如图,AB=AC,AD=AE,BD=CE,BD与CE相交于点O.求证:∠CAB=∠EAD=∠BOC

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如图,AB=AC,AD=AE,BD=CE,BD与CE相交于点O.求证:∠CAB=∠EAD=∠BOC

如图,AB=AC,AD=AE,BD=CE,BD与CE相交于点O.求证:∠CAB=∠EAD=∠BOC
如图,AB=AC,AD=AE,BD=CE,BD与CE相交于点O.求证:∠CAB=∠EAD=∠BOC

如图,AB=AC,AD=AE,BD=CE,BD与CE相交于点O.求证:∠CAB=∠EAD=∠BOC
证明:
∵AB=AC,AD=AE,BD=CE
∴△ABD≌△ACE (SSS)
∴∠BAD=∠CAE,∠B=∠C
∵∠CAB=∠BAD-∠CAD,∠EAD=∠CAE-∠CAD
∴∠CAB=∠EAD
∵∠BFC=∠CAB+∠B,∠BFC=∠BOC+∠C
∴∠CAB+∠B=∠BOC+∠C
∴∠CAB=∠BOC
∴∠CAB=∠EAD=∠BOC

∵AB=AC,AD=AE,BD=CE
∴△ACE≌△ABD
∴∠BAD=∠CAE, ∠ABD=∠ACE
∴∠CAB+∠CAD=∠CAD+∠EAD
即∠ CAB=∠EAD
∵∠BFA=∠CFO
∠ABD=∠ACE即∠ABF=∠FCO
∴△ABF∽△OCF(相似)
∴∠FOC=∠FAB
即∠BOC=∠CAB
∴∠CAB=∠EAD=∠BOC