No, it only follows that$\angle BAC=60$. It is well-known that the circumcircle of $\triangle AFE$ and $BFEC$ are orthogonal. So, $ \angle DFE =\angle DEF=\angle FAE $, hence $ \angle FDE=180-2 \angle BAC$. Therefore $ \angle FDE=60\Longleftrightarrow\angle BAC=60$

Or, obvious, $ED=EF=\frac{a}{2}$ ( medians in right-angled triangles), We have $\triangle{AEF}$ and $\triangle{ABC}$ similar, so $\frac{EF}{a}=\frac{AE}{AB}=cos\widehat{BAC}$, but $EF=\frac{a}{2}$, so $ cos\widehat{BAC}=\frac{1}{2}$, so $\widehat{BAC}=60$. It's indeed easy to prove that $\widehat{BAC}=60$ is enough for $\triangle{DEF}$ to be equilateral.