如图在△ABC和△ade中AB

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如图,在△ABC和△ADE中,AB=AC,AD=AE,∠BAC=∠DAE=90°.①求证:C

1∠CAD=∠DABCD=ABAE=AD△ACD≌△ABDCE=BD2由上题全等得∠ACE=∠ABD所以∠ACB+∠ABC=∠ECB+∠DBC所以∠COB=∠CAB=90°O为CE,BD交点再答:虽然

如图,在△ABC和△ADE中,AB=AC,AD=AE,∠BAC=∠DAE,求证:△ABD≌△ACE.

证明:∵∠BAC=∠DAE,…(3分)∴∠BAC+∠CAD=∠DAE+∠CAD,即∠EAC=∠DAB,…(4分)在△AEC和△ADB中AD=AE∠DAB=∠EACAB=AC,∴△AEC≌△ADB(SA

如图,在三角形abc和三角形ade中,∠bad=∠cae,∠abc=∠ade,求证,ab比ad=ac比ae

∠dae=∠dac+∠cae又∵∠bad=∠cae∴∠bac=∠dae,∠abc=∠ade∴三角形△abc和△ade两个角相等∴△abc∽△ade∴ab/ad=ac/ae(相似三角形相等角的两夹边成比

如图,在△ABC和△ADE中,AC=AB,AE=AD,∠CAB=∠EAD.求证CE=BD

证明:∵∠CAB=∠EAD∠CAE=∠CAB-∠EAB∠BAD=∠EAD-∠EAB∴∠CAE=∠BAD又∵AC=ABAE=AD∴△CAE≌△BAD∴CE=BD

如图,在△ABC和△ADE中,∠BAD=∠CAE,∠ABC=∠ADE.

(1)△ABC∽△ADE,△ABD∽△ACE(2分)(2)①证△ABC∽△ADE,∵∠BAD=∠CAE,∠BAD+∠DAC=∠CAE+∠DAC,即∠BAC=∠DAE.(4分)又∵∠ABC=∠ADE,∴

如图,在△ABC中,BD,CE分别是AC,AB上的高,BD,CE相交于点F,△ABC与△ADE相似吗?

BC中点O为圆心BO为半径作圆,ED在圆上∵BD⊥AC,CE⊥AB,∴∠EBD=∠DCE,∠DEC=∠DBC,∠ADE=∠DEC+∠DCE=∠DBC+∠EBD=∠ABC,又∠A为公共角,∴△ADE∽△

如图,在△ABC和△ADE中,AC=AB,AE=AD,∠CAB=∠EAD=90°

第一问三角形AEC和ADB全等这个很简单AE=ADAC=AB而且角EAC=90+BAE=角BAD所以EC=DB第二问设ABCE交于PECDB交于O看三角形ACP和BOP根据上一问全等角ACP=角OBP

如图,在△ABC中,DE∥BC,EF∥AB,△ADE和△EFC的面积分别为4和9,求△ABC的面积.

∵DE∥BC,EF∥AB∴∠C=∠AED,∠FEC=∠A(4分)∴△EFC∽△ADE(5分)而S△ADE=4,S△EFC=9∴(ECAE)2=94(6分)∴ECAE=32∴ECAC=35(8分)∴S△

如图,在△ABC和△ADE中,∠BAD=∠CAE,∠ABC=∠ADE

相似因为∠BAD=∠CAE,所以∠BAC=∠DAE又因为∠ABC=∠ADE所以△ABC∽△ADE所以AD/AE=AB/AC在△ABD和△ACE中AD/AE=AB/AC,∠BAD=∠CAE所以△ABD∽

如图,在△ABC中,DE∥BC,EF∥AB,求证:△ADE∽△EFC.

证明:∵DE∥BC,∴DE∥FC,∴∠AED=∠C.又∵EF∥AB,∴EF∥AD,∴∠A=∠FEC.∴△ADE∽△EFC.

已知,如图8,在△ABC中,AB=AC,DE‖BC交AB于D,交AC于E.试说明:△ADE是等腰三角形.

因为AB=AC所以∠B=∠C又因为DE//BC所以∠ADE=∠B∠AED=∠C所以∠ADE=∠AED所以等腰三角形ADE

如图,在△ABC中,BD⊥AC,CE⊥AB.求证:△ADE∽△ABC.

证明:∵BD⊥AC,CE⊥AB,∴∠ADB=∠AEC=90°,∵∠A=∠A,∴△ABD∽△ACE,∴ADAE=ABAC,∴ADAB=AEAC,∴△ADE∽△ABC.

如图,在△ABC和△ADE中,AB=AC,AD=AE,若BD=CE,求证∠ABD=∠ACE

证明:在△ABD与△ACE中,∵AB=ACBD=CEAD=AE∴△ABD≌△ACE(SSS)∴∠ABD=∠ACE

如图,在△ABC中BC=AC,CD⊥AB,DE∥BC,试说明△ADE和△CED都是等腰三角形.

∵BC=AC,∴∠A=∠B,∵DE∥BC,∴∠EDA=∠B,∴∠A=∠EDA,∴EA=ED,∴△ADE是等腰三角形,∵DE∥BC,∴∠EDC=∠DCB,∵BC=AC,CD⊥AB,∴CD平分∠ACB,∴

如图,在△ABC和△ADE中,AB=AC,AD=AE,∠BAC=∠DAE=90°.

①∵AB=ACAD=AE∠BAD=∠CAE=90∴△ABD≌△ACEBD=CE∠EBF=∠ACE延长BD交CE于F∠BFC=∠BEF+∠EBF=∠BEF+∠ACE=90∴BD与CE有长度相等、位置垂直

如图,△ABC中,AB=AC,点D在AB上,点E在AC上,BE和CD相交于点O.若∠EBC=∠DCB,则△ADE是什么形

等腰三角形易证,△DCB和,△EBC全等,所以BD=EC,因为AB=AC,所以AD=AE所以是等腰三角形

如图,在△ABC和△ADE中,点E在BC边上,∠BAC=∠DAE,∠B=∠D,AB=AD.

(1)证明:在△ABC和△ADE中∠BAC=∠DAEAB=AD∠B=∠D,∴△ABC≌△ADE;(2)∵△ABC≌△ADE,∴AC=AE,∴∠C=∠AEC=75°,∴∠CAE=180°-∠C-∠AEC

如图,在△ABC中,AB>AC,过AC上一点D作直线DE,交AB于E,使△ADE和△ABC相似,这样的直线可作_____

如图,①作DE∥BC,则△ADE∽△ACB;②作AE:AC=AD:AB,因为∠A=∠A,所以△ADE∽△ABC;所以这样的直线有两条.

1.如图,在△ABC和△ADE中,∠BAD=∠CAE,∠ABC=∠ADE.

(1)∵∠BAD=∠CAE,∠DAC=∠DAC.∴∠BAC=∠DAE,又∵∠ABC=∠ADE.∴△ABC∽△ADE,(AA)∴AB:AC=AD:AE°∵∠BAD=∠CAE∴△ABD∽ACE(SAS)(