐸𝑀= 70°£
(2)•BÍ£
(3)ćĈ
𝐸𝑀2 +𝐸𝐻2 = 𝐻𝑁2X&•BÍ´
[Í-1/>8SB횣
-2/Š‚8>[(S@>’(0ÑҍŽ∠𝑁𝐸𝐹= ∠𝐸𝑁𝐹XìU80ƅ8‹UŽ∠𝐸𝑀𝑁= ∠𝐹𝐸𝑀X
U8ŷUTŽ𝐸𝐹= 𝑀𝐹= 𝐹𝑁X획G£
-3/ÄÅ𝑀𝐻XȜ𝑀𝐻= 𝐻𝑁XìƆ횴
ÕB-1/B
 ∵ ∠𝐵𝐸𝑁= 20°X∠𝑀𝐸𝑁= 90°X
∴∠𝐴𝐸𝑀= 180°−∠𝐵𝐸𝑁−∠𝑀𝐸𝑁= 180°−20°−90° = 70°£
-2/GH
 ∵ 𝐸𝑁>[∠𝐵𝐸𝐹X
∴∠𝐵𝐸𝑀= ∠𝑁𝐸𝐹X
ô ∵ 𝐴𝐵∥𝐶𝐷X
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∴∠𝐵𝐸𝑁= ∠𝐸𝑁𝐹X
∴ ∠𝑁𝐸𝐹= ∠𝐸𝑁𝐹X
∴𝐸𝐹= 𝑁𝐹X
∵ ∠𝑀𝐸𝑁= 90°X
∴∠𝐸𝑀𝐹+ ∠𝐸𝑁𝑀= 90°X∠𝑀𝐸𝐹+ ∠𝐹𝐸𝑁= 90°X
∴ ∠𝐸𝑀𝑁= ∠𝐹𝐸𝑀X
∴𝐸𝐹= 𝑀𝐹X
∴ 𝑀𝐹= 𝐹𝑁X
íÐ𝐹b𝑀𝑁04У
-3/ćĈ
𝐸𝑀2 +𝐸𝐻2 = 𝐻𝑁2X&Y
Y" 2XÄÅ𝑀𝐻´
∵ 𝐹𝐻⊥𝐶𝐷XÐ𝐹u𝑀𝑁04д
∴ 𝐻𝐹u𝑀𝑁04đ(´
∴ 𝑀𝐻= 𝐻𝑁´
¿Rt △𝐸𝑁𝐻4X∠𝑀𝐸𝐻= 90°´
&𝐸𝑀2 +𝐸𝐻2 = 𝐻𝑀2´
∴𝐸𝑀2 +𝐸𝐻2 = 𝐻𝑁2´
ÐØéÚÛõ80@AXUĐ98#05›ÑÒXXđ7>[(0ÑÒXÜÝ1rĪ
ÑÒƆbB0à´
“† 5-3-24-25 ¶·¸·áââƇ·Ĩ)Ýĩ/ýśƈŷ8(Ź‹đ70ÓT#deđŭÓT#´
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-1/Y"①XžŸÓT#𝐴𝐵𝐶𝐷bđŭÓT#Xĉ !<ŒŷT𝐴𝐵2 +𝐶𝐷2›𝐴𝐷2 +𝐵𝐶2~0,€X
ĕČH&£
-2/Y"②X[\rRt △𝐴𝐶𝐵078T𝐴𝐶@WT𝐴𝐵uTÁ¦ÂÃ?#𝐴𝐶𝐹𝐺@Ã?#𝐴𝐵𝐷𝐸XÄÅ
𝐵𝐸,𝐶𝐺,𝐸𝐺XžŸ𝐴𝐶= 4,𝐴𝐵= 5X'𝐺𝐸2´
ËÌ-1/Ƅy
𝐴𝐷2 +𝐵𝐶2 = 𝐴𝐵2 +𝐶𝐷2´&•BÍ£-2/73
[ÍéÚÛÃ?#0ÑÒÀÀđŭÓT#0SUŸê´
-1/ćĈ
𝐴𝐷2 +𝐵𝐶2 = 𝐴𝐵2 +𝐶𝐷2´íšGH£
-2/ÄÅ𝐶𝐺X𝐵𝐸XřiGHÓT#𝐶𝐺𝐸𝐵bđŭÓT#X-2/4ćĈíšBCF´
ÕBB
-1/Ƅy
𝐴𝐷2 +𝐵𝐶2 = 𝐴𝐵2 +𝐶𝐷2´&Y
∵ÓT#𝐴𝐵𝐶𝐷bđŭÓT#X
∴𝐴𝐶⊥𝐵𝐷X
∴∠𝐴𝑂𝐷= ∠𝐴𝑂𝐵= ∠𝐵𝑂𝐶= ∠𝐶𝑂𝐷= 90°X
&X𝐴𝐷2 +𝐵𝐶2 = 𝐴𝑂2 +𝐷𝑂2 +𝐵𝑂2 +𝐶𝑂2X
𝐴𝐵2 +𝐶𝐷2 = 𝐴𝑂2 +𝐵𝑂2 +𝐶𝑂2 +𝐷𝑂2X
∴𝐴𝐷2 +𝐵𝐶2 = 𝐴𝐵2 +𝐶𝐷2£
-2/ÄÅ𝐶𝐺X𝐵𝐸XY"
∵∠𝐶𝐴𝐺= ∠𝐵𝐴𝐸= 90°X
∴∠𝐶𝐴𝐺+ ∠𝐵𝐴𝐶= ∠𝐵𝐴𝐸+ ∠𝐵𝐴𝐶Xí∠𝐺𝐴𝐵= ∠𝐶𝐴𝐸X
¿ △𝐺𝐴𝐵@ △𝐶𝐴𝐸4X𝐴𝐺= 𝐴𝐶X∠𝐺𝐴𝐵= ∠𝐶𝐴𝐸X𝐴𝐵= 𝐴𝐸X
∴△𝐺𝐴𝐵≌△𝐶𝐴𝐸(SAS)X
∴∠𝐴𝐵𝐺= ∠𝐴𝐸𝐶X
ô∵∠𝐴𝐸𝐶+ ∠𝐴𝑀𝐸= 90°X
∴∠𝐴𝐵𝐺+ ∠𝐴𝑀𝐸= 90°X
ô∵∠𝐵𝑀𝐶= ∠𝐴𝑀𝐸X
∴∠𝐴𝐵𝐺+ ∠𝐵𝑀𝐶= 90°X
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∴𝐶𝐸⊥𝐵𝐺X
∴ÓT#𝐶𝐺𝐸𝐵bđŭÓT#X
&-1/šŸ𝐶𝐺2 +𝐵𝐸2 = 𝐶𝐵2 +𝐺𝐸2X
∵𝐴𝐶= 4X𝐴𝐵= 5X
∴&X𝐶𝐵2 = 9X𝐶𝐺2 = 32X𝐵𝐸2 = 50X
∴𝐺𝐸2 = 𝐶𝐺2 +𝐵𝐸2−𝐶𝐵2 = 73´
【题型6  利用勾股定理解决折叠问题】
µ 6-24-25 ¶·¸·úûRƉ·½4/Y"X¿Rt △𝐴𝐵𝐶4X∠𝐶= 90°X𝐴𝐶= 4,𝐵𝐶= 6XċĿ0Ɗ8 A
ƋDXƌÐ A ƍ¿T𝐵𝐶04Ð D ƎXDƏÏ𝐴𝐶TVÐ EXÏ𝐴𝐵TVÐ FXÈ𝐷𝐸0*u-   /
A´2
B´3
C´
25
9
D´
25
8
ËÌD
[ÍéÚÛõDE0ÑÒÀX&¡Ê𝐶𝐷=
