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Suppose $O$ is the centre of the circumscribing circle of triangle $ABC$ and $H$ is its orthocentre. Prove that vector $OH$ is equal to the sum of the vectors $OA$, $OB$ and $OC$.

An answer I found on the Internet.

Let's start with the following observation. For any vectors $U$ and $V$, if $\lvert U\rvert = \lvert V\rvert$ then $U+V$ is perpendicular to $U-V$. We show this using the dot product:

$$(U+V)\cdot(U-V) = \lvert U\rvert^2 - U\cdot V + V\cdot U - \lvert V\rvert^2 = 0$$

Now for simplicity we will use different notation from the first proof. Let the vectors from the center of the circumscribed circle to the vertices of the triangle be $A$, $B$ and $C$. Note that $\lvert A\rvert = \lvert B\rvert = \lvert C\rvert$ and vectors $A-B$, $B-C$ and $C-A$ represent the three sides of the triangle. We want to show that the orthocenter is the point $A+B+C$.

Now construct the orthocenter as the intersection of two altitudes of the triangle. Since $B+C$ is perpendicular to $B-C$ the altitude from $A$ is on the line with parametric equation $$L_1 = A + a(B+C),$$ where $a$ is the parameter. Also, since $A+B$ is perpendicular to $A-B$ the altitude from $C$ is on the line with parametric equation $$L_2 = C + b(A+B),$$ where $b$ is the parameter. When $L_1$ and $L_2$ intersect, $$ A + a(B+C) = C + b(A+B)$$ and $$ a(B+C) = C - A + b(A+B).$$

Now take the dot product of both sides with $A-B$. Note that this causes $b$ to drop out, which allows us to solve for $a$: $$ a(B+C)\cdot (A-B) = (C-A)\cdot (A-B)$$ $$a(B\cdot A - \lvert B\rvert^2 + C\cdot A - C\cdot B) = C\cdot A - C\cdot B - \lvert A\rvert ^2 + A\cdot B.$$

Since $\lvert B\rvert = \lvert A\rvert$ the quantity in the parentheses is equal to the right hand side. It can be shown that this quantity is non-zero, so we can divide by it to arrive at $$ a = 1.$$

The orthocenter is the value of $L_1$ when $a = 1$. Therefore, the orthocenter is the point $A+B+C$ as was to be shown, so the proof is complete.

However, I understood EVERY WORD OF it except this:

Since $B+C$ is perpendicular to $B-C$ the altitude from $A$ is on the line with parametric equation $$ L_1 = A + a(B+C).$$

Would someone guide me please? What do they mean by this line?

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  • $\begingroup$ Hi @Fringo! $$\color{red}{\Large\text{Welcome to Math.SE!}}$$ Don't worry about it now (since you're new) but you might like to know that we use MathJax here (e.g. $\theta$ for $\theta$). $\endgroup$
    – Shaun
    Jan 7 '14 at 22:27
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We have $OB=OC$, so the parallelogram they span is a rhombus that has orthogonal diagonals.

And, $L_1$ is the line with the property that it goes through $A$ and perpendicular to side $BC$ (i.e. to vector $\vec{OC}-\vec{OB}$), consequently, its direction is $\vec{OB}+\vec{OC}$.

If you are familiar with scalar product, this one would also convince you: $$|\vec{OB}|=|\vec{OC}| \ \iff\ \vec{OB}\cdot\vec{OB} = \vec{OC}\cdot\vec{OC} \ \iff \\ \vec{OB}^2-\vec{OC}^2=0\ \iff\ (\vec{OB}-\vec{OC})(\vec{OB}+\vec{OC})=0\ \iff \\ (\vec{OB}-\vec{OC})\ \perp\ (\vec{OB}+\vec{OC}) $$

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  • $\begingroup$ Yes i basically understood why (B+C) | (B-C), But what i didn't get is this : the altitude from A is on the line with parametric equation L1 = A + a(B+C), can you explain further please? How did they even manage to get the equation? $\endgroup$
    – Fringo
    Jan 7 '14 at 20:16
  • $\begingroup$ Yes, extended. $B+C$ is a direction vector of the altitude, because it is orthogonal to $BC$ side. $\endgroup$
    – Berci
    Jan 7 '14 at 20:19
  • $\begingroup$ Thank you! But how did they get the L1 equation? $\endgroup$
    – Fringo
    Jan 7 '14 at 20:21
  • $\begingroup$ For a direction vector $v$, exactly the points $av$ are on the line with that direction starting from origin where $a\in\Bbb R$. If the line doesn't pass through origin, but passes through a point $A$, the points of the line at origin are all translated by vector $A$, yielding: $L_1=\{A+av\mid a\in\Bbb R\}$. $\endgroup$
    – Berci
    Jan 7 '14 at 20:25

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