AB=AD,AC=AE,BC=DE.求证∠BAD=∠CAE
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/26 20:40:42
![AB=AD,AC=AE,BC=DE.求证∠BAD=∠CAE](/uploads/image/z/6142727-47-7.jpg?t=AB%3DAD%2CAC%3DAE%2CBC%3DDE.%E6%B1%82%E8%AF%81%E2%88%A0BAD%3D%E2%88%A0CAE)
xۊP_%cOI&}هduqRrm2s#"*T]J1M@HV5Ӗv]sݠӘy^|>m/p:MqjVl&G|ȳ̭%{٩tkq5Mވ4YCD=n4NO `fщDBш)q`)! ͈AFmHp!H`$lrE: D؍VB{)+C0 Xض)E.3m5$Ƴ|ll2,.?=6y-\|ʹrVrp6=df)?EsU_WY>tJG\ N1Y-+jAL,`"D28R!bA7D"GYCǭ^2h(@bC/7V˚=+Տh=7α^+Ot%Y.Y%o_?/_IDm@\yBUZx{4
AB=AD,AC=AE,BC=DE.求证∠BAD=∠CAE
AB=AD,AC=AE,BC=DE.求证∠BAD=∠CAE
AB=AD,AC=AE,BC=DE.求证∠BAD=∠CAE
因为AB=AD,AC=AE,BC=DE 所以三角形BAC与三角形DAE全等 所以∠BAC=∠DAE
所以∠BAC-∠DAC=∠DAE-∠DAC 所以∠BAD=∠CAE
标准过程见图
3边相同 三角形全等 ∠BAD=∠BAC+∠CAD=∠CAD+∠DAE=∠CAE
∵AB=AC,AC=AE,BC=DE
∴△ABC≌△ADE
∴∠BAC=∠DAE
又∵∠BAD+∠DAC=∠DAC+∠EAC
∴∠BAD=∠CAE
已知:如图,AC/AD=AB/DE=BC/AE.求证AB=AE
已知,如图,AD/AC=DE/AB=AE/BC,求证AB=AE
已知,如图,AD:AC=DE:AB=AE:BC 求证:AB=AE
初一的1题几何:已知AB=AC,AD=AE,求证:DE平行于BC已知AB=AC,AD=AE,求证:DE // BC
如图AB=AE,AC=AD,ED=BC,AE⊥AB 求证BC⊥DE
AB=CD.DE垂直AC.BF垂直AC.AE=CF.求证BC=AD.BC∥AD
AB=AD,AC=AE,BC=DE.求证∠BAD=∠CAE
如图,AB/AD=BC/DE=AC/AE,求证:∠BAD=∠CAE
如图AB/AD=BC/DE=AC/AE,求证∠ABD=∠ACE
如图,在△ABC中,DE∥BC,AD/BD=AE/EC,求证:AD/AB=AE/AC
如图,AB=AD,AC=AE,角BAD=角CAE,连接BC,DE,求证:BC=DE.
如图,AB=AD,AC=AE,∠BAD=∠CAE,连接BC、DE,求证:bc=de
如图,DE//BC,请问AD:AC=AE:AB?AD:AE=AB:AC?
如图,AB垂直于AC,AD垂直于AE,AB=AD,BC=DE.求证:AM=AN
AB⊥AC,AD⊥AE,AB=AD,BC=DE,求证AM=AN图片图片
已知:如图,AB/AD=AC/AE=BC/DE,求证,AB乘EC=AC乘BD
如图,AB=AE,AB⊥AE,AD=AC,AD⊥AC,M为BC中点,求证:DE=2AM
如图,在△ABC中,AD⊥AB,AD=AB,AE⊥AC,AE=AC,M为BC中点 求证:2AM=DE