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!

Answer 1

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.