1
2𝐵𝐶= 3X&DE0ÑҚ𝐴𝐸= 𝐷𝐸Xó
𝐴𝐸= 𝐷𝐸= 𝑥XÈ𝐶𝐸= 4−𝑥Xì&®‡íšÊËÌ´
ÕBB
 ∵Ð D u𝐵𝐶04ÐX
∴𝐶𝐷=
1
2𝐵𝐶= 3X
&DE0ÑҚ𝐴𝐸= 𝐷𝐸X
ó𝐴𝐸= 𝐷𝐸= 𝑥XÈ𝐶𝐸= 𝐴𝐶−𝐴𝐸= 4−𝑥X
&𝐶𝐸2 +𝐶𝐷2 = 𝐷𝐸2X
∴(4−𝑥)2 + 32 = 𝑥2X
B
𝑥=
25
8 X
∴𝐷𝐸=
25
8 X
Ö×
D´
“† 6-1-24-25 ¶·¸·žſƐƑ·Ĩ)Ýĩ/Y"¿Ô#𝐴𝐵𝐶𝐷4X𝐵𝐶= 8X𝐶𝐷= 6Xċ △𝐵𝐶𝐷ƒŷ
8(𝐵𝐷ƋDXÐ𝐶ƍ¿Ð𝐶′ƎX𝐵𝐶′Ï𝐴𝐷VÐ𝐸XÈ △𝐵𝐷𝐸0$%u          
 ´
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ËÌ
75
4
[ÍƓijŠ‚¡Ž𝐵𝐸= 𝐷𝐸XĜĝŠ‚ŽV()𝐴𝐵À𝐴𝐸À𝐵𝐸0?­XB?­íšB
CF´
ÕBB
 ∵ÓT#𝐴𝐵𝐶𝐷uÔ#X
∴𝐴𝐷∥𝐵𝐶X𝐴𝐷= 𝐵𝐶= 8X𝐴𝐵= 𝐶𝐷= 6X∠𝐴𝐵𝐶= ∠𝐶= ∠𝐶𝐷𝐴= ∠𝐴= 90°X
∴∠𝐸𝐷𝐵= ∠𝐷𝐵𝐶X
ó𝐸𝐷= 𝑥XÈ𝐴𝐸= 8−𝑥X
Š‚DEšŸ
∠𝐸𝐵𝐷= ∠𝐷𝐵𝐶X 
∴∠𝐸𝐷𝐵= ∠𝐸𝐵𝐷X
∴𝐸𝐵= 𝐸𝐷= 𝑥X
&
𝐵𝐸2 = 𝐴𝐵2 +𝐴𝐸2X
í𝑥2 = 62 + (8−𝑥)2X
B
𝑥=
25
4 X
∴𝐸𝐷=
25
4 X
∴△𝐵𝐷𝐸0$%u
1
2𝐷𝐸⋅𝐴𝐵=
1
2 ×
25
4 × 6 =
75
4 ´
ÖËÌu
75
4 ´
ÐØéhiÚÛõÔ#›DEFXXU8ŷUTX>’(0ÑÒXB0àbŠ‚Ƌ
D“t0ÑÒXU©ªŸêXù±’56À[ÍÀƆƔBË´
“† 6-2-24-25 ¶·¸·ºŐƕƖ·½4/Y"X¿Ô#𝐴𝐵𝐶𝐷4X𝐴𝐵= 12X𝐵𝐶= 10XÐ EÀF [\
¿𝐴𝐵À𝐷𝐶ĪXį𝐴𝐵= 3𝐵𝐸X𝐶𝐷= 3𝐶𝐹XÐ P u7(𝐴𝐵Ī–ƗÐXÄÅ𝐷𝑃Xċ △𝐷𝐴𝑃ƒ𝐷𝑃a¿7(ƋD
Ž △𝐷𝐴′𝑃XƘÐ𝐴′ƙƚƍ¿7(𝐸𝐹ĪÿX𝐴𝑃0*u    ´
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ËÌ5 Ɣ 20/20 Ɣ 5
[ÍéÚÛõƋD“t-DEF/XÔ#0ÑÒX0IXäƛÊ"#XeŽ,#ć
RXbBCF0à´ 
&Ô#0ÑҍŽ𝐷𝐶= 𝐴𝐵= 12X∠𝐵𝐴𝐷= 90°X𝐴𝐷= 𝐵𝐶= 10XŠ‚žŸ=ƜŽ𝐵𝐸= 𝐶𝐹= 4XƆÊÓ
T#𝐵𝐶𝐹𝐸bÔ#XÓT#𝐴𝐷𝐹𝐸bÔ#XŽ∠𝐴𝐸𝐹= ∠𝐷𝐹𝐸= ∠𝐵𝐸𝐹= ∠𝐶𝐹𝐸= 90°X𝐴𝐷= 𝐸𝐹= 10XŠ
‚DE0ÑÒÀ획ŽćĈ´
ÕBB
∵ÓT#𝐴𝐵𝐶𝐷bÔ#X
∴𝐷𝐶= 𝐴𝐵= 12X∠𝐵𝐴𝐷= 90°X𝐴𝐷= 𝐵𝐶= 10X
∵𝐴𝐵= 3𝐵𝐸X𝐶𝐷= 3𝐶𝐹X
∴𝐵𝐸= 𝐶𝐹= 4X𝐴𝐸= 𝐷𝐹= 8X
∵𝐵𝐸∥𝐶𝐹X
∴ÓT#𝐵𝐶𝐹𝐸bÔ#X
ŒÓT#𝐴𝐷𝐹𝐸bÔ#X
∴∠𝐴𝐸𝐹= ∠𝐷𝐹𝐸= ∠𝐵𝐸𝐹= ∠𝐶𝐹𝐸= 90°X𝐴𝐷= 𝐸𝐹= 10X
∵ċ △𝐷𝐴𝑃ƒ𝐷𝑃a¿7(ƋDŽ △𝐷𝐴′𝑃X
∴𝐷𝐴′ = 𝐷𝐴= 10X𝐴𝑃= 𝐴′𝑃X
ƘÐ𝑃¿T𝐴𝐵ĪÿXÐ𝐴′ƙƚƍ¿7(𝐸𝐹ĪXY"a^
¿Rt △𝐴′𝐷𝐹4X
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∵𝐷𝐹2 + 𝐴′𝐹2 = 𝐴′𝐷2X
∴82 + 𝐴′𝐹2 = 102X
∴𝐴′𝐹= 6X
∴𝐴′𝐸= 4X
¿Rt △𝐴′𝑃𝐸4X𝐴′𝐸2 +𝑃𝐸2 = 𝐴′𝑃2X
í42 + (8−𝐴𝑃)2 = 𝐴′𝑃2X
∴𝐴𝑃= 5£
ƘÐ𝑃¿T𝐴𝐵Î*(ĪÿXÐ𝐴′ƙƚƍ¿7(𝐸𝐹ĪXY"a^
Œš'𝐴′𝐹= 6X
∴𝐴′𝐸= 10 + 6 = 16X
¿Rt △𝐴′𝑃𝐸4X𝐴′𝐸2 +𝑃𝐸2 = 𝐴′𝑃2X
í162 + (𝐴𝑃−8)2 = 𝐴′𝑃2X
∴𝐴𝑃= 20£
ÖËÌu 5 Ɣ 20´
