如图AD·AB=AE·AC.求证OC·OD=OB·OE

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如图,已知AD//BC,AD=BC,AE⊥AD,AF⊥AB,AE=AD,AB=AF.求证:AC=EF.

∵AD‖BC,AD=BC∴ADCB为平行四边形∴AD=BC=AE∵AE⊥AD,AF⊥AB∴∠BAF=∠DAE=90度∴∠EAF+∠DAB=∠DAB+∠B∴∠EAF=∠B在△AEF与△BCA中AE=BC

如图,已知AD/AE=DB/EC,AD=15,AB=35,AC=30,求AE

DE为三角形ABC的中线?这怎么可能?你回去课本,看看中线是怎么定义的按题意,没错的话,就是说D在边AB上,E在边AC上.不用相似就不用呗,那就用比例的基本性质来解答,不会扯到三角形相似.AD/AE=

如图 已知AD/DB=AE/EC=3/2,求AB/DB,EC/AC,AB/AD

设AD=3X,则DB=2X,AB=5X,AB/DB=5/2,AB/AD=5/3设AE=3Y,则EC=2Y,EC/AC=2/5

如图在三角形abc中ad/db=ae/ec,AD=15,AB=40,AC=28求AE的长

∵AD/AB=AE/AC,∴AD/AB=AE/AC,∴15/40=AE/28,AE=10.5.

如图,AD⊥AE,AB垂直AC,AD=AE,AB=AC,AD⊥AE ,AB=AC,求证:三角形ABD≌三角形ACE

证明:∵AB⊥AC,AD⊥AE∴∠BAC=∠DAE=90∵∠BAD=∠BAE+∠DAE,∠CAE=∠BAE+∠BAC∴∠BAD=∠CAE∵AB=AC,AD=AE∴△ABD≌△ACE(SAS)数学辅导团

如图,⊙O是△ABC的外接圆,AB=AC,求证:AB²=AE·AD

证明:∵AB=AC∴∠B=∠ACB连接CD,则ABCD四点共圆∴∠ADC+∠B=180º∵∠ACE+∠ACB=180º∴∠ADC=∠ACE又∵∠DAC=∠CAE∴⊿ADC∽⊿ACE

如图,已知△ABC内接于圆O,AE为直径,AD为BC上的高.求证:AB·AC=AE·AD

因为角aeb=角acb因为ae直径AD为BC上的高所以角aeb=角aec=角acb所以三角形abe和adc相似所以AB/AE=AD/AC得AB·AC=AE·AD

如图,AB⊥AC,AD⊥AE,AB=AC,AD=AE.求证BE⊥CD

∠BAC=∠DAE=90度所以∠BAE=∠CAD又AB=AC,AD=AE所以⊿BAE与⊿CAD全等所以∠C=∠B令BE交AC于O则∠BOD=∠C+∠COE=∠B+∠AOB=90度所以BE⊥CD

如图,AD/DB=AE/EC=2/3,求AB/DB、AE/AC的值.

如图,ad/db=2/3则AB/DB=5/3AE/EC=2/3则AE/AC=5/3

如图,三角形ABC中AB=AC、角BAD=30度、AD=AE.求角EDC.

设∠EDC=x,∠B=∠C=y∠AED=∠EDC+∠C=x+y又因为AD=AE,所以∠ADE=∠AED=x+y则∠ADC=∠ADE+∠EDC=2x+y又因为∠ADC=∠B+∠BAD所以2x+y=y+3

如图,A,B,C三点在⊙O上,且AB=AC,弦AE交BC于D,求证:AB²=AD·AE.

∵AB=AC∴∠ACB=∠ABC=∠AEB又∠BAE公共.所以△ABD和△AEB相似即AB/AD=AE/AB即AB²=AD·AE

如图,AB=AC,AD=AE,∠BAC=∠DAE=α,求∠AOE.

∵∠BAC=∠DAE=α∴∠BAE=∠CAD∵AB=AC,AD=AE,∠BAE=∠CAD∴△ABE≌△ACD(SAS)∴S△ABE=S△ACD,AB=AC,∠AEB=∠ADC∴∠DOE=180°-(∠

已知:如图,AE=AC,AB=AD,∠EAB=∠CAD 已知:如图,AE=AC,AB=AD,∠EAB=∠CAD

因为∠EAB=∠CAD所以∠EAB+∠BAD=∠DAC+∠BAD=∠EAD=∠BAC又因为AE=AC,AB=AD,所以有定理两边一角所以:①△ABC全等于△ADE②是什么?!

已知:如图,AD/AB=AE/BC求证:AD/AE=DB/EC和AB/DB=AC/EC

由AD/AB=AE/AC,且夹角∠A是公共角,∴△ADE∽△ABC,即DE∥BC.(1)∵AD/AB=AE/AC∴AB/AD=AC/AEAB/AD-1=AC/AE-1,(AB-AD)/AD=(AC-A

如图△ABC为等腰直角三角形,AB=AC,AD⊥AE且AD=AE

因为△ABC为等腰直角三角形所以∠CAB=90因为AD⊥AE所以∠DAE=90所以∠CAD=∠BAE因为AB=AC,AD=AE所以△ACD与△ABE全等所以BE=CD

如图,已知AD/DB=AE/EC=2/3,求AD/AB和EC/AC

以为大家是神啊,你的图呢!AD/AB=AD/(AD+DB)=2/(2+3)=2/5EC/AC=EC/(AE+EC)=3/(2+3)=3/5

如图,已知ad:bd=ae:ec=5:2,求ab:bd,ae:ac的值

AB:BD=(AD+BD):BD=7:2AE:AC=AE:(AE+EC)=5:7

如图,已知AD/DB=AE/EC,求证 :AD/AB=AE/AC.

已知条件AD/DB=AE/EC取个倒数,BD/AD=EC/AE两边+1,BD/AD+1=EC/AE+1通分(BD+AD)/AD=(EC+AE)/AE也就是AB/AD=AC/AE再取个倒数,AD/AB=

已知:如图,AC/AD=AB/DE=BC/AE.求证AB=AE

∵AC/AD=AB/DE=BC/AE,∴ΔABC∽ΔDEA,∴∠B=∠DEA,∴AB=AE.再问:老师,能详细点吗??再答:三边对应成比例,两个三角形相似,相似三角形的对应角相等,等角对等边。三个步骤