How to prove $\Delta DEF$ is an equilateral triangle? 
$\Delta ABC$ is an equilateral triangle and $AD = BE = CF$. Prove that $\Delta DEF$ is an equilateral triangle.
 A: Draw circles centered at A, B, and C, with radius AD=BE=CF.
Triangle $\triangle$DEF must lie in the region between these 3 circles, touching each circle with a vertex.
Draw the rotationally symmetric solution as $\triangle$D'E'F'.
(Side exercise: Show it must exist.)
Now suppose $\triangle$DEF $\ne$ $\triangle$D'E'F'.
Up to here, everything I've said is very straightforward, but you might want to stop for a second and see if you can solve it from here.  If you can't, one more hint is hidden in the gray box below.

 DE is either outside $\triangle$D'E'F', or crosses it.
 Either way, can you say something similar about DF?  About EF?

A: Suppose we fix $\ell = AD=BE=CF$, and let $x=\angle{BAD}$, $y=\angle{CBE}$, $z=\angle{ACF}$.
Then we can define $f : [0,\pi/6]\to[0,\pi/6]$ such that $y = f(x)$, $z = f(y)$, and $x = f(z)$. Note that $f$ is strictly decreasing: one good way to see this is by drawing the circles centered at $A,B,C$ with radius $\ell$. But then $f^3(x) = x$ forces $f(x) = x$ (and thus $x=y=z$); otherwise, if WLOG $x<f(x)$, then $$f(x)>f^2(x)\implies f^2(x)<f^3(x)=x\implies x=f^3(x)>f(x),$$ contradicting $x<f(x)$.
Note that this solution assumes that $x,y,z<\pi/6$, but the problem still holds as long as $x,y,z<\pi/3$. In particular, I think you can show that if one of $x,y,z$ is larger than $\pi/6$ (which requires $\ell>1/2$), the other two must be as well, so $f$ is still well-defined. Anyway, all of this is equivalent to the rotation method Matt mentioned in his answer, but I thought it would be good to include this more algebraic perspective.
fedja also has a long post about this kind of problem on AoPS.