“† 6-3-24-25 ģ·¸·áâƝƞ·Ĩ)Ýĩ/Y"X¿Rt △𝐴𝐵𝐶4X∠𝐵= 90°X𝐴𝐵= 5X𝐵𝐶= 8X
Ð𝐷u𝐴𝐵Ī–ÐXÄÅ𝐶𝐷Xċ △𝐵𝐶𝐷ƒ𝐶𝐷ƋDŽ △𝐵′𝐶𝐷XkÐ A Â𝐴𝐹∥𝐵𝐶Ï𝐷𝐵′VÐ EXÏ𝐶𝐵′VÐ
FXÆ𝐴𝐸= 𝐵′𝐸XÈ𝐵𝐷0*u      
´
ËÌ
40
11
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[ÍéÚÛõDE0ÑÒXæU98#05@ÑÒX´Â𝐹𝐺⊥𝐵𝐶VÐ𝐺XGH
△𝐷𝐸𝐴≌△𝐹𝐸𝐵′(ASA)XŽ𝐴𝐷= 𝐵′𝐹X𝐷𝐸= 𝐸𝐹Xó𝐵𝐷0*u𝑥X¿Rt △𝐹𝐺𝐶4X𝐹𝐺= 5X𝐶𝐹= 𝑥+ 3X
𝐶𝐺= 8−𝑥XІ®‡íš'B´
ÕBB
ÂFG ⊥BCVÐGX
∵∠B = 90°XAF∥BCX
∴ÓT#ABGFb*?#X∠BAF = 90°X
∴FG = AB = 5X
&DE0Ñҍ∠B′ = ∠B = 90°XD′B = DBXB′C = BCX
∵∠DAE = ∠B′ = 90°X∠DEA = ∠B′EF = 90°XAE = B′EX
∴△DEA≌△FEB′(ASA)X
∴AD = B′FXDE = EFX
óBD0*uxXÈB′F = AD = 5−xXCF = B′C−B′F = 8−(5−x) = x + 3XB′D = BD = xX
∴AF = AE + EF = B′E + DE = B′D = xX
¿Rt △FGC4XFG = 5XCF = x + 3XCG = BC−BG = BC−AF = 8−xX
∴FG2 +CG2 = CF2Xí52 + (8−x)2 = (x + 3)2X
Bx = 40
11X
∴BD =
40
11´
ÖËÌu
40
11´
【题型7  勾股定理的证明】
µ 7-24-25 ¶·¸·žſžä·½4/¿0ĩk­4XýśžľƟõ1"#gGƠơ0
XĞ×Ŝ40"#XÉGH0b-   /
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A´①②③
B´①②④
C´②③④
D´①②③④
ËÌD
[ÍéhiÚÛ0GHk­X[\ŋå"#$%0<qēŒ0®‡?mXìſƢU†X
ìöíš56´
ÕBB
¿"①4Xöå"#0$%UV<å98#0$%ƣÃ?#0$%XƤUV<åĢÃ?#
0$%ƣĪ<å7898#0$%X
∴𝑐2 +2 ×
1
2𝑎𝑏= 𝑎2 + 𝑏2 +2 ×
1
2𝑎𝑏X
ö𝑎2 + 𝑏2 = 𝑐2X
Ö①šrGH£
¿"②4XÃ?#0$%UVÓå98#0$%ƣĢÃ?#0$%X
∴4 ×
1
2𝑎𝑏+ 𝑐2 = (𝑎+ 𝑏)2X
ö𝑎2 + 𝑏2 = 𝑐2X
Ö②šrGH£
¿"③4X&"šŸ9å98#0$%0@UVĔ#0$%X
∴
1
2𝑎𝑏+
1
2𝑎𝑏+
1
2𝑐2 =
1
2(𝑎+ 𝑏)(𝑎+ 𝑏)X
öš𝑎2 + 𝑏2 = 𝑐2X
Ö③šrGH£
¿"④4XÄÅ𝐵𝐷X
Ù"ƤšrƥoRt △𝐵𝐸𝐴ĶO78ŴÐħÿĸĹs90°XìÁ>ƃŽ´–?$XÓT#𝐴𝐵𝐶𝐷0$%U
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V △𝐴𝐵𝐶@Rt △𝐴𝐶𝐷0$%~@X¥–?$XÓT#𝐴𝐵𝐶𝐷0$%UVRt △𝐴𝐵𝐷@ △𝐵𝐶𝐷0$%~@X
ar𝑆△𝐴𝐵𝐶+ 𝑆△𝐴𝐶𝐷= 𝑆△𝐴𝐵𝐷+ 𝑆△𝐵𝐶𝐷X
í
1
2𝑏2 +
1
2𝑎𝑏=
1
2𝑐2 +
1
2𝑎(𝑏−𝑎)X
ö
𝑏2 +𝑎𝑏= 𝑐2 +𝑎(𝑏−𝑎)X
𝑏2 +𝑎𝑏= 𝑐2 +𝑎𝑏−𝑎2X
∴𝑎2 + 𝑏2 = 𝑐2X
Ö④šrGH£
∴ÉGH0b①②③④´
Ö×
D´
“† 7-1-24-25 ¶·¸Ī·źƦä·½¾/Y"a^X¡ƠơƛĀŤ▪Ƨƨ–ƩƪƫƬlÊē–
ů0ƭƮXGHõ´Æó" 1 4ƭƯ}[-<åÃ?#@<å7898#Œo/0$%u𝑆1Xľ
krưƬXƋsXlÊ" 2XO4ƭƯ}[0$%u𝑆2XƱƲŒÊõrÓåćĈ
①𝑆1 = 𝑎2 + 𝑏2
