On the sides $[AB]$ and $[BC]$ of the parallelogram $ABCD$ are constructed the equilateral triangles $ABE$ and $BCF,$ so that the points $D$ and $E$ are on the same side of the line $AB$, and $F$ and $D$ on different sides of the line $BC$. If the points $E,D$ and $F$ are collinear, then prove that $ABCD$ is rhombus.
Observe $\triangle ABC \cong \triangle BEF$. In fact, a $60^\circ$ rotation about $B$ maps the two triangles. This rotation sends $\overline{AC}$ to $\overline{BF}$. But $D \in BF$ so $D$ is in the image of this map, so there exists $D' \in AC$ for which the rotation sends $D'$ to $D$.
But then for such $D'$, $ADD'$ is equilateral. So, $D'$ lies on perp. bisector $l$ of $BD$.
Now, if $ABCD$ were not a rhombus, then intersection of $AC$ and $l$ is $BD \cap AC$, but intersection of diagonals cannot form an equilateral triangle with $B$ and $D$. Thus $ABCD$ is a rhombus.
Remark: Using the rotation observation, the converse follows easily.