# Fermat Points and Concurrent Euler Lines

Let F_{1} and F_{2} denote the (inner and outer) *Fermat-Toricelli* points of a given ΔABC. We prove that the Euler lines of the 10 triangles with vertices chosen from A, B, C, F_{1}, F_{2} (three at a time) are concurrent at the centroid of ΔABC [Beluhov]. This is obviously true for ΔABC itself. The applet below illustrates the case where the triangles are formed by one of the Fermat points and a pair of vertices A, B, C. There are six such triangles. The other three triangles are considered separately.

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Copyright © 1996-2018 Alexander Bogomolny

Suffice it to consider one of the six triangles. We will actually consider two at once, viz., ΔBCF, where F is one of the Fermat points.

In any triangle, the Euler line passes through the centroid, the circumcenter, the orthocenter and the nine-point center. The circumcircle of ΔBCF, coincides with the circumcircle of the equilateral ΔBCA'', implying that A', the center of ΔBCA'', is the circumcenter of ΔBCF. What we are going to show is that the line GA' passes through the centroid, say, G' of ΔBCAF.

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Let M_{a} be the midpoint of side BC. Observe that _{a}:CG = 1:2_{a}:A'A'' = 1:2_{a} and divides it in the same ratio: G'M_{A}:FG' = 1:2. On the other hand, FM_{a} is cut by GA' in the ratio 1:2. It follows that G' lie at the intersection of GA' and FM_{a} and, in particular, on GA' making it the Euler line of ΔBCF.

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Copyright © 1996-2018 Alexander Bogomolny