Let \(\triangle ABC\) be a triangle and \(A'\) be the midpoint of the side \(BC\). The segment \(AA'\) is a called a median of \(\triangle ABC\). Similarly, there are two more medians constructed from \(B\) and \(C\). Show that the three medians intersect at a point and give a formula for that point in terms of the three vertices. This point is called the centroid of \(\triangle ABC\).
