Brazil

First we'll prove the following well-known Lemma. Let $ABP$, $BCM$ and $CAN$ be the equilateral triangles constructed externally to triangle $ABC$. The lines $AM$, $BN$ and $CP$ concur on the Fermat point of $ABC$. Proof. Let $F$ be the intersection point of $BN$ and $CP$. Since triangles $BAN$ and $PAC$ are congruent, $\angle APE=\angle ABF$. So the quadrilateral $APBF$ is cyclic, and thus $\angle AEP=\angle ABP=60^{\circ }$ and $\angle BFP=\angle BAP=60^{\circ }$. Hence $\angle AFB=\angle BFC=120^{\circ }$, which proves that $F$ is the Fermat point of triangle $ABC$. Analogously, $AM$ and $BN$ pass through the same point $F$, and the lemma is proved.

Now, let $O_A$, $O_B$ and $O_C$ be the centers of these equilateral triangles (and also the circumcenters of triangles $BCF$, $ACF$ and $ABF$, respectively); $G_A$, $G_B$ and $G_C$ are the centroids of $BCF$, $ACF$ and $ABF$, respectively. It is immediate that the Euler lines $O_A{G}_A$, $O_B{G}_B$, $O_C{G}_C$ are parallel to $AF$, $BF$ and $CF$, respectively.

Now consider the homothety with center $F$ and ratio $3/2$, which transforms $G_A$, $G_B$ and $G_C$ into the midpoints of $BC$, $CA$ and $AB$, respectively. This homothety also transforms the Euler lines of $BCF$, $ACF$ and $ABF$ into the lines $l_A$, $l_B$ and $l_C$, parallel to $AF$, $BF$ and $CF$ and passing through the midpoints of $BC$, $CA$ and $AB$, respectively. Hence these Euler lines are concurrent iff $l_A$, $l_B$ and $l_C$ are.

Finally, consider the homothety with center $G$ (centroid of $ABC$) and ratio $-2$. The midpoints of $BC$, $CA$ and $AB$ are transformed into the vertices $A$, $B$ and $C$, respectively, so $l_A$, $l_B$ and $l_C$ are transformed into $AF$, $BF$ and $CF$, respectively. Since the latter are concurrent at $F$, it follows that $l_A$, $l_B$ and $l_C$ are too, and we are done.