有等腰三角形时常用的辅助线

【MeiWei_81重点借鉴文档】

有等腰三角形时常用的辅助线⑴作顶角的平分线，底边中线，底边高线

例：已知，如图，AB = AC, BDLAC于D,

求证：/ BAC = 2/DBC

证明：（方法一）作Z BAC的平分线AE,交BC于E,则Z 1 = Z 2 = 又v AB = AC 1

-Z BAC 2

••• AEL BC

•••/ 2+Z ACB = 90°

••• BD L AC

•••/ DBCM ACB = 90°

•••/ 2 = / DBC

•••/ BAC = 2 / DBC

（方法二）过A作AEL BC于E （过程略）

（方法三）取BC中点E,连结AE （过程略）

⑵有底边中点时，常作底边中线

例：已知，如图，△ ABC中, AB = AC，D 为BC 中点，DEL AB 于E，DFL AC于F, 求证：DE = DF

证明：连结AD.

v D为BC中点，

••• BD = CD

又v AB =AC

••• AD平分/ BAC

v DEL AB, DFL AC

••• DE = DF

⑶将腰延长一倍，构造直角三角形解题

例：已知，如图，△ ABC中, AB = AC在BA延长线和AC上各取一点E、F,使AE = AF, 求证：EFL BC

证明：延长BE至U N,使AN = AB,连结CNJ则AB = AN = AC •••/ B = / ACB, / ACN = / ANC

vZ B+Z ACB^Z ACNbZ ANC = 180°

••• 2Z BC外2Z ACN = 180°

•••Z BCAbZ ACN = 90°

即Z BCN = 90°

•NCL BC

v AE = AF

•Z AEF = Z AFE

又vZ BAC = Z AEF + Z AFE

Z BAC = Z ACN + Z ANC

•Z BAC =2Z AEF = 2 Z ANC

•Z AEF = Z ANC

•EF// NC

•EFL BC

⑷常过一腰上的某一已知点做另一腰的平行线

例：已知，如图，在△ ABC中，AB = AC, D在AB上，E在AC延长线上，且BD = CE,连结DE 交BC于F

求证：DF = EF

证明：（证法一）过D作DN/ AE,交BC于N,则/ DNB= /

/ NDE = / E，

••• AB = AC，

•••/ B = / ACB

•••/ B = / DNB

••• BD = DN

又••• BD = CE

••• DN = EC

在厶DNF^P^ ECF中

/ 1 = Z 2

/ NDF =Z E

DN = EC

•••△ DNF^A ECF

••• DF = EF

（证法二）过E作EM/ AB交BC延长线于M,则/ EMB =/ B （过程略）⑸常过一腰上的某一已知点做底的平行线

例：已知，

求证：

如图，△ ABC中, AB =AC E在AC上, D在BA延长线上，且AD = AE，连结DE DEL BC

证明：（证法一）过点E作EF// BC交AB于F,贝U

Z AFE =Z B

/ AEF =Z C

••• AB = AC

•••/ B = / C

•••/ AFE =Z AEF

••• AD = AE

•••/ AED =Z ADE

又•••/ AFE^Z AE阡/ AED^Z ADE = 180

••• 2Z AE阡2Z AED = 90°

即Z FED = 90° ••• DEI FE

又v EF/ BC ••• DEI BC

（证法二）过点D作DN/ BC交CA的延长线于N,（过程略）（证法三）过点A作AM/ BC交DE于M,（过程略）

⑹常将等腰三角形转化成特殊的等腰三角形------等边三角形

例：已知，如图，△ ABC 中，AB = AC，Z BAC = 80 ° ,P 为形内一点，若Z PBC = 1C0 Z PCB = 3C0 求Z PAB的度数•

解法一：以AB为一边作等边三角形，连结CE

则Z BAE =Z ABE = 60°

AE = AB = BE

v AB = AC

N D

E

••• AE = AC / ABC =Z ACB •••/ AEC =Z ACE

vZ EAC =Z BAC-Z BAE

=80 0— 60° = 20 0

1

••• Z ACE = (180° — Z EAC)=

2

1

vZ ACB= - (180 °—Z BAC)= 50°

2

•••Z BCE =Z ACE-Z ACB

0 0 0

=80 — 50 = 30

vZ PCB = 30°

•Z PCB = Z BCE

vZ ABC =Z ACB = 5O0, Z ABE = 60。

•Z EBC =Z ABE-Z ABC = 600— 500 =10

vZ PBC = 100

•Z PBC = Z EBC 在^ PBM3 EBC中

Z PBC = Z EBC

BC = BC

ZPCB = Z BCE

•△ PBC^A EBC

•BP = BE v AB = BE

•AB = BP

•Z BAP =Z BPA

vZ ABP =Z ABC-Z PBC = 500— 100 = 40

1

• Z PAB = (1800—Z ABP)= 700

2

解法二：以AC为一边作等边三角形，证法同一。

解法三：以BC为一边作等边三角形△ BCE连结AE,则

EB = EC = BC ,Z BEC =Z EBC = 600 v EB = EC

•E在BC的中垂线上同理A在BC的中垂线上

•EA所在的直线是BC的中垂线

•EAL BC

1

Z AEB = - Z BEC = 300 = Z PCB

2

由解法一知：Z ABC = 500

•Z ABE = Z EBC-Z ABC = 100 = Z PBC

•• Z ABE =Z PBC,BE = BC, Z AEB =Z PCB ••△ ABE^A PBC

•• AB = BP •/ BAP =Z BPA

•• Z ABP =ZABC-Z PBC = 500— 100 = 40

800 E