Orthocenter Vector Formula

A vector proof that, with the circumcenter at the origin, the orthocenter is the sum of the three vertex vectors: OH = OA + OB + OC. The sum is built with two parallelograms and shown to land on every altitude.

Keywords: orthocenter, circumcenter, vector addition, dot product, Euler line

Prerequisites: orthocenter-proof, circumcenter-proof · Difficulty: advanced

Take the circumcenter $O$ as the origin, so each vertex is a vector $O A$ , $O B$ , $O C$ , all of length $R$ . With that one choice of origin, the orthocenter has a strikingly simple address:

O H = O A + O B + O C

Strategy: build this sum with two parallelograms, then check the result is perpendicular to every side using the dot product, so it lands on all three altitudes.

Step 1: Triangle and Circumcircle

We take the circumcenter $O$ as our reference point. The three vertices lie on the circumcircle, so $O A = O B = O C = R$ .

Step 2: Position Vectors from $O$

With $O$ as the origin, read each vertex as a vector: $O A$ , $O B$ , $O C$ . They add by the parallelogram rule, and two vectors are perpendicular exactly when their dot product is zero. Here all three are radii, so $∣ O A ∣ = ∣ O B ∣ = ∣ O C ∣ = R$ .

Step 3: $O H = O A + O B + O C$

Claim. With $O$ at the origin, the orthocenter is the vector sum

O H = O A + O B + O C .

We build that sum with two parallelograms, then verify the result lies on every altitude — so it can only be the orthocenter.

Step 4: Add the Vectors

Complete the parallelogram on $O A$ and $O B$ to reach $Q$ , so $O Q = O A + O B$ . Add $O C$ the same way to reach $H$ , giving $O H = O Q + O C = O A + O B + O C$ .

Step 5: The Sum Lies on the Altitudes

Why is $H$ the orthocenter? Since $O H = O A + O B + O C$ , subtracting $O A$ gives $A H = O B + O C$ , simply $B + C$ since $O$ is the origin. Compare it with the side $B C = O C - O B$ :

A H \cdot B C = ∣ O C ∣^{2} - ∣ O B ∣^{2} = R^{2} - R^{2} = 0.

So $A H ⊥ B C$ , and $H$ is on the altitude from $A$ . The same at $B$ and $C$ places $H$ on all three altitudes.

Step 6: The Euler Line, for Free

So the vector sum is exactly the orthocenter,

O H = O A + O B + O C .

The centroid, already the average of the vertices, is a third of this sum, so it lies one-third of the way from $O$ to $H$ , giving $O G : G H = 1 : 2$ . This is the Euler line, which its own chapter proves again geometrically. Halving the sum instead gives the nine-point center $N$ .

With the circumcenter at the origin,

O H = O A + O B + O C .

The centroid is the plain average of the vertices, proved earlier, so with the circumcenter still at the origin it is one-third of this same sum:

O G = \frac{1}{3} (O A + O B + O C) = \frac{1}{3} O H .

The centroid therefore lies one-third of the way from $O$ to $H$ , so $O$ , $G$ , $H$ are collinear with $O G : G H = 1 : 2$ . This collinearity is the Euler line, proved again geometrically when the line is introduced. Halving the sum instead gives the nine-point center $N$ , the midpoint of $O H$ . ∎

Notes

A little 2D-vector language

Two facts are all we need. Addition is the parallelogram rule: $O A + O B$ is the diagonal of the parallelogram built on the two vectors. Perpendicularity is the dot product: $u \cdot v = 0$ exactly when $u ⊥ v$ , and $v \cdot v = ∣ v ∣^{2}$ .

Why the sum is the orthocenter

Let $H$ be the point with $O H = O A + O B + O C$ . Then $A H = O B + O C$ and $B C = O C - O B$ , so

A H \cdot B C = (O B + O C) \cdot (O C - O B) = ∣ O C ∣^{2} - ∣ O B ∣^{2} = R^{2} - R^{2} = 0.

Hence $A H ⊥ B C$ : $H$ lies on the altitude from $A$ . The same computation at $B$ and $C$ puts $H$ on all three altitudes, so $H$ is the orthocenter.