+𝑎𝑏£②𝑆2 = 𝑐2 +𝑎𝑏£③𝑆1 = 𝑆2£④𝑎2 + 𝑏2 = 𝑐2´ÈO4ä0œ-   /
A´1 å
B´2 å
C´3 å
D´4 å
ËÌD
[ÍéÚÛ0GHX7898#0ÑÒUŸêXB0àbƳƴ"Ƶƶ´Š‚
X7898#rNÃ?#0$%†®‡XíšBCF´
ÕBB
&
𝑎2 + 𝑏2 = 𝑐2X
&¡
𝑆1 = 𝑆2 = 𝑎2 + 𝑏2 +2 ×
1
2𝑎𝑏= 𝑎2 + 𝑏2 +𝑎𝑏= 𝑐2 +𝑎𝑏X
Ö①X②X③X④äX
Ö×
D´
“† 7-2Y"Xƈ*ÀƷÀŷ8(0*[\b aÀbÀc 0Ô#ƒŷ8(ƬƸX›–å78T*u c 0U
Đ7898#lÅoĎT0"#X$%ƹzmÉʼnŽ0–åU†b  ´
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ËÌa2+b2ƺc2
[Í98#0$%@ÀĔ#0$%ƒ]^på"#0$%XŶ‘ĞÊU†XŊ…T›T~0€´
ÕBB
Ù"šrpůBXœ9å Rt△O$%[\u 
1
2abX
1
2ab @ 
1
2c2´
ƻœ–å78Ĕ#XO$%u 
1
2-a+b/-a+b/´
&"#šŸ
1
2-a+b/-a+b/ƺ
1
2ab+
1
2ab+
1
2c2X
ö-a+b/2ƺ2ab+c2Xa2+b2+2abƺ2ab+c2X
∴a2+b2ƺc2´
ÖËÌu
a2+b2ƺc2´
ÐØÙÚÛ0ŸêÐb0GHXhiõ98#0$%†
Ƽ×ƽ÷2X@Ĕ#0$%
†
-ĪƼ+Ƽ/×ƽ÷2´
“† 7-3-24-25 Ĭ·¸·ºŐūä·Ĩ)Ýĩ/ƾö†0ƿm‰–ǀĩ4Xýśǁgõ“r#ļ,X
r,B#”0ǂ!ǃ´pč[ǁ…õ,4“,#ćR”p–,xy?m0ĆiÑ´DŽDžĬ·¸,Dždž
ĢŒjk$%LJU0?mŷ7898#9T€
．．．．．．．．．’õ !´
Lj˜ !
  
-1/Y"-1/X7898#ƪƫ9=T*[\u aXbXc-𝑏< 𝑎< 𝑐/XĢŒŒÓåpů0ƪƫlo
õ–åÃ?#X4ƭ–åĢÃ?#-ljNJ}[/´
①–å7898#ƪƫ0$%u____XĢÃ?#T*u_____´-¨ aXb 0Ġ,†]^/
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②ĉ<qēŒ0?m]^ÊljNJ}[-ĢÃ?#/0$%XŶ‘ !Ê aXbXc 9Nj~0€´-nj．
Ǎǎ
．．/
ćĈ1
-2/Y" 2XžŸX △𝐴𝐵𝐶b7898#X∠𝐶= 90°´ĉĪ$Ž0ćĈ'B´
①Æ𝐴𝐶= 3X𝐵𝐶= 4X'𝐴𝐵0*´
②Æ𝐴𝐶= 6X𝐴𝐵0*Ǐ𝐵𝐶0* 2X'𝐴𝐵0*´
IǐǑ
-3/Y" 3XžŸX¿ △𝐴𝐵𝐶4X𝐴𝐵= 13X𝐴𝐶= 15X𝐵𝐶= 14Xĉ'Ê △𝐴𝐵𝐶0$%´
  
ËÌ-1/①
1
2𝑎𝑏£𝑎−𝑏£②ĢÃ?#$%u(𝑎−𝑏)2Ɣ𝑐2−2𝑎𝑏X𝑎2 + 𝑏2 = 𝑐2£-2/①5£②10£
-3/84
[ÍéhiÚÛõ@0GHXĞĠ,†XܟNOGH?mbB0
à´
-1/①Š‚98#$%®‡†š¢–ƭËÌXì&"#~0€šĢÃ?#$%UV7898#
0*78T0*ǒńK78T0*X‚Ùš¢ğƭ0ËÌ£②Š‚Ã?#$%®‡†šĢÃ?#$
%u(𝑎−𝑏)2XŠ‚ĢÃ?#0$%UVT*u c 0Ã?#$%ǒń 4 å7898#0$%šĢÃ?#0
$%u𝑐2−2𝑎𝑏XÈ(𝑎−𝑏)2 = 𝑐2−2𝑎𝑏X‚ÙšËÌ£
-2/①Š‚-1/š𝐴𝐵2 = 𝐴𝐶2 +𝐵𝐶2X‚Ù®‡'B횣②Š‚-1/š𝐴𝐵2 = 62 + (𝐴𝐵−2)2X
‚Ù'B횣
-3/kÐ A Â𝐴𝐷⊥𝐵𝐶V DXó𝐵𝐷= 𝑥XÈ𝐶𝐷= 14−𝑥XȚGH𝐴𝐷2 = 𝐴𝐵2−𝐵𝐷2 = 𝐴𝐶2−𝐶𝐷2Xí132−𝑥2
= 152−(14−𝑥)2XB?­'Ê𝐴𝐷0*획ŽËÌ´
ÕBB
-1/①&¡X–å7898#ƪƫ0$%u
1
2𝑎𝑏XĢÃ?#0T*u𝑎−𝑏£
②∵ĢÃ?#0T*u𝑎−𝑏X
∴ĢÃ?#0$%u(𝑎−𝑏)2£
∵ĢÃ?#0$%UVT*u c 0Ã?#$%ǒń 4 å7898#0$%X
