Euler line (circumcenter O, centroid G, orthocenter H are collinear)

Dependencies: Three medians meet (centroid, 2:1) (centroid exists + divides each median in $2:1$), Three perpendicular bisectors meet (circumcenter) (circumcenter = intersection of perpendicular bisectors of the sides), Three altitudes meet (orthocenter) (orthocenter = intersection of the three altitudes), Triangle midsegment theorem (midsegment = half the base), Homothety (central similarity) (basic properties of homothety).

Statement

Let $\triangle ABC$ be any non-degenerate triangle, and write

• $O$ for its circumcenter (by Three perpendicular bisectors meet (circumcenter)),
• $G$ for its centroid (by Three medians meet (centroid, 2:1)),
• $H$ for its orthocenter (by Three altitudes meet (orthocenter)).

Then the three points $O$, $G$, $H$ are collinear, and the centroid divides the segment from circumcenter to orthocenter in the ratio $1 : 2$:

In other words, the circumcenter $O$ and the orthocenter $H$ are "doubly internally divided" through the centroid $G$. This common line is called the Euler line of $\triangle ABC$.