We have $\angle FAD = \angle BAD - \angle BAF = 90^\circ - 60^\circ = 30^\circ$. Since $AF = AB = AD$, triangle $AFD$ is isosceles, which means that $\angle ADF = \angle AFD = \frac{180^\circ - \angle FAD}{2} = 75^\circ$ and $\angle CDF = \angle CDA - \angle ADF = 90^\circ - 75^\circ = 15^\circ$. By symmetry, we also have $\angle ADE = 15^\circ$, thus $\angle EDF = 90^\circ - \angle ADE - \angle CDF = 60^\circ$.

Again by symmetry (with respect to the diagonal $BD$), $DE = DF$, so $DEF$ is an isosceles triangle with an angle of $60^\circ$. Therefore, $DEF$ is indeed an equilateral triangle.

Let $G$ and $H$ be the midpoints of $AB$ and $CD$ respectively, and let $M$ be the centre of the square. We denote the side length of the square and the two equilateral triangles by $a$. By Pythagoras' Theorem,
$$GF^2 = AF^2 - AG^2 = a^2 - \left(\frac{a}{2}\right)^2 = \frac{3a^2}{4},$$
so $GF = \frac{\sqrt{3}a}{2}$. Next we find $FH = GH - GF = a - \frac{\sqrt{3}a}{2} = \frac{(2-\sqrt{3})a}{2}$, $FM = GF - GM = \frac{\sqrt{3}a}{2} - \frac{a}{2} = \frac{(\sqrt{3}-1)a}{2}$ and by symmetry $EM = FM = \frac{(\sqrt{3}-1)a}{2}$.

Applying Pythagoras' Theorem again, we obtain
$$DF^2 = DH^2 + FH^2 = \left(\frac{a}{2}\right)^2 + \left(\frac{(2-\sqrt{3})a}{2}\right)^2 = a^2 \left(\frac{1}{4} + \frac{4-4\sqrt{3}+3}{4}\right) = a^2(2-\sqrt{3})$$
and
$$EF^2 = EM^2 + FM^2 = 2 \left( \frac{(\sqrt{3}-1)a}{2} \right)^2 = 2a^2 \cdot \frac{3-2\sqrt{3}+1}{4} = a^2(2-\sqrt{3}).$$
Thus $DF = EF$, and by symmetry $DE = EF$. This means that $DEF$ is an equilateral triangle.