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∴ĢÃ?#0$%u𝑐2−4 ⋅
1
2𝑎𝑏= 𝑐2−2𝑎𝑏X
∴(𝑎−𝑏)2 = 𝑐2−2𝑎𝑏X
∴𝑎2−2𝑎𝑏+ 𝑏2 = 𝑐2−2𝑎𝑏X
∴𝑎2 + 𝑏2 = 𝑐2£
-2/①&-1/š𝐴𝐵2 = 𝐴𝐶2 +𝐵𝐶2X
∵𝐴𝐶= 3X𝐵𝐶= 4X
∴𝐴𝐵2 = 32 + 42 = 25X
∴𝐴𝐵= 5Ɣ𝐴𝐵= −5-Ǔń/£
②∵𝐴𝐵0*Ǐ𝐵𝐶0* 2X
∴𝐵𝐶= 𝐴𝐵−2X
ô∵𝐴𝐵2 = 𝐴𝐶2 +𝐵𝐶2X𝐴𝐶= 6X
∴𝐴𝐵2 = 62 + (𝐴𝐵−2)2X
∴𝐴𝐵= 10£
-3/Y"a^XkÐ A Â𝐴𝐷⊥𝐵𝐶V DX
ó𝐵𝐷= 𝑥XÈ𝐶𝐷= 14−𝑥X
¿Rt △𝐴𝐵𝐷4X𝐴𝐷2 +𝐵𝐷2 = 𝐴𝐵2X
¿Rt △𝐴𝐶𝐷4X𝐴𝐷2 +𝐶𝐷2 = 𝐴𝐶2X
∴𝐴𝐷2 = 𝐴𝐵2−𝐵𝐷2 = 𝐴𝐶2−𝐶𝐷2X
∴132−𝑥2 = 152−(14−𝑥)2X
B𝑥= 5X
∴𝐴𝐷2 = 𝐴𝐵2−𝐵𝐷2 = 152−𝑥2 = 152−52 = 144X
∴𝐴𝐷= 12X
∴𝑆△𝐴𝐵𝐶=
1
2𝐵𝐶⋅𝐴𝐷=
1
2 × 14 × 12 = 84´
  
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【题型8  勾股定理的应用】
µ 8-24-25 ¶·¸·ǔǕºǖǗǘ·Ĩ)Ýĩ/ǙǚǛbƝǜĀǝǞ0–Ŝ±ƗXǟǚǛđ7ą$ÿ
ǠǡMą$0LM𝐴𝐶u0.5ǢXċǠǡǣ>ƆƗ 3 Ǣ-𝐵𝐸= 3Ǣ/XÙÿǠǡ›ą$0LM𝐵𝐷u1.5ǢXÆ
ƆƗk­4ŦǤǥǦŦǧ7XÈǚǛ0ŦǤ𝑂𝐴0*+u     Ǣ´
ËÌ5
[ÍéhiÚÛõ0IXB0àbÜÝÞß´
ǨóǩŸ,XíšBËÙ´
ÕBB
&"šŸXÓT#𝐸𝐶𝐷𝐵bÔ#X
∴𝐸𝐶= 𝐵𝐷= 1.5X
∴𝐴𝐸= 𝐸𝐶−𝐴𝐶= 1.5−0.5 = 1X
Ǩó𝑂𝐴0*+u𝑥XÈ𝑂𝐸= 𝑂𝐴−𝐴𝐸= 𝑥−1X𝑂𝐵= 𝑥X
¿Rt △𝑂𝐵𝐸4X&X
𝑂𝐸2 +𝐵𝐸2 = 𝑂𝐵2X
í(𝑥−1)2 + 32 = 𝑥2X
BX𝑥= 5X
ÖËÌu
5´
“† 8-1Y"a^XǟǪǫƀVâï?Á0Ǭǭ(Ī´“Ǯǯ”	À“Ǭǰ”	ęDZŒÿMƸǪǫX|Dzƒ
–dz?Áǯ’X“Ǯǯ”	ŋĢÿǯ’16ǬŗX“Ǭǰ”	ŋĢÿǯ’12Ǭŗ´ĿśMƸǪǫ
3
2Ģÿĝ‹L30
Ǭŗ´YZŸǴ“Ǯǯ”	ƒâ?Áǯ’XɟǴ“Ǭǰ”	ƒǵå?Áǯ’ǶǷ
ËÌï?Á
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[ÍéÚÛõ0PÀ?ƀ8UŸêXB0àbB¡Xù±1aŸêBC
FXǸV4ڔÚ´
Š‚ǹ­ = Ǻ+ × ÿX[\'𝑃𝑄À𝑃𝑅0*X쐖˜Š‚0PšrGH98#𝑃𝑄𝑅b
7898#XŶ‘'B´
ÕBB
Š‚¡X𝑃𝑄= 16 × 1.5 = 24-Ǭŗ/X𝑃𝑅= 12 × 1.5 = 18-Ǭŗ/X𝑄𝑅= 30-Ǭ
ŗ/X
∵242 + 182 = 302X
í𝑃𝑄2 +𝑃𝑅2 = 𝑄𝑅2X
∴∠𝑄𝑃𝑅= 90°´
&“Ǯǯ	”ƒâ?Áǯ’šŸX∠𝑄𝑃𝑆= 45°XÈ∠𝑆𝑃𝑅= 45°X
í“Ǭǰ”	ƒï?Áǯ’´
“† 8-2-24-25 ¶·¸·ǻäǼâĮ·Ĩ)Ýĩ/ǟǽǾǿ¿Ȁǹ𝐴𝐵ĪȁſȂėȃ𝐸Xr?ȄȀǹ𝐴𝐵<
ȅ0𝐶X𝐷<Ɩ0ȆDŽʒ´Y"X𝐶ƖŽȀǹ𝐴𝐵0LM𝐴𝐶= 20kmX𝐷ƖŽȀǹ𝐴𝐵0LM𝐷𝐵= 60kmX
𝐴𝐵= 100kmXľǽȇȈ›Ȁǹ}ƑȉŻJĝ¤¿Ž𝐶X𝐷<ƖLM‹U0𝐸ƎȁſȂėȃX'𝐴𝐸X𝐵𝐸0
*´
ËÌ𝐴𝐸= 66kmX𝐵𝐸= 34km
[ÍjkóǩŸ,X[\]^Ê𝐶𝐸@𝐷𝐸X슂𝐶𝐸= 𝐷𝐸ſƢ?­'B´éhiÚÛ
õ0IXÜÝÞßXŠ‚LM‹UſƢ?­bB0à´
ÕBB
ó𝐴𝐸= 𝑥kmXÈ𝐵𝐸= (100−𝑥)km´
Š‚¡X𝐶𝐸= 𝐷𝐸´
∴202 + 𝑥2 = (100−𝑥)2 + 602X
B𝑥= 66´
∴100−𝑥= 34´
