Prove that, in a triangle $ABC$, $\vec{OA} + \vec{OB} + \vec{OC} = \vec{OO'}$, where is $O$ is the circumcenter and $O'$ is the orthocenter

Question

Problem. If $O$ is the circumcentre and $O'$ the orthocenter of a triangle $ABC$, then prove that:

$$\vec{OA} + \vec{OB} + \vec{OC} = \vec{OO'}$$

I would like to know, if there is a more elementary approach to prove this identity, using purely the rules of vector addition and some geometry? Any inputs or suggestions of a simpler/alternative approach would be awesome.

I was able to prove this identity using the facts.

$(1)$ the altitude $\vec{AO'}\perp \vec{BC}$.

\begin{align} \vec{AO'}\cdot \vec{BC} &= 0\\ (\vec{o'} - \vec{a})\cdot(\vec{c}-\vec{b}) &= 0 \end{align}

$(2)$ the perpendicular bisector $\vec{DO}\perp \vec{BC}$.

\begin{align} \vec{DO}\cdot \vec{BC} &= 0\\ (\vec{o} - \vec{d})\cdot(\vec{c}-\vec{b}) &= 0\\ \left(\vec{o} - \frac{\vec{b}+\vec{c}}{2}\right)\cdot(\vec{c}-\vec{b}) &= 0\\ (\vec{2o} - \vec{b} - \vec{c})\cdot(\vec{c}-\vec{b}) &= 0 \end{align}

Adding $(1)$ and $(2)$, we obtain:

\begin{align} (\vec{o'} + \vec{2o} - \vec{a} - \vec{b} - \vec{c})\cdot(\vec{c}-\vec{b}) &= 0 \end{align}

$\vec{c} - \vec{b}\neq 0$. Therefore, either the first term is 0, or it is a proper vector orthogonal to $\vec{BC}$.

\begin{align} (\vec{o'} + \vec{2o} - \vec{a} - \vec{b} - \vec{c}) &= 0\\ \vec{o'} &= \vec{a} + \vec{b} + \vec{c} - \vec{2o}\\ \vec{o'} - \vec{o} &= \vec{a} + \vec{b} + \vec{c} - \vec{3o}\\ \vec{OO'} &= \vec{a} - \vec{o} + \vec{b} - \vec{o} + \vec{c} - \vec{o}\\ \vec{OO'} &= \vec{OA} + \vec{OB} + \vec{OC}\\ \end{align}

Answer 1

Hint:

Let $X$ such that $OA+OB+OC=OX$. What can we say about $X$?

Let $\vec{OA}+\vec{OB} =\vec{OD}$. Then OADB is a rombous, so $AB\bot OD$ and thus $CX||OD$, so $CX$ is an altitude. The same is true for $AX$ and $BX$ so $X=O'$ and we are done.

Answer 2

For a point $P$ such that $$\vec{OA}+\vec{OB}+\vec{OC}=\vec{OP}$$ we get $AP\perp BC$ since $$\begin{align}\vec{AP}\cdot\vec{BC}&=(\vec{OP}-\vec{OA})\cdot (\vec{OC}-\vec{OB}) \\\\&=(\vec{OC}+\vec{OB})\cdot (\vec{OC}-\vec{OB}) \\\\&=|\vec{OC}|^2-|\vec{OB}|^2 \\\\&=0\end{align}$$ Similarly, we get $BP\perp CA$ and $CP\perp AB$, so we see that $P=O'$ from which $$\vec{OA}+\vec{OB}+\vec{OC}=\vec{OO'}$$ follows.