Let $H'$ be a point on $DC$ such that $\angle H'BC=60^{\circ}-\angle FBA$, denote $\angle FBA=a$ and $\angle H'BC=b$. We have $a+b=60^{\circ}$ so, $\angle FBH'=60^{\circ}$, also $\angle FBD=b$ and $\angle DBH=a$. Now, $\angle H'GB=b+60^{\circ}$, in fact, $\angle AFB=60^{\circ}+b$ therefore $AF=DH'$. But $AH=FG$ since $\angle AFG=120^{\circ}$ and $\angle FAG=30^{\circ}$. Thus $DFGH'$ is a parallelogram so $H'=H$. However $DFBH$ is cyclic, hence $\angle FHB=\angle FDB=60^{\circ}=\angle FBH$. Therefore $BFH$ is equilateral. q.e.d.