一道关于三角形的题

显示全部楼层 · 2012-7-31 18:46:35

AB=AC,AD=AE,BD=CE, 推出三角形全等所以∠DAB=∠EAC 所以∠DAB-∠DAC=∠EAC-∠DAC 所以 ∠CAB=∠EAD∠B=∠C 再加上对顶角相等所以 ∠CAB=∠BOC所以结论成立...

千问 · 2012-7-31 18:46:35

令AD,EO交于G∵AB=AC,AD=AE,BD=CE∴△BAD≌△CAE∴∠D=∠E,∠BAD=∠CAE∵∠DGO=∠EFA∴△DOG∽△EAG∴∠DOG=∠EAD∵∠DOG=∠COF∴∠COB=∠DAE∵,∠BAD=∠CAE∴∠BAD-∠CAD=∠CAE-∠CAD即∠BAC=∠DAE∴∠CAB=∠EAD=∠...

千问 · 2012-7-31 18:46:35

AB=AC,AD=AE,BD=CE三角形ACE全等三角形ABD角CAE=角BAD 同时减去角CAD 即角DAE=角BAC角ABD=角ACE 角AFB=角CFO 所以角BAC=角BOC所以∠CAB=∠EAD=∠BOC...

千问 · 2012-7-31 18:46:35

∵AB=AC,AD=AE,BD=CE∴三角形AEC≌三角形ADB∴∠B=∠C∴∠EAC＝∠BAD又∵∠EAD+∠DAC=∠BAC+∠DAC∴∠EAD=∠BAC∵∠BAC+∠B=∠BFC
∠BOC+∠C=∠BFC∴∠BAC=∠BOC∴∠CAB=∠EAD=∠BOC...

千问 · 2012-7-31 18:46:35

证明：∵AB=AC,AD=AE,BD=CE∴△ABD≌△ACE∴∠B=∠C,∠D=∠E,∠BAD=∠CAE又∠BAD∠CAB+∠CAD=∠CAE=∠CAD+∠EAD即∠CAB=∠EAD又∠AFB与∠OFC为对顶角，∠B=∠C∴∠BOC=180°-∠C-∠OFC ∠BAC=180°-∠B-∠AFB即∠BOC=∠BAC∴∠...