∴𝐴𝐸= 66kmX𝐵𝐸= 34km´
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“† 8-3-24-25 ¶·¸·ºïȊä·½¾/kƾ‰ĝXȋ,DždžĢŒ0ȌȍśƒŽȎȏ
ĪȐȑȒ𝐴𝐵0ƽ+XjkȐŽYƶ
①ȐŶȑȒŴȓđ7Ȕƒ0ȕȑ0ǤȖǏȑȒ* 3 Ǣ-Y" 1/£
②ƘċǤȖŦ7ÿXȐÙÿŦǤȖ0{Žą$0LM𝐶𝐷u 1 ǢXŽȑȒ0LM𝐶𝐸u 12 Ǣ-Y" 2/´
Š‚rĪƶXB˝ĞF
(1)óȑȒ𝐴𝐵= 𝑥ǢXÈ𝐴𝐶= ______ǢX𝐴𝐸= ______Ǣ-¨𝑥0†Ȗ]^/
(2)'ȑȒ𝑥0Ç´
ËÌ(1)(𝑥+ 3)£(𝑥−1)
(2)17 Ǣ
[Í-1/Š‚¡Ğ†]Ťíš´
-2/óȑȒ0ƽu x ǢXÈǤȖ*u(𝑥+ 2)mǢX®‡íš´
éÚÛõ0IXÜÝÞßbB0à´
ÕB-1/B
Š‚¡XŶȑȒŴȓđ7Ȕƒ0ȕȑ0ǤȖǏȑȒ* 3 Ǣ
ÖǤ*u(𝑥+ 3)Ǣ£
Š‚¡XŽÓT#𝐶𝐷𝐵𝐸bÔ#XŽ𝐶𝐷= 𝐵𝐸= 1ǢX
Ö𝐴𝐸= (𝑥−1)ǢX
ÖËÌu
(𝑥+ 3)£(𝑥−1)´
-2/B
¿Rt △𝐴𝐸𝐶4X∠𝐴𝐸𝐶= 90°
∴𝐴𝐸2 +𝐸𝐶2 = 𝐴𝐶2    
í(𝑥−1)2 + 122 = (𝑥+ 3)2    
B
𝑥= 17    
Ë
ȑȒ𝑥0Çu 17 Ǣ´
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【题型9  利用勾股定理求最短距离】
µ 9-2025 ¶·¸·æþ·ȗÝĩ/Y"XȘș"#B˝$0F
(1)Ù"#0ơŔu_______£
(2)ĉĊ›ŒȚ–țe–åpů0Ƣǁ"#XĕƈĿ0ď$ƒ𝐴𝑆ƬƸXȜ¿ȝ$ĪXÈĿ0ď$ǑƸ"b–
å_______£
(3)YZÐ𝐶b𝑆𝐴04ÐX¿𝐴Ǝœ–řȞȟX¿𝐶ƎƙƚœȞȟyȠ0ȡȢXįĿřÉĶÙƢǁ"#0ď$
ȣ’–ȤŽ𝐶Ǝ´ĊÉ¿ď$ǑƸ"4ƛÊȞȟȣ’0JKǹ(ǶǷ
(4)𝑆𝐴0*u 10Xď$ǑƸ"0ȥȦ8u90°XĉĊ'ÊȞȟȣ’0JKǹ­0>?´
ËÌ(1)ȥȧ
(2)Ȩ#
(3)•BÍ
(4)125
[ÍéÚÛIÀȥȧ0ď$ǑƸ"X“Ǎȩ$u>$”UŸêXB0àbBȨ#0Ȫ
*UVȥȧƼ$Ȥ*XȨ#0ȫȬUVȥȧ0ȭ(*´
-1/Š‚©ªǁ0ŰК56Ù"#uȥȧ£
-2/ȥȧ0ď$ǑƸ"bȨ#£
-3/i'Ȟȟȣ’0JKLMXnjċȥȧ0ď$ǑƸX‘Š‚“<Ð~()JK”ÊćZ£
-4/7ÅB7898#횴
ÕB-1/B
Ù"#0ơŔuȥȧ;
ÖËÌu
ȥȧ£
-2/Ɨ{Ȏ´Ŀ0ď$ǑƸ"b–åȨ#£
ÖËÌu
Ȩ#£
-3/ƈÙƢǁ"#0ď$ǑƸXYË"a^XÄÅ𝐴𝐶XÈ𝐴𝐶uȞȟȣ’0JKǹ(´
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-4/&ȮŸ𝑆𝐶= 5´
¿Rt △𝐴𝑆𝐶4X&X𝐴𝐶2 = 102 + 52 = 125´
ÖȞȟȣ’0JKǹ­0>?u 125´
“† 9-1Y"a^b–å9¸ȯĨXĿ0ŋ–¸0*ÀƷÀƽ[\UV 7cmÀ6cmÀ2cmXA @ B bp
<åȯĨ0<å‹ŷ0ȓÐXȖřȰȱŶÐ A ÊŊľkȯĨȣŽÐ B 0JKǹ(œȲ*Ƿ
ËÌ25cm
[ÍǑƸĝŽ7898# ACBXŠ‚¡'Ê ACÀBCXŠ‚'Ê AB 횴
ÕBB
Y"XċȯĨǑƸX
&¡£ACƺ6×3+2×3ƺ24XBCƺ7X´
ar&
AB2ƺAC2+BC2ƺ625X
í ABƺ25-cm/X
Ë
Ȱȱȣ’0JK(ǹu 25cm´
ÐØéhiÚÛŷX>$ǑƸ——JKǹȬFUŸêÐ0B@ÞßXÉB¡ŸǴb
'Ê7898# ABC 0WT AB 0*bBÙ0à´
“† 9-2-24-25 ¶·¸·æþ·Ǩ½Âȳ/Y"X*?ǁ0ȴ*𝐴𝐵= 𝐵𝐶= 6cmX𝐴𝐴1 = 𝐵𝐵1 = 𝐶𝐶1 = 14
cmXǨóȵȶȷŶ*?ǁòŴÐ𝐶1r2cm/s0Ǻ+¿*?ǁ0ò}ƒȴ𝐶1𝐶Áȣ’´ŒÿXȵȶȸŶ*?
