The Euler Line (e): O, G, N, H are Collinear

A proof that the circumcenter O, centroid G, orthocenter H, and nine-point center N all lie on the Euler line, with OG:GH = 1:2 and ON:NH = 1:1.

Keywords: Euler line, collinearity, circumcenter, centroid, orthocenter, nine-point center, similar triangles

Prerequisites: circumcenter-proof, centroid-proof, orthocenter-proof · Difficulty: intermediate

For any non-equilateral triangle, the circumcenter $O$, centroid $G$, and orthocenter $H$ are collinear, lying on the Euler line (Euler, 1765).

Strategy: extend $OG$ past $G$ to $H^{\prime }$ with $G{H}^{\prime }=2\cdot OG$. The median and perpendicular bisector make $△OG{M}_A\sim △H^{\prime }GA$, which forces $A{H}^{\prime }\perp BC$, so $H^{\prime }$ sits on the altitude from $A$. Repeating on a second altitude gives $H^{\prime }=H$, and the nine-point center $N$ follows as the midpoint of $OH$.

Step 1: Locate $O$ and $G$

Start with triangle $ABC$. Locate the circumcenter $O$ (intersection of perpendicular bisectors) and the centroid $G$ ($2/3$ along the median from $A$).

Step 2: Construct $H^{\prime }$ on Line $OG$

Extend segment $OG$ past $G$ to a point $H^{\prime }$ such that $G{H}^{\prime }=2\cdot OG$. We will show that $H^{\prime }=H$ (the orthocenter).

Step 3: Similar Triangles: $△OG{M}_A\sim △H^{\prime }GA$

Consider triangles $OG{M}_A$ and $H^{\prime }GA$:

• Vertical angles at $G$
• $OG:G{H}^{\prime }=1:2$ (by construction)
• $G{M}_A:GA=1:2$ (centroid divides median in ratio $2:1$)

By SAS similarity, $△OG{M}_A\sim △H^{\prime }GA$ with ratio $1:2$.

Step 4: $A{H}^{\prime }\perp BC$

Since the triangles are similar, corresponding sides are parallel: $A{H}^{\prime }\parallel O{M}_A$.

But $O{M}_A\perp BC$ (the perpendicular bisector of $BC$).

Therefore $A{H}^{\prime }\perp BC$ — so $H^{\prime }$ lies on the altitude from $A$!

Step 5: $H^{\prime }=H$: The Euler Line

By the same argument with another median and perpendicular bisector, $H^{\prime }$ also lies on a second altitude. The intersection of two altitudes is the orthocenter, so $H^{\prime }=H$.

Therefore $O$, $G$, and $H$ are collinear — they lie on the Euler line!

Step 6: The Nine-Point Center $N$

The nine-point center $N$ also lies on the Euler line — at the midpoint of segment $OH$.

This gives the complete ordering $O\to G\to N\to H$ with ratios $OG:GH=1:2$ and $ON:NH=1:1$.

The circumcenter $O$, centroid $G$, orthocenter $H$, and nine-point center $N$ all lie on the Euler line, with:

The four centers appear in order $O\to G\to N\to H$. The key insight is that the median and perpendicular bisector create similar triangles that force $H$ onto the line through $O$ and $G$; $N$ follows as the midpoint of $OH$. ∎