如图AD·AB=AE·AC.求证OC·OD=OB·OE
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/19 07:46:23
![如图AD·AB=AE·AC.求证OC·OD=OB·OE](/uploads/image/f/3627713-65-3.jpg?t=%E5%A6%82%E5%9B%BEAD%C2%B7AB%3DAE%C2%B7AC.%E6%B1%82%E8%AF%81OC%C2%B7OD%3DOB%C2%B7OE)
∵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
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=
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
∵AD/AB=AE/AC,∴AD/AB=AE/AC,∴15/40=AE/28,AE=10.5.
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
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
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
∠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=2/3则AB/DB=5/3AE/EC=2/3则AE/AC=5/3
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
∵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
∵∠BAC=∠DAE=α∴∠BAE=∠CAD∵AB=AC,AD=AE,∠BAE=∠CAD∴△ABE≌△ACD(SAS)∴S△ABE=S△ACD,AB=AC,∠AEB=∠ADC∴∠DOE=180°-(∠
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?!
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
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
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
AB:BD=(AD+BD):BD=7:2AE:AC=AE:(AE+EC)=5:7
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=
∵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