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ǁòŴÐ𝐴r‹Œ0Ǻ+¿*?ǁ0ď$Īȣ’X_`ȵȶȸȹȺnjiȲ*ÿȻÉȼȽŽȵȶȷǷ
ËÌȵȶȸȹȺnji
85
14sȻÉȼȽŽȵȶȷ
[ÍéÚÛ0I´ċ*?ǁ0ď$}[ǑƸXÆȵȶȸ¿Ð𝐹ƎȼȽŽȵȶȷXÄÅ𝐴𝐹Ï
𝐵𝐵1VÐ𝐸XÈƘ𝐴𝐹= 𝐶1𝐹ÿXÿJK´óȵȶȷŶŴÐ𝐶1ƒȴ𝐶1𝐶ÁŴÐ𝐶ȣ’0ŒÿXȵȶȸŶŴÐ𝐴
ķǹȬ𝐴→𝐸→𝐹ȣ’ȼȽŽȵȶȷnji𝑥s´¿Rt △𝐴𝐶𝐹4XŠ‚«¬?­X'B횴
ÕBB
ċ*?ǁ0ď$}[ǑƸXY"a^´
Æȵȶȸ¿Ð𝐹ƎȼȽŽȵȶȷXÄÅ𝐴𝐹Ï𝐵𝐵1VÐ𝐸X
ÈƘ𝐴𝐹= 𝐶1𝐹ÿXÿJK´
óȵȶȷŶŴÐ𝐶1ƒȴ𝐶1𝐶ÁŴÐ𝐶ȣ’0ŒÿXȵȶȸŶŴÐ𝐴ķǹȬ𝐴→𝐸→𝐹ȣ’ȼȽŽȵȶȷnji𝑥sX
È𝐴𝐹= 𝐶1𝐹= 2𝑥cm´
&¡šŸ∠𝐴𝐶𝐹= 90°X𝐴𝐶= 𝐴𝐵+ 𝐵𝐶= 6 + 6 = 12(cm)X𝐶𝐹= 𝐶𝐶1−𝐶1𝐹= 14−2𝑥(cm)
∵¿Rt △𝐴𝐶𝐹4X𝐴𝐹2 = 𝐴𝐶2 +𝐶𝐹2´
∴(2𝑥)2 = 122 + (14−2𝑥)2X
B𝑥=
85
14´
∴ȵȶȸȹȺnji
85
14sȻÉȼȽŽȵȶȷ´
“† 9-3-24-25 ģ·¸·æþ·ıIJÝĩ/Y"u–å*4cmXƷ2cmXƽ1cm0„Ȧ*?ǁX–řȰ
ȱŶ„Ȧ*?ǁ0ŴÐ A ÊŊXƒ*?ǁ0]$ȣŽŷ$ŴÐ𝐶1ƎXFȾůȿǹ(JKǷJKǹ(*uȲ
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ȺǷ
Ë̃"①0?†ȣ’ǹ(JKXJKǹ(*u5cm
[ÍéÚÛ>$ǑƸ-JKǹȬFXƈ“Ƣǁ"#”sǍu“>$"#”XĜĝŠ‚®‡BË´
ÕBB
Ȱȱ&Ð A ƒ*?ǁ0]$ȣ’ŽÐ𝐶1Xœ9q?†X[\Ǒo>$"#Y
Y" 1X¿Rt △𝐴𝐵𝐶14X
𝐴𝐶2
1 = 𝐴𝐵2 +𝐵𝐶2
1 = 42 + 32 = 52 = 25´
Y" 2X¿Rt △𝐴𝐶𝐶14X
𝐴𝐶2
1 = 𝐴𝐶2 +𝐶𝐶2
1 = 62 + 12 = 37´
Y" 3X¿Rt △𝐴𝐵1𝐶14X
𝐴𝐶2
1 = 𝐴𝐵2
1 + 𝐵1𝐶2
1 = 52 + 22 = 29´
∵25 < 29 < 37X
∴ƒ"①0?†ȣ’ǹ(JKXJKǹ(*u5cm´
【题型10  勾股定理及其逆定理的综合】
µ 10-24-25 ¶·¸·źĮɀä·Ĩ)Ýĩ/FɁɂY"X¿ △𝐴𝐵𝐶4X𝐴𝐷u𝐵𝐶TĪ0ƽ´
۵ǂ!-1/Æ𝐶𝐷= 1X𝐴𝐷= 2X𝐵𝐷= 4X'G
𝐴𝐵⊥𝐴𝐶£
ƄyGH-2/Š‚-1/40ćĈXĢHƄy
ƘœĒ𝐴𝐷2 = 𝐵𝐷⋅𝐶𝐷XNOPXš
GH △𝐴𝐵𝐶b7898#XĉĊgGĢH0ƄybŖä´
ËÌ-1/•BÍ-2/ćĈoƢX•BÍ
[ÍéÚÛNOPXnæ>?†XÜÝÞßNOPbB0à´
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-1/ij¿Rt △𝐴𝐶𝐷4X&𝐴𝐶2 = 𝐴𝐷2 +𝐶𝐷2 = 5XŒ𝐴𝐵2 = 𝐴𝐷2 +𝐵𝐷2 = 20Xš𝐴𝐵2 +𝐴𝐶2 = 𝐵
𝐶2XÈ △𝐴𝐵𝐶u7898#XíšGH£
-2/ij¿Rt △𝐴𝐶𝐷4X&𝐴𝐶2 = 𝐴𝐷2 +𝐶𝐷2XŒ𝐴𝐵2 = 𝐴𝐷2 +𝐵𝐷2X&𝐵𝐶2 = (𝐶𝐷+ 𝐵𝐷)2 = 𝐶
𝐷2 +2𝐶𝐷⋅𝐵𝐷+ 𝐵𝐷2X𝐴𝐷2 = 𝐵𝐷⋅𝐶𝐷XÊ𝐵𝐶2 = 𝐶𝐷2 +2𝐴𝐷2 +𝐵𝐷2 = 𝐴𝐵2 +𝐴𝐶2XÈ △𝐴𝐵𝐶u7898
#XíšGH´
ÕBB
-1/∵𝐴𝐷⊥𝐵𝐶X
∴∠𝐴𝐷𝐵= ∠𝐴𝐷𝐶= 90°X
∵𝐶𝐷= 1X𝐴𝐷= 2X
∴¿Rt △𝐴𝐶𝐷4X&𝐴𝐶2 = 𝐴𝐷2 +𝐶𝐷2 = 5X
Œ𝐴𝐵2 = 𝐴𝐷2 +𝐵𝐷2 = 20X
∵𝐵𝐶2 = (𝐶𝐷+ 𝐵𝐷)2 = 25X
∴𝐴𝐵2 +𝐴𝐶2 = 𝐵𝐶2X
∴△𝐴𝐵𝐶u7898#X
∴𝐴𝐵⊥𝐴𝐶£
-2/ćĈoƢX&Y
∵𝐴𝐷⊥𝐵𝐶X
∴∠𝐴𝐷𝐵= ∠𝐴𝐷𝐶= 90°X
∴¿Rt △𝐴𝐶𝐷4X&𝐴𝐶2 = 𝐴𝐷2 +𝐶𝐷2X
Œ𝐴𝐵2 = 𝐴𝐷2 +𝐵𝐷2X
∵𝐵𝐶2 = (𝐶𝐷+ 𝐵𝐷)2 = 𝐶𝐷2 +2𝐶𝐷⋅𝐵𝐷+ 𝐵𝐷2X𝐴𝐷2 = 𝐵𝐷⋅𝐶𝐷X
∴𝐵𝐶2 = 𝐶𝐷2 +2𝐴𝐷2 +𝐵𝐷2 = 𝐴𝐵2 +𝐴𝐶2X
∴△𝐴𝐵𝐶u7898#X
∴𝐴𝐵⊥𝐴𝐶´
“† 10-1-24-25 ¶·¸·ÓɃɄɅ·Ĩ)Ýĩ/Y" 1XÓT#𝐴𝐵𝐶𝐷X∠𝐴= 90°X𝐴𝐵= 3X𝐵𝐶= 12X
𝐶𝐷= 13X𝐷𝐴= 4´
