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$ABC$ and $ADE$ are equilateral triangles such that $B-A-E$. If $M$ and $N$ are midpoints of $EC$ and $BD$, prove that $AMN$ is an equilateral triangle.

I assumed that $D$ and $C$ lie on the same side of the line $BAE$ and proved it with vector algebra, but the question hints that we use some sort of a rotation-transformation to solve it. Any ideas?

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$1)$ $\Delta DAB \equiv \Delta EAC$ because $EA=DA$, $AC=AB$ and $\angle EAC= \angle DAB$ and then $EC=BD$

$2)$ Once $EM=DN$ (because $EM=EC/2$ and $DN = BD/2$) and $\Delta DAB \equiv \Delta EAC$, we also can conclude that $\Delta EAM \equiv \Delta DAN$ and then $AM=AN$.

$3)$ Once $\Delta EAM \equiv \Delta DAN$ then

$$\angle EAM =\angle DAN \Rightarrow 60º+\angle DAM= \angle DAM +\angle MAN \Rightarrow \angle MAN=60º$$

So we got that $AM=AN$ and $\angle MAN=60º$ so the triangle $AMN$ is equilateral.

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