Shows that $EFD$ is also equilateral. In the following figure, it is known that ABC is equilateral and that $AE = BD = CF$, shows that $EFD$ is also equilateral.

Attemp:
I got a little lost in proving these triangles to be similar, but if you can do that, it's simple. Call the angles EAB, DBC and FCA x and the angles DBA, FCB and EAC y. (We know they are the same because we assume the triangles to be similar.) We know that x + y + x + y + x + y = 180 °, so x + y = 60 °. DEF is the outer angle of AEB, so DEF = x + y = 60 °. Similarly the other angles are 60 °, thus being DEF equilateral.
I also think this looks like Morley's triangle. But I don't know if DBC is 1/3 ABC:

 A: Let us embed the construction in the complex plane by taking $A=0, B=1, C=\omega=\frac{1+i\sqrt{3}}{2}$.
Given $E=z$, the positions of $D$ and $F$ are fixed via
$$ D = B+|z|\frac{E-B}{|z-1|} = 1+\frac{|z|}{|z-1|}(z-1) $$
$$ F = C+|z|\frac{D-C}{|D-C|} = \omega+\frac{|z|}{|1+|z|(z-1)-\omega|}(1+|z|(z-1)-\omega)$$
and $F$ is a real multiple of $z$, since $F\in AE$. This implies that
$$\omega+\frac{|z|}{|1+|z|(z-1)-\omega|}(1-|z|-\omega)$$
is a real multiple of $z$ too. After a few manipulations, this leads to the fact that $z$ and $\omega(1-z)$ are real multiples of each other, i.e. to the fact that  $\widehat{AEB}=120^\circ$. This gives that $AEB,BDC$ and $CFA$ are similar, hence congruent and such that
$$\widehat{FED}=\widehat{EDF}=\widehat{DFE} $$
QED.
A: A different approach. Let $\alpha=\widehat{BAE},\beta=\widehat{DBC}$ and $\gamma=\widehat{DCA},r=\frac{AE}{AB}$.
By the sine theorem $DC=r\sin\beta\sin(60^\circ-\gamma)$, so the sum of the distances of $D$ from the sides of $ABC$ is given by
$$ s_D = r\sin(60^\circ-\alpha)+r\sin\beta+r\sin\beta\sin(60^\circ-\gamma)\sin\gamma.$$
By Viviani's theorem $s_D=s_E=s_F$, so the following expressions have equal values:
$$\sin(60^\circ-\alpha)+\sin\beta+\sin\beta\sin(60^\circ-\gamma)\sin\gamma$$
$$\sin(60^\circ-\beta)+\sin\gamma+\sin\gamma\sin(60^\circ-\alpha)\sin\alpha$$
$$\sin(60^\circ-\gamma)+\sin\alpha+\sin\alpha\sin(60^\circ-\beta)\sin\beta$$
where $0<\alpha,\beta,\gamma<60^\circ$. It should not be difficult to prove through inequalities / convexity tricks that $\alpha=\beta=\gamma$ is the only chance.
A: COMMENT:
1- Proof by reverse argument:
Consider equilateral triangle DEF and  construct triangle ABC as follows:
Extend EF to a certain value and mark it as A. Extend DE and FD to same value and mark them as B and C respectively. Join point A, B and C. Triangles AFC, AED and CBD are equal for two sides and angle between equal, therefore :
$AB=BC=AC$
That is triangle ABC is equilateral. Now we have a equilateral triangle in which, $AE=BD=FC$ and $DE=FE=DF$, that is triangle DEF is equilateral.
2-Proof by contradiction:
Suppose $DE≠FE≠DF$, then $AE≠EB≠CD$.
Therefore :
$\triangle AEB ≠ \triangle ACF ≠\triangle BCD$
And: $AB≠ BC≠ AC$ which contradicts $AB= BC= AC$.
A: 
EDIT!:
Assume that each side is rotated by $\delta$ and when lines are drawn in the way shown there is an angle $ (60-\delta) $  at $A$ and the angle of reduced amount $\delta$ at $C$ as shown.
In the triangle $CAF$  external angle equals sum of two opposite angles.
$$\angle DFE= \delta + 60-\delta = 60\, deg $$ and the same holds for other two vertices $E,D$ by rotational symmetry, thus creating a smaller equilateral triangle with scaled down ratio $\dfrac{CF}{CA}$ that can be found by Sine Law.
$$ \dfrac{CF}{CA}=\dfrac{\sin (60-\delta)}{\sin 60}= \cos\delta- \sin \delta/\sqrt 3 $$
$$ \dfrac{AF}{CA}=\dfrac{\sin \delta}{\sin 60}$$
scale ratio
$$ \rho= \dfrac{CF-AF}{CA}= f(\delta) $$
which is a single variable monotone function of $\delta.$
It is given that a constant proportion is valid for 
$$ AE:BD:CF = AB: BC : CA $$
or
$$\dfrac{AE}{AB}=\dfrac{BD}{BC}=\dfrac{CF}{CA} = constant = \rho , say$$
When this proposition is shown by converse logic it can be concluded that each line must rotate by a constant amount,say $\delta$, establishing scaling/similarity.
