I know how to do it in normal Euclid geometry, but is it possible to do it with vector algebra?

  • $\begingroup$ The translation of the claim is: Given $a,b,c$ there exists $h$ such that $\langle h-a,c-b\rangle=\langle h-b,a-c\rangle=\langle h-c,b-a\rangle=0$. Or: If $h$ is such that $\langle h-a,c-b\rangle=\langle h-b,a-c\rangle=0$, then also $\langle h-c,b-a\rangle=0$. Can you find a proof? You may assume that $h=xa+yb+zc$ with $x,y,z\in\mathbb R$ and $x+y+z=1$. $\endgroup$ Commented Dec 10, 2012 at 16:44
  • $\begingroup$ @HagenvonEitzen Thanks, but I don't quite follow your notation, could you please explain further? $\endgroup$
    – qed
    Commented Dec 10, 2012 at 17:27
  • $\begingroup$ Ah, maybe the use of $\langle,\rangle$ for scalar (or dot) product is unusual for you ... $\endgroup$ Commented Dec 10, 2012 at 17:59
  • $\begingroup$ @HagenvonEitzen, Oh, that's dot product, i see. Thanks! $\endgroup$
    – qed
    Commented Dec 10, 2012 at 18:48

5 Answers 5


Denote the three vertices of the triangle by three column vectors $\mathbf{u}, \mathbf{v}, \mathbf{w} \ (\in\mathbb{R}^2)$ and the orthocenter by $\mathbf{x}$. By definition, $\mathbf{x}$ must satisfies $$ \begin{cases} (\mathbf{v}-\mathbf{w})\cdot(\mathbf{x}-\mathbf{u})=0\\ (\mathbf{w}-\mathbf{u})\cdot(\mathbf{x}-\mathbf{v})=0\\ (\mathbf{u}-\mathbf{v})\cdot(\mathbf{x}-\mathbf{w})=0 \end{cases} \ \Rightarrow \begin{cases} (\mathbf{v}-\mathbf{w})\cdot\mathbf{x}=(\mathbf{v}-\mathbf{w})\cdot\mathbf{u}\\ (\mathbf{w}-\mathbf{u})\cdot\mathbf{x}=(\mathbf{w}-\mathbf{u})\cdot\mathbf{v}\\ (\mathbf{u}-\mathbf{v})\cdot\mathbf{x}=(\mathbf{u}-\mathbf{v})\cdot\mathbf{w} \end{cases} $$ This can be rewritten in the matrix form of $A\mathbf{x}=\mathbf{b}$: $$ \underbrace{\begin{bmatrix}(\mathbf{v}-\mathbf{w})^T\\ (\mathbf{w}-\mathbf{u})^T\\ (\mathbf{u}-\mathbf{v})^T\end{bmatrix}}_{A} \ \mathbf{x} = \underbrace{\begin{bmatrix} (\mathbf{v}-\mathbf{w})\cdot\mathbf{u}\\ (\mathbf{w}-\mathbf{u})\cdot\mathbf{v}\\ (\mathbf{u}-\mathbf{v})\cdot\mathbf{w} \end{bmatrix}}_{\mathbf{b}} $$ So, the remaining question is, does there exist a unique solution for $A\mathbf{x}=\mathbf{b}$? Note that $A$ is a 3x2 matrix and $\mathbf{x}$ is a 2-vector, so the above system has three equations in two unknowns, i.e it is overdetermined. Can you show that each one of these three equations is a linear combination of the other two? Is it possible to remove one row from $A$, so that the remaining 2x2 submatrix is invertible?

  • $\begingroup$ While translation and rotation techniques might streamline the solution, I was in search of a more generalized approach. Your response provides an intuitive perspective on the problem. Instead of stating directly that the equation $(\mathbf{u} - \mathbf{v})\cdot \mathbf{x} = (\mathbf{u} - \mathbf{v})\cdot \mathbf{w}$ could be derived from the initial two equations. $+1$ $\endgroup$
    – Darshan P.
    Commented Jan 30 at 6:56

I am giving below a rather simple solution using vectors. Please make your own figure as I could not put in the same. Also treat small letters $a$, $b$, and $c$ as vectors. Write to me if you need more explanation.

Let $ABC$ a be triangle, let the perpendicular from $A$ to $BC$ meet $BC$ at $D$, and let the perpendicular from $B$ to $AC$ meet $AC$ at $E$. Let the lines $AD$ and $BE$ intersect at $O$. Let $O$ be the origin.

To prove: if we join $C$ and $O$ and extend it up to meet $AB$ at $F$, then $CF$ is perpendicular to $AB$.

Now, $CB$ is perpendicular to $OA$ (part of $AD$).

Therefore (1) $(c-b) \cdot a = 0$, since $(c-b)$ is the vector $CB$ and $OA$ is the vector $a$.

Similarly, $CA$ is perpendicular to $OB$ (part of $BE$).

Therefore, (2) $(c-a) \cdot b = 0$.

From (1), we obtain (3) $c \cdot a - b \cdot a = 0$.

From (2), we obtain (4) $c \cdot b - a \cdot b = 0$.

