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

Euler's Theorem (1765): For any non-equilateral triangle, the circumcenter $O$, centroid $G$, and orthocenter $H$ are collinear. The line through these three points is the Euler line.

Proof Strategy

1. Draw median $AM_A$ (where $M_A$ is the midpoint of $BC$).
2. Extend $OG$ past $G$ to a point $H'$ with $GH' = 2 \cdot OG$.
3. Show $\triangle OGM_A \sim \triangle H'GA$ (AA similarity, ratio $1:2$).
4. Since $OM_A \perp BC$ (perpendicular bisector) and the similar triangles give $AH' \perp BC$, the point $H'$ lies on the altitude from $A$.
5. Repeat for another altitude — $H'$ lies on two altitudes, so $H' = H$.

Therefore $O$, $G$, $H$ are collinear, and by construction $OG:GH = 1:2$. The nine-point center $N$ — midpoint of $OH$ — also lies on this line.

The Proof

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 $H'$ on Line $OG$

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

Step 3: Similar Triangles: $\triangle OGM_A \sim \triangle H'GA$

Consider triangles $OGM_A$ and $H'GA$:

• Vertical angles at $G$
• $OG : GH' = 1 : 2$ (by construction)
• $GM_A : GA = 1 : 2$ (centroid divides median in ratio $2:1$)

By SAS similarity, $\triangle OGM_A \sim \triangle H'GA$ with ratio $1:2$.

Step 4: $AH' \perp BC$

Since the triangles are similar, corresponding sides are parallel: $AH' \parallel OM_A$.

But $OM_A \perp BC$ (the perpendicular bisector of $BC$).

Therefore $AH' \perp BC$ — so $H'$ lies on the altitude from $A$!

Step 5: $H' = H$: The Euler Line

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

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 $ON : NH = 1 : 1$.

Conclusion

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

OG : GH = 1 : 2, \qquad ON : NH = 1 : 1
    

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 $OH$.