Euler's line : a projective geometry proof

Question

In a given triangle $ABC$, let

$G$ be the common point to the three medians,
$H$ be the common point to the three altitudes, and
$M$ be the common point to the perpendicular bissectors of the three sides.

These three points lie on a line. It is not difficult to prove this result using barycenters, or some transformations of the plane. I remember I once saw a very short and elegant proof using projective geometry arguments. Something like, "such or such line can be decided to be at infinity, so...". Perhaps using Desargues' Theorem? I was trying to rebuild that proof, but I failed. Could anyone help? Hints / suggestions?

Thanks!

There are non-euclidean versions (which is not the same as your query but it might help) of the Euler line : the Euler-Wildberger line like here p. 49... or the Euler-Akopyan line... But I do not know much on the subject... — Jean Marie, Jun 05 '23 at 18:22

score 2 · Answer 1 · answered Jun 27 '23 at 20:27

Let $A_1$ be the midpoint of $BC$, $B_1$ the midpoint of $AC$. Let $A_2$ be the ideal point of the lines that are perpendicular to $BC$, and $B_2$ be the ideal point of the lines that are perpendicular to $AC$.

Then the triangles $AA_1A_2$ and $BB_1B_2$ are perspective from a point (the ideal point of $AB$): $AB$ and $A_1B_1$ are parallel, so they intersect at an ideal point. Therefore their intersection is on the ideal line, which is $A_2B_2$.

So we can use Desargues' theorem for these triangles and we get that $G$, $H$ and $M$ are collinear.

Euler's line : a projective geometry proof

1 Answers1