$解:正确.$
$证明:由△ABD≌△ACE得∠ABD=∠ACE, $
$∵∠AGB=∠CGF,$
$∴∠BFC=∠BAC=60°.$
$∴∠BFE=120°.$
$如图,过点A作BD,CF的垂线,$
$垂足分别为M,N.$
$∵△ABD≌△ACE,BD=CE,$
$∴由面积相等可得AM=AN.$
$在\mathrm{Rt}△AFM和\mathrm{Rt}△AFN中,$
$\begin{cases}{AF=AF,}\\{AM=AN,}\end{cases}$
$∴\mathrm{Rt}△AFM≌△\mathrm{Rt}△AFN(\mathrm{HL}).$
$∴∠AFM=∠AFN=60°.$
$∴∠BFC=∠AFB=∠AFE=60°. $