如图11,角bad=角cae=90度,
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![如图11,角bad=角cae=90度,](/uploads/image/f/3617836-52-6.jpg?t=%E5%A6%82%E5%9B%BE11%2C%E8%A7%92bad%3D%E8%A7%92cae%3D90%E5%BA%A6%2C)
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
(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,∴
△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
因为全等三角形,所以角BAC=角DAE;所以角BAC-角DAC=角DAE-角DAC;即角BAD=角CAE再答:给好评啊
∵∠BAD=∠CAE∴∠BAD-∠CAD=∠CAE-∠CAD即∠BAC=∠DAE在△BAC和△DAE中{AB=AD{∠BAC=∠DAE{AC=AE∴△BAC≌△DAE(SAS)∴BC=DELZ的图有点
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∽
相等因为旋转后∠CAB=∠EAD如果旋转的角度<∠CAB:∵∠CAE+∠EAB=∠CAB∠BAD+∠EAB=∠CAB∴∠CAE=∠BAD如果旋转角>∠CAB∵∠CAB=∠EAD∠CAE=∠CAB+∠B
第一个应该是求证:△ABE≌△ACD1、证明∵∠BAD=∠CAE=90∴∠CAD=∠CAB+∠BAD=∠CAB+90,∠BAE=∠CAB+∠CAE=∠CAB+90∴∠CAD=∠BAE∵AB=AD,AC
因为AD=AE,AB=AC,∠BAD=∠CAE所以△ADB≌△AEC所以∠ADB=∠AEC,BD=CE因为BD=CE,DE=BC所以四边形BCED是平行四边形所以BD=CE所以∠BDE+∠DEC=18
∵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
∵∠BAD=∠CAE=90∴∠CAD=∠CAB+∠BAD=∠CAB+90,∠BAE=∠CAB+∠CAE=∠CAB+90∴∠CAD=∠BAE∵AB=AD,AC=AE∴△ABE全等于△ACD∴∠BEA=∠
∵AD=AE(已知)∴角ADE=角AEB(等边对等角)∵角BAD=角CAE(已知)∴角BAD+角DAE=角CAE+角DAE(加法法则)即角BAE=角CAD又∵AD=DE,角ADE=角AEB(已证)∴△
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.
∵△ABE≌△ACD(已知)∴BD+DE=CE+DE(全等三角形的性质)又∵BD=BE-DE,CE=CD-DE∴BD=CE(等量代换)∴∠BAD=∠CAD(全等三角形的性质)又∵∠BAD=∠BAE-∠
证明:因为AD=AE所以角ADE=角AED因为角ADE+角ADB=180度角AED+角AEC=180度所以角ADB=角AEC因为AB=AC所以角B=角C因为AB=AC所以三角形BAD和三角形CAE全等
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