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 \(OQ.\)
- Show that \(\overrightarrow{AQ}\) is perpendicular to \(\overrightarrow{BC}\)
and deduce that the three altitudes of \(\triangle ABC\) are concurrent.
- Show that the midpoints of \(AQ,BQ\) and \(CQ\), and the midpoints of
the sides of \(\triangle ABC\) are all equidistant from \(N\).
{[}The
centroid of \(\triangle ABC\) is the point \(G\) such
that \(\overrightarrow{OG}=\frac{1}{3}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}).\)
The
altitudes of the triangle are the lines through the vertices
perpendicular to the opposite sides.{]}