(by Polymath) This page is part of the Advanced High School Math Project.
This page proves the existence of the Euler Line of a triangle. That is, it proves that the orthocenter, centroid, and circumcenter of a triangle are all collinear. It does not use the standard analytic proof, where you find equations for the various lines and show that the intersection of two of them lies on the third. Rather, this proof is synthetic (non-analytical), which means it is structured more like a standard geometry proof. I found this proof on the Internet several years ago, but I regret that I cannot recall the source. If you find (or are) the source of this proof, please let me know, and I will gladly cite it.
Consider the diagram below:
The upper diagram illustrates two easy preparatory lemmas:
- Since the segment connecting the midpoints of two sides of a triangle is half the length of the third side, the triangle formed by all three of those segments (the medial triangle) is similar to the original triangle in a 2:1 ratio.
- Because the sides of the medial triangle are each parallel to a side of the original triangle, the lines that serve as the perpendicular bisectors of the sides of the original triangle are clearly altitudes of the medial triangle. Therefore, the circumcenter of the original triangle coincides with the orthocenter of the medial triangle.
In the lower diagram, AH is an altitude containing the orthocenter O. AM is a median containing the centroid D. MR is the perpendicular bisector of BC, and it contains the circumcenter R. Connecting D to O and to R gives 2 segments that we want to prove to belong to the same line. I aim to show that that they do so by proving angle ADO congruent to angle MDR. (This could only happen if O, D, R were collinear, since O and R must fall on opposite sides of the median.)
To prove those angles congruent, I will show that the triangles AOD and MRD are similar:
- Since the altitude and the perpendicular bisector are clearly parallel, angle DAO and angle DMR are congruent.
- AD and MD stand in a ratio of 2:1, since that's one of the basic properties of a centroid.
- If you consider the medial triangle of ABC in light of lemma 2, it is clear that AO and MR are precisely corresponding segments of the original and medial triangles (since R is the orthocenter of the medial triangle). Thus by lemma 1, AO:MR = 2:1.
This proves the triangles similar, which means the desired angles are congruent, which means the points O, D, R are collinear.