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(1)'ÓT#𝐴𝐵𝐶𝐷0$%£
(2)Y" 2Xr𝐴uɆɇōÐXr𝐴𝐵À𝐴𝐷a¿7(u𝑥ŽÀ𝑦ŽſƢ78Ɇɇ€XÐ𝑃¿𝑦ŽĪXÆ𝑆△𝑃𝐵𝐷=
1
6
𝑆四边形𝐴𝐵𝐶𝐷X'𝑃0Ɇɇ´
ËÌ(1)36
(2)(0,0)Ɣ(0,8)
[ÍéÚÛÀ0PXɆɇ›"#X98#0$%UŸêÐXBËé0à
bH¤¡XŪÊa'Fnji0=ƜX,#ćR0xyBË´
-1/ÄÅ𝐵𝐷XŠ‚šr'𝐵𝐷0*XĜĝŠ‚0Pšr56 △𝐵𝐷𝐶0#ɈXŶ‘
šrBËé£
-2/ijŠ‚𝑆△𝑃𝐵𝐷=
1
6𝑆四边形𝐴𝐵𝐶𝐷X'Ê𝑃𝐷0*+X슂𝐷Ð0Ɇɇíš'B´
ÕB-1/B
ÄÅ𝐵𝐷X
∵¿ △𝐴𝐵𝐷4X∠𝐷𝐴𝐵= 90°X
∴ 𝐵𝐷2 = 𝐴𝐵2 +𝐴𝐷2 = 32 + 42 = 25X
∴𝐵𝐷= 5X
∵¿ △𝐷𝐵𝐶4X𝐷𝐵2 +𝐵𝐶2 = 52 + 122 = 25 + 144 = 169X𝐶𝐷2 = 132 = 169X
∴ 𝐷𝐵2 +𝐵𝐶2 = 𝐶𝐷2X
∴△𝐷𝐵𝐶b7898#X
∴∠𝐷𝐵𝐶= 90°X
∴ 𝑆四边形𝐴𝐵𝐶𝐷= 𝑆△𝐷𝐴𝐵+ 𝑆△𝐷𝐵𝐶=
1
2 × 3 × 4 +
1
2 × 5 × 12 = 36´
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-2/B
 ∵ 𝑆△𝑃𝐵𝐷=
1
6𝑆四边形𝐴𝐵𝐶𝐷X
∴ 
1
2𝑃𝐷× 𝐴𝐵=
1
6 × 36 = 6X
∴ 
1
2𝑃𝐷× 3 = 6X
∴𝑃𝐷= 4X
∵ 𝐷(0,4)XÐ𝑃¿𝑦ŽĪX
∴𝑃0Ɇɇu(0,0)Ɣ(0,8)´
“† 10-2-24-25 ¶·¸·ɉĮ·½¾/Y"X¿ △𝐴𝐵𝐶4X𝐴𝐵= 𝐴𝐶X𝐵𝐶= 10XÐ𝐷bT𝐴𝐶Ī0
–ÐX𝐵𝐷= 8X𝐶𝐷= 6´
(1)'G
 △𝐵𝐷𝐶b7898#£
(2)'()𝐴𝐷0*´
ËÌ(1)GH•BÍ
(2)
7
3
[Í-1/Š‚𝐵𝐷= 8X𝐶𝐷= 6X𝐵𝐶= 10XŊ…b–Œ”•0,XÖȮG △𝐵𝐷𝐶b7898#£
-2/&-1/Ȯ △𝐴𝐷𝐵b7898#X슂ĞÊV7898#09T€†XćR
𝐴𝐵= 𝐴𝐶= 𝐴𝐷+ 𝐷𝐶Xšrċ€†subV𝐴𝐷0?­XBÊ횴
ÕB-1/B
 ∵𝐵𝐷= 8X𝐶𝐷= 6X𝐵𝐶= 10X
∴𝐵𝐷2 +𝐶𝐷2 = 𝐵𝐶2X
∴ △𝐵𝐷𝐶b7898#´
-2/B
 ∵△𝐵𝐷𝐶b7898#X
∴∠𝐴𝐷𝐵= ∠𝐵𝐷𝐶= 90°X
∴¿Rt △𝐴𝐷𝐵4X𝐵𝐷2 +𝐴𝐷2 = 𝐴𝐵2X
∵𝐴𝐵= 𝐴𝐶X
∴𝐴𝐵= 𝐴𝐷+ 𝐷𝐶= 𝐴𝐷+ 6X
∴82 +𝐴𝐷2 = (𝐴𝐷+ 6)2X
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B𝐴𝐷=
7
3X
Ö𝐴𝐷0*u
7
3´
ÐØéÚÛõN0PXàŪŽ–å98#0<T0>?@UV¢9T0>
?bBé0à´
“† 10-3-24-25 Ĭ·¸·ÓɃoë·½¾/Y"X¿ △𝐴𝐵𝐶4X𝐶𝐷b𝐴𝐵TĪ0ƽXžŸ𝐴𝐶= 9X
𝐵𝐶= 12X𝐴𝐵= 15´
(1)'𝐶𝐷0*£
(2)kÐ A Â∠𝐶𝐴𝐵08>[(Ï𝐵𝐶TVÐ EX' △𝐴𝐸𝐵0$%´
ËÌ(1)
36
5 £
(2)
135
4 ´
[ÍéÚÛõ0PX8>[(0ÑÒXæU98#05@ÑÒX´
-1/ijGH △𝐴𝐵𝐶b7898#į∠𝐴𝐶𝐵= 90°X슂U$%m®‡íš£
-2/kÐ E Â𝐸𝐹⊥𝐴𝐵Ï𝐴𝐵VÐ FXŠ‚8>[(0ÑҍŽ𝐶𝐸= 𝐸𝐹XGH △𝐴𝐶𝐸≌△𝐴𝐹𝐸Xó
𝐶𝐸= 𝐸𝐹= 𝑥XŠ‚'Ê𝑥=
9
2XŠ‚98#$%†®‡íš´
ÕB-1/B
∵𝐴𝐶2 +𝐵𝐶2 = 92 + 122 = 225 = 152 = 𝐴𝐵2X
∴△𝐴𝐵𝐶b7898#į∠𝐴𝐶𝐵= 90°X
∵𝑆△𝐴𝐵𝐶=
1
2 × 𝐴𝐶× 𝐵𝐶=
1
2 × 𝐴𝐵× 𝐶𝐷X
∴𝐶𝐷=
𝐴𝐶×𝐵𝐶
𝐴𝐵
=
9×12
15 =
36
5 £
-2/B
Y"XkÐ E Â𝐸𝐹⊥𝐴𝐵Ï𝐴𝐵VÐ FX
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∵𝐴𝐸>[∠𝐶𝐴𝐵X
∴𝐶𝐸= 𝐸𝐹X
∴△𝐴𝐶𝐸≌△𝐴𝐹𝐸(HL)
∴𝐴𝐶= 𝐴𝐹= 9X
∴𝐵𝐹= 6X
ó𝐶𝐸= 𝐸𝐹= 𝑥XÈ𝐵𝐸= 12−𝑥X
∴𝑥2 + 62 = (12−𝑥)2X
B
𝑥=
9
2X
∴𝑆△𝐴𝐸𝐵=
1
2 × 𝐴𝐵× 𝐸𝐹=
1
2 × 15 ×
9
2 =
135
4 ´
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