Not solution: Proof for $\angle A=120^{\circ}$ Let $G$ be the circumcenter of $AEF$. $\angle BED=120^{\circ}$ $\angle CFD=120^{\circ}$ $\implies\angle EDF=120^{\circ}$ $EF$ perpendicularly bisects $AD$ so $\angle AEF=120^{\circ}\implies \angle EGF=120^{\circ}\implies \angle GEF=30^{\circ}$ so $\angle GDF=150^{\circ}$ and combining with $\angle FDC=30^{\circ}$ makes $G$ on $BC$

Denote by $O$ the circumcenter of $AEF$ and by $O_1$ the circumcenter of $DEF$. $\angle{BAD}=\angle{DAC}=\frac{\angle{BAC}}{2}< \frac{\pi}{2}\implies E, F$ and $A$ on the same side of $BC$. $\angle{EDB} = \frac{\pi-\angle{BED}}{2} = \frac{\pi-2\angle{BAD}}{2} = \frac{\pi-\angle{BAC}}{2} = \frac{\pi-2\angle{DAC}}{2} = \frac{\pi-\angle{DFC}}{2}=\angle{FDC}$ $\implies BC$ - external bisector $\angle{EDF}$ $\implies O$ on the midpoint perpendicular to $EF$ and on external bisector $\angle{EDF}$ $\implies O$ - midpoint of arc $EDF\implies O_{1}O=O_{1}E=O_{1}F$. $\triangle {AEF} = \triangle {DEF} (AE=DE, AF=DF, EF=EF)\implies OE = O_{1}E = O_{1}F = OF$ $ \implies \triangle{EOO_{1}}$ and $ \triangle{FOO_{1}}$ are equilateral. $\implies \angle{EDF}=\angle{EOF} = \angle{EOO_{1}} + \angle{O_{1}OF} = \frac{2\pi}{3} \implies \angle{EDB} = \frac{\pi -\angle{EDF}}{2} =\frac{\pi}{6}\implies\angle{BAC} = \angle{BED}= \pi - 2\angle{BED}=\frac{2\pi}{3}$