Relation between Vectors from Orthocentre and Circumcentre in an acute angle triangle

Question

Given an acute-angled triangle $\Delta ABC$ having it's Orthocentre at $H$ and Circumcentre at $O$. Prove that $\vec{HA} + \vec{HB} + \vec{HC} = 2\vec{HO}$

I realise that $\vec{HO} = \vec{BO} + \vec{HB} = \vec{AO} + \vec{HA} =\vec{CO} + \vec{HC}$ which leads to $3\vec{HO} = (\vec{HA} + \vec{HB} + \vec{HC}) + (\vec{AO} + \vec{BO} + \vec{CO})$

How can I prove that $(\vec{AO} + \vec{BO} + \vec{CO}) = \vec{HO}$ in order to solve the problem?

Thank you for answering!

Answer 1

If $G$ is the centroid of the triangle, that relation follows from $$ \vec{AO} + \vec{BO} + \vec{CO}= \vec{AG} + \vec{BG} + \vec{CG}+3\vec{GO} =3\vec{GO}=\vec{HO} $$ where the last one is a well-known equality, arising in the proof for the Euler line, see e.g. https://en.wikipedia.org/wiki/Euler_line#A_vector_proof .

Answer 2

Write $ \vec{v}:=\vec{OA} + \vec{OB} + \vec{OC}$, Since we can translate vectors and they don't change we can assume that $\vec{v}$ starts at $O$ and let it end point be $X$. So $\vec{v}= \vec{OX}$. Now what can we say for $X$? Let $C'$ halves $AB$. Since $$\vec{OA} + \vec{OB} ={1\over 2}\vec{OC'}$$ and $$\vec{OA}+\vec{OB}= \vec{OC}-\vec{OX} = \vec{XC}$$ we see that $\vec{XC}\bot \vec{AB}$. So $X$ is on altitude from $C$. In the same manner we see that $X$ is also on other altitudes, so $X=H$.