如图,AD评分∠CAE,CF AD,∠1=80°,∠2=多少度
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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
∠dae=∠dac+∠cae又∵∠bad=∠cae∴∠bac=∠dae,∠abc=∠ade∴三角形△abc和△ade两个角相等∴△abc∽△ade∴ab/ad=ac/ae(相似三角形相等角的两夹边成比
(1)三角形外角的性质得:∠D=∠DAE-∠B=55°-30°=25°;(2)∵AD是△ABC的外角∠CAE的平分线,∴∠CAD=∠DAE=55°,∴∠ACD=180°-∠D-∠CAD=180°-25
证明:∵BD平分∠ABC∴∠ABD=∠DBC∵AB=AC∴∠ABC=∠ACB∵∠CAE=∠ABC+∠ACB=2∠ACB∴∠CAD=½∠CAE=∠ACB∴AD//BC∴∠D=∠DBC=∠ABD
∵AD∥BC,∴∠1=∠B,∠2=∠C,∵∠1=∠2,∴∠B=∠C,∴AB=AC.
∵AD∥BC∴∠1等于∠ABC∠2=∠ACB∵AD平分∠EAC∴∠1=∠2∴∠ABC=∠ACB∴△ABC为等腰三角形
EF垂直平分AD所以AE=ED所以在三角形EAD中,∠EDA=∠EAD又∠EAD=∠EAC+∠CAD,∠EDC=∠B+∠DAB所以∠EAC+∠CAD=∠B+∠DAB又AD平分∠BAC所以∠DAB=∠C
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
△ABC与△ADE相似,∠ADE=∠ABC,∠AED=∠ACB,∠DAE=∠BAC,∵BAD=BAC-DAC,CAE=DAE-DAC∴BAD=CAE∴△ABD与△ACE相似(两边夹一角)∴∠ABd=∠
延长CB至F使BF=DE连接AF∠ABC+∠AED=180°所以∠AED=∠ABF又AB=AEBF=DE△ABF≌△AEDAD=AF,∠F=∠ADE连接ACCF=BC+DE=CDAC=ACAD=AF△
∵∠B=30°,∠ACD=70°∴∠CAB=80°又∵AE平分∠BAC∴∠CAE=1/2∠BAC=1/2*80°=40°
∵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.
1、延长BP和AD交于E∵AP平分∠DAB BD平分∠ABC ∴∠PAB=1/2∠DAB,∠PBA=1/2∠ABCAD∥BC即∠DBA+∠ABC=180°∴∠PAB+∠PBA=90
证明:∵AD平分∠CAE,∴∠EAD=∠CAD,∵AD∥BC,∴∠EAD=∠B,∠CAD=∠C,∴∠B=∠C,∴AB=AC.故△ABC是等腰三角形.
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
在AB上截取AF=AC,连接DF∵AD平分∠BAC∴∠CAD=∠BAD∵AD=AD,AC=AF∴⊿ACD≌⊿AFD﹙SAS﹚∴∠C=∠AFD∵AB=2AC=AF+BF∴BF=AF∵AD=BD∴DF⊥A