等腰三角形經(jīng)典練習(xí)題

1、等腰三角形經(jīng)典練習(xí)題等腰三角形練習(xí)題一、計(jì)算題：1 .如圖，AABCx>AB=AC,BC=BD,AD=DE=EB求/A的度數(shù)設(shè)/AB防x,則/A為2x由8x=180得/A=2x=452 .如圖，CA=CB,DF=DB,AE=AD求/A的度數(shù)設(shè)/A為X,由5x=180得/A=36°3.如圖，ABCxxAB=ACD在BCxx,DHAB于E,DNBC交AC于點(diǎn)F,若/EDF=70,求/AFD的度數(shù)ZAFD=1604 .如圖，AABCxxAB=AC,BC=BD=ED=EA求/A的度數(shù)設(shè)/A為x/A=5 .如圖，AABCxxAB=ACD在BCxx,/BAD=30,在AC上取點(diǎn)E,使AE=

2、AD,求/EDC勺度數(shù)設(shè)/ADE為x/EDCMAED-/C=156 .如圖，zABCxx/C=90°,D為ABxx一點(diǎn)，作DHBC于E,若BE=AC,BD=,DE+BC=1,求/ABC的度數(shù)延長DE到點(diǎn)F,使EF=BC可證得：zAB竽ABFE所以/1=/F由/2+/F=90°,得/1+/F=90在RtADBFxx,BD=,DF=1所以/F=Z1=307 .如圖，ABCxxADf分/BAC若AC=AB+BD求/B:ZC的值在AC上取一點(diǎn)E,使AE=AB可證AB乎AADE所以/B=/AED等腰三角形經(jīng)典練習(xí)題由ac=ab+bD|de=ec,所以/AED=ZC故/B:/C=2:1

3、二、證明題：8 .如圖，zABCxx/ABC,/CAB的平分線交于點(diǎn)P,過點(diǎn)P作DE/ZAB,分別交BGAC于點(diǎn)D.E求證：DE=BD+AE證明PBDffiAPEA是等腰三角形9 .如圖，DEFxx/EDF=2E,FAIDE于點(diǎn)A,問：DRADAE間有什么樣的大小關(guān)系DF+AD=AE在AE上取點(diǎn)B,使AB=AD等腰三角形經(jīng)典練習(xí)題10 .如圖，zABCxx/B=60°,角平分線ADCE交于點(diǎn)O求證：AE+CD=AC在AC上取點(diǎn)F,使AF=AE易證明AO國AAOF,得/AOEMAOF由/B=60°,角平分線ADCE,得/AOC=120所以/AOE=AOFhCOFMCOD=60

4、ftACODizCOF得CF=CtDf以AE+CD=AC11 .如圖，AABCxxAB=AC,/A=100,BD平分/ABC求證：BC=BD+AD延長BD到點(diǎn)E,使BE=BC連結(jié)CE在BC上取點(diǎn)F,使BF=BA易證AB乎AFBD得AD=DF再證CD國ACDF得DE=DF故BE=BC=BD+AD也可:在BC上取點(diǎn)E,使BF=BD連結(jié)DF在BF上取點(diǎn)E,使BF=BA連2DE先證DE=DCg由AAB*4EBD得AD=DEg后證明DE=DF5|3可12 .如圖，ABCxx,AB=AC,的zABC外一點(diǎn)，且/ABDWACD=60求證：CD=AB-BD在AB上取點(diǎn)E,使BE=BD在AC上取點(diǎn)F,使CF=C

5、D得BDE與CDF均為等邊三角形，只需證AD降A(chǔ)AED13 .已知：如圖，AB=AC=BECD為AABCxxABi上的xx線求證：CD=CE延長CDIU點(diǎn)E,使DE=C型結(jié)AE證明AC自ABCE14 .如圖，AABCxx/1=/2,/EDC=BAC求證：BD=ED在CE上取點(diǎn)F,使AB=AF易證AB乎AADF,得BD=DF/B=/AFD由/B+/BAC廿C=/DEC廿EDC廿C=180所以/B=/DEC所以/DECMAFD所以DE=DF故BD=ED15 .如圖，zABCxxAB=AC,BE=CF,EFBC于點(diǎn)G求證：EG=FG16 .如圖，zABCxx/ABC=ZC,AD是BC邊上的高，B到點(diǎn)

6、E,使BE=BD求證：AF=FC17 .如圖，zABCxxAB=AC,A冷口BE兩條高，交于點(diǎn)H,且AE=BE求證：AH=2BD由4AH自4BCE得BC=AH18 .如圖，AABCxxAB=AC,/BAC=90,BD=AB,/ABD=30求證：AD=DC作AF,BD于F,DE±AC于E可證得/DAF=DAE=15,所以AAD自AADF得AF=AE,由AB=2AF=2AE=AC,以AE=EC,因此DE是AC的中垂線，所以AD=DC等腰三角形經(jīng)典練習(xí)題19 .如圖，等邊ABCxx分別延長BA至點(diǎn)E,延長BC至點(diǎn)D,使AE=BD求證：EC=ED延長BD到點(diǎn)F,使DF=BC,可得等邊BEF,只需證明BC自4FDE即可20 .如圖，四邊形ABCDxx/BAD廿BCD=180,ADBC白xx交于點(diǎn)F,DCAB的xx交于點(diǎn)E,/E、/