Problem source

Let $A,B,C$ be three non-collinear points in the plane. Explain briefly
why it is possible to choose an origin equidistant from the three
points. Let $O$ be such an origin, let $G$ be the centroid of the
triangle $ABC,$ let $Q$ be a point such that $\overrightarrow{GQ}=2\overrightarrow{OG},$
and let $N$ be the midpoint of $OQ.$ 
\begin{questionparts}
\item Show that $\overrightarrow{AQ}$ is perpendicular to $\overrightarrow{BC}$
and deduce that the three altitudes of $\triangle ABC$ are concurrent. 
\item Show that the midpoints of $AQ,BQ$ and $CQ$, and the midpoints of
the sides of $\triangle ABC$ are all equidistant from $N$. 
\end{questionparts}
{[}The \textit{centroid }of $\triangle ABC$ is the point $G$ such
that $\overrightarrow{OG}=\frac{1}{3}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}).$
The \textit{altitudes }of the triangle are the lines through the vertices
perpendicular to the opposite sides.{]}