OBM 2005 Problem 5

Let $ABC$ be a triangle with all angles $\leq 120^{\circ}$. Let $F$ be the Fermat point of triangle $ABC$, that is, the interior point of $ABC$ such that $\angle AFB = \angle BFC = \angle CFA = 120^\circ$. For each one of the three triangles $BFC$, $CFA$ and $AFB$, draw its Euler line - that is, the line connecting its circumcenter and its centroid.

Prove that these three Euler lines pass through one common point.

Remark. The Fermat point $F$ is also known as the first Fermat point or the first Toricelli point of triangle $ABC$.

Floor van Lamoen