TITLE: Constructions | Olympiad Geometry
QUESTION [6 upvotes]: Let $ABCD$ be a parallelogram with no angle equal to $60^\circ$. Find all pairs of points $(E,F)$ such that $AE = BE$, $BF = CF$, and triangle $DEF$ is equilateral.

So far after making a couple of diagrams, I've figured out that $DEF$ is equilateral only when $ABE$ and $BFC$ are equilateral as well. But I don't know how to prove it. 
We could use similarity or rotation somewhere here, but I don't see how. 
We know that the locus of points E and F lie on the perpendicular bisectors of AB and BC respectively.

REPLY [2 votes]: If $\Delta ABE$ and $\Delta BFC$ are equilaterals, so since to rotate a vector in the plane it's to rotate this vector around his  tail, we obtain:
$$R^{60^{\circ}}\left(\overrightarrow{DE}\right)=R^{60^{\circ}}\left(\overrightarrow{DA}+\overrightarrow{AE}\right)=R^{60^{\circ}}\left(\overrightarrow{CB}+\overrightarrow{AE}\right)=\overrightarrow{CF}+\overrightarrow{AB}=\overrightarrow{DC}+\overrightarrow{CF}=\overrightarrow{DF},$$
which says that $\Delta DFE$ is an equilateral triangle.