Subtracting (3) from (4), we get

$(c \cdot b - a \cdot b) – (c \cdot a - b \cdot a) = 0$

$\Rightarrow (c \cdot b - c \cdot a) – (a \cdot b - b \cdot a) = 0$

$\Rightarrow (c \cdot b - c \cdot a) = 0$, since $a \cdot b = b \cdot a$

$\Rightarrow (b - a) \cdot c = 0$.

Therefore $BA$ is perpendicular to $OC$ or $FC$. Therefore $O$ is the orthocenter of the triangle $ABC$ from which the three altitudes pass.


enter image description here

Let's use the figure above in our approach.

Let be $|\vec{b}|=b$, $|\vec{c}|=c$, and $\vec{b}\cdot\vec{c}= m$. The orthocenter can be calculated in two ways: $$O=A+\overrightarrow{AB}+\lambda\overrightarrow{HB}\quad (1)$$ or $$O=A+\overrightarrow{AC}+\mu\overrightarrow{JC}\quad (2)$$ But $$\overrightarrow{HB}=\vec{c}-[\frac{(\vec{b}\cdot\vec{c})}{b}]\frac{\vec{b}}{b}\quad (3)$$ and $$\overrightarrow{JC}=\vec{b}-[\frac{(\vec{b}\cdot\vec{c})}{c}]\frac{\vec{c}}{c}\quad (4)$$ If we substitute $(3)$ and $(4)$ in $(1)$ and $(2)$, we get: $$O=A+\vec{c}+\lambda[\vec{c}-(\frac{m}{b^2})\vec{b}]\quad (5)$$ and $$O=A+\vec{b}+\mu[\vec{b}-(\frac{m}{c^2})\vec{c}]\quad (6)$$ Using equations $(5)$ and $(6)$ we get: $$\vec{c}+\lambda[\vec{c}-(\frac{m}{b^2})\vec{b}]=\vec{b}+\mu[\vec{b}-(\frac{m}{c^2})\vec{c}]\Rightarrow$$


As $\vec{b}$ and $\vec{c}$ are linearly independent we can solve equation $(7)$ and we get: $$\mu=\frac{(m-b^2)}{c^2b^2-m^2}c^2$$

Now we can express $\overrightarrow{AO}$ as


If $\overrightarrow{AO}\cdot\overrightarrow{BC}=0$ then we can conclude that the three altitudes are concurrent.

So $$\overrightarrow{AO}\cdot\overrightarrow{BC}=(\vec{b}+\frac{(m-b^2)}{c^2b^2-m^2}c^2[\vec{b}-(\frac{m}{c^2})\vec{c}])\cdot(\vec{b}-\vec{c})\Rightarrow$$

$$\Rightarrow \overrightarrow{AO}\cdot\overrightarrow{BC}=b^2+\frac{(m-b^2)}{c^2b^2-m^2}c^2(b^2-\frac{m^2}{c^2})-m+\frac{(m-b^2)}{c^2b^2-m^2}c^2(-m+m)\Rightarrow$$

$$\Rightarrow \overrightarrow{AO}\cdot\overrightarrow{BC}=b^2+(m-b^2)-m+0\Rightarrow$$

$$\Rightarrow \overrightarrow{AO}\cdot\overrightarrow{BC}=0$$

Therefore the three altitudes are concurrent at point $O$.


Assume that the altitudes $AI$ and $CJ$ intersect at point O. We need to prove that $BO$ is perpendicular to $AC$.

$\vec{BJ}$ is the projection of $\vec{BO}$ onto $\vec{BA}$. Hence:

$$ \vec{BJ} = \frac{\vec{BO}\cdot\vec{BA}}{||\vec{BA}||^2}\vec{BA} \quad (1) $$

However, $\vec{BJ}$ is also the projection of $\vec{BC}$ onto $\vec{BA}$. Hence:

$$ \vec{BJ} = \frac{\vec{BC}\cdot\vec{BA}}{||\vec{BA}||^2}\vec{BA} \quad (2) $$

From (1) and (2) we have that $\vec{BO}\cdot\vec{BA} = \vec{BC}\cdot\vec{BA}$. In the same way working with $\vec{BO}$ on $\vec{BC}$ we get $\vec{BO}\cdot\vec{BC} = \vec{BA}\cdot\vec{BC}$. Hence we have $\vec{BO}\cdot\vec{BA} = \vec{BC}\cdot\vec{BA} = \vec{BO}\cdot\vec{BC}$. Now:

$$ \vec{BO} \cdot \vec{AC} = \vec{BO} \cdot (\vec{AB} + \vec{BC}) = \vec{BO} \cdot \vec{BC} - \vec{BO} \cdot \vec{BA} = 0 $$

Hence $\vec{BO}$ and $\vec{AC}$ are perpendicular.


Triangle vertices O, vector a and vector b Intersection of perpendiculars from a and b is H, vector h from O, say. We need to show that h is perpendicular to vector b - a, the side of the triangle opposite to O.

Now $(h - a).b = 0$ as line from vertex A to H is perpendicular to b and $(h - b).a = 0$ as line from vertex B to H is perpendicular to a

Removing brackets: $h.a = h.b = a.b$ So$ h.(b-a) = h.b - h.a =0$ and$ h$ is perpendicular to $b - a$, the third side of the triangle. QED.


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