First note that reflecting the figure around the line $BD$ doesn't change the figure (the equilateral triangle $AEB$ reflects to the triangle $CBF$). This means that $DE$ and $DF$ are symmetric about the line $BD$, which means that $DF = DE$. If we can show that $\angle FDE = 60^\circ$, then triangle $DEF$ is an isosceles triangle with one $60^\circ$ angle, which means it's equilateral.

Now, since $DA = AB = AE$, triangle $ADE$ is isosceles. Next,
$$\angle DAE = 90^\circ - \angle EAB = 90^\circ - 60^\circ = 30^\circ \text{ and so } \angle ADE = \frac{1}{2}(180^\circ - \angle DAE) = \frac{1}{2}(180^\circ - 30^\circ) = 75^\circ.$$
Similarly, $\angle DFC = 75^\circ$.