Let $O$ be the circumcenter of triangle $AEF$ and denote $\alpha = \angle BAC$. Since $\angle BAD$ and $\angle CAD$ are acute (Fig. 2), points $E$, $F$ lie in the half-plane $BCA$ and the Inscribed angle theorem yields
$$\angle BED = 2 \cdot \angle BAD = \alpha = 2 \cdot \angle DAC = \angle DFC.$$

Fig. 2

The isosceles triangles $BED$ and $DFC$ are thus similar and we easily compute that $\angle EDF = \alpha$ and that $BC$ is the external $D$-angle bisector in triangle $DEF$.

Point $O$ lies on $BC$ and on the perpendicular bisector of $EF$. Framed with respect to triangle $DEF$, it lies on the external $D$-angle bisector and on the perpendicular bisector of the opposite side $EF$. Thus it is the midpoint of arc $EDF$ and $\angle EOF = \angle EDF = \alpha$.
Quadrilateral $AEDF$ is a kite, hence $\angle EAF = \alpha$. Moreover, line $EF$ separates points $A$ and $O$, thus the Inscribed angle theorem implies that the size of the reflex angle $EOF$ is twice the size of the convex angle $EAF$. This yields $360^\circ - \alpha = 2 \cdot \alpha$ and $\alpha = 120^\circ$.