[asy][asy] draw((0,0)--(0,4)--(4,4)--(4,0)--cycle); label("$A$",(0,4),NW); label("$B$",(4,4),NE); label("$C$",(4,0),SE); label("$D$",(0,0),SW); label("$F$",(2,4-2*sqrt(3)),S); draw((0,4)--(2,4-2*sqrt(3))--(4,4)); label("$E$",(4-2*sqrt(3),2),W); draw((4,4)--(4-2*sqrt(3),2)--(4,0)); [/asy][/asy]

We use coordinates. Let $A=(0,a),B=(a,a),C=(a,0)$ and $D=(0,0)$. Since $\triangle ABF$ and $\triangle BCE$ are equilateral, we find $E=(a-\tfrac{a\sqrt{3}}{2},\tfrac{a}{2})$ and $F=(\tfrac{a}{2},a-\tfrac{a\sqrt{3}}{2})$. Thus, $DF=DE = \sqrt{\frac{a^2}{4} + \frac{a^2}{4} ( 7-4\sqrt{3})} = a\sqrt{ 2-\sqrt{3}} = \frac{a ( \sqrt{6}-\sqrt{2} )}{2}$. Likewise, $EF = \frac{a(\sqrt{6}-\sqrt{2} ) }{2}$. Thus, $DE=EF=FD$, and $\triangle DEF$ is equilateral, as desired.

Since $\triangle BEC$ is equilateral, and $ABCD$ is a square, we have $EC=CD$. Likewise, $EB=BA$. Also, $\angle EBA = \angle ECD = 30^\circ$, so $\angle AEB = \angle DEC = 75^\circ$. Thus, $\angle DEA = 360^\circ - 60^\circ -75^\circ - 75^\circ = 150^\circ$. Clearly, $AE=DE$, so $\angle ADE = 15^\circ$, and, by the same logic, $\angle CDF = 15^\circ$. Thus, $\angle EDF = 90^\circ - 15^\circ - 15^\circ = 60^\circ$. However, $DE=DF \implies \angle DEF = \angle DFE = 60^\circ$, so $\triangle DEF$ is equilateral.