如图,已知∠BAD=∠CAE,∠ADE=∠AED,BD=CE,求证:AB=AC
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楼主硬盘上的图我们是无法看到的,估计是这样的是和图,解答在图上:
因为∠CAE=∠BAD所以∠CAB=∠EAD因为AB=AD,∠CAB=∠EAD,AC=AE(边角边原则)所以△EAD≌△CAB
证明:∵∠BAD=∠CAE∴∠BAD+∠DAC=∠CAE+∠DAC即∠BAC=∠DAE又AB=AD,AC=AE∴△BAC≌△DAE(SAS)∴∠C=∠E
因为,在△ABC和△ADE中,AB=AD,∠BAC=∠BAD+∠CAD=∠CAE+∠CAD=∠DAE,AC=AE,所以,△ABC≌△ADE,可得:∠ABC=∠ADE,所以,∠BOD=180°-∠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,∴
第一题,因为AD=AE所以三角形DAE是等腰三角形,于是,∠ADE=∠AED又因为∠ADE=∠B+∠BAD∠AED=∠C+∠CAE而,已知∠BAD=∠CAE所以∠B=∠C从而三角形BAC是等腰三角形因
△ABD∽△ACE你已经证明△ABC∽△ADE那么得AB/AC=AD/AE∠BAD=∠CAE△ABD∽△ACE(边角边)
AB=AC证明:∵∠BAE=∠BAD+∠DAE,∠CAD=∠CAE+∠DAE,∠BAD=∠CAE∴∠BAE=∠CAD∵AD=AE,AB=AC∴△ABE≌△ACD(SAS)
证明:∵∠BAE=∠BAD+∠DAE,∠CAD=∠CAE+∠DAE,∠BAD=∠CAE∴∠BAE=∠CAD∵AD=AE∴∠ADC=∠AEB∴△ABE≌△ACD(ASA)数学辅导团解答了你的提问,理解请
证明:因为AD=AE所以∠ADE=∠AED所以∠ADB=∠AEC又因为∠BAD=∠CAE所以△ABC≌△ABC所以AB=AC
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∽
因为三角形全等,所以角bac等于角dae所以角bad等于角cae
20°因为△ABC≌△ADE,所以∠BAC=∠DAE∠BAD=∠BAC-∠DAC∠CAE=∠DAE-∠DAC=20
∵AB=AC,AD=AE,∠BAD=∠CAE∴△ABD≌ACE,∠ADE=∠AED∴BD=CE,∠ADB=∠AEC∴∠BDE=∠CED∵DE=BC∴四边形BCED是平行四边形∴BD∥CE∴∠BDE+∠
∵AB/AD=BC/DE=AC/AE,∴△ADE∽△ABC,∴∠BAC=∠DAE,∴∠BAC-∠DAC=∠DAE-∠DAC,∴∠BAD=∠CAE.
利用相似三角形的性质做:证明:因为∠BAD=∠CAE,又因为,∠DAC=∠DAC,所以,∠BAD+∠DAC=∠CAE+∠DAC,即∠BAC=∠DAE,又根据题意知道:AB=AD,AC=AE,由相似三角
楼主你好∵AB分之AE=BC分之ED=AC分之AD∴△ABC∽△ADE,∴∠BAC=∠DAE,∴∠BAD=∠CAE.满意请点击屏幕下方“选为满意回答”,谢谢.
证明:作AF⊥BC于F,∵AD=AE∴∠FAD=∠FAE(三线合一)又∵∠BAD=∠CAE∴∠FAD+∠BAD=∠FAE+∠CAE即∠BAF=∠CAF又∵AF=AF,∠BFA=∠CFA=90°∴△BA
因为AB/AD=BC/DE=AC/AE所以三角形ABC相似三角形ADE所以角BAC=角DAE又因为角BAC=角BAD+角DAC,角DAE=角CAE+角DAC所以角BAD=角CAE
(1)∵∠BAD=∠CAE,∠DAC=∠DAC.∴∠BAC=∠DAE,又∵∠ABC=∠ADE.∴△ABC∽△ADE,(AA)∴AB:AC=AD:AE°∵∠BAD=∠CAE∴△ABD∽ACE(SAS)(