如图∠BAO=∠CAE=90°,AB=AD,AE=AC,AF垂直CE,垂足为F
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∠AFD=∠AFE.理由:过A作AM⊥DC于M,AN⊥BE于N.∵∠BAD=∠CAE=90°,∴∠BAD+∠BAC=∠CAE+∠BAC,即∠DAC=∠BAE;在△ABE和△ADC中,AB=AD(已知)
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
DE与AB的交点为O.∠D+∠DOA+∠DAO=180,∠B+∠BOE+∠BEO=180..因为∠D=∠B,∠BOE=∠DOA,所以∠DAO=∠BEO..因为∠DEB=∠CAE,所以∠DAB=∠EAC
证明:∵AF平分∠CAE,∴∠CAF=∠EAF,在△ACF和△ADF中∵AC=AD∠CAF=∠EAFAF=AF,∴△ACF≌△ADF(SAS),∴∠ADF=∠ACF,∵∠ACB=90°,∴∠ACF+∠
(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,∴
你的第一题还漏了一个条件,不过应该是从证明△abc和△ade全等来证明bc=de的ae=cd,且ae⊥cd延长ae交cd于f,则只需要证明af⊥cd即可因为ab=cb,eb=db,∠abc=∠cbd,
△ABD∽△ACE你已经证明△ABC∽△ADE那么得AB/AC=AD/AE∠BAD=∠CAE△ABD∽△ACE(边角边)
∵AD∥BC,∴∠1=∠B,∠2=∠C,∵∠1=∠2,∴∠B=∠C,∴AB=AC.
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)
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
∠BAC=∠DAE所以∠CAE=∠BAD再问:等于多少度
∵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.
设角EAD=5x则角CAE=8x因为ED为中垂线所以EA=EB角B=∠EAD=5x因为∠C=90°则∠CAE+∠B=90°解得x=5°则∠CAE=40°在RT△CAE中可得∠CEA=50°
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)(