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Let $BCE$ and $ECF$ be the equilateral triangles such that the points $A$ and $E$ are at the same side of the line $BC$, and $B$ and $F$ at different sides of the line $CE$. We will assume that $E$ and $A$ are different points, otherwise the statement holds.



Since $\angle FBC=30^{\circ }$, and then $\angle FBC=\angle DBC$, we conclude that $B$, $F$ and $D$ are collinear. Triangles $DFC$ and $DFE$ are congruent (SAS), so $|DE|=|DC|=|DA|$. Since $AE$ is perpendicular to $BC$ and then also to $AD$, from the right triangle $DAE$ we see that $|DE|>|DA|$, which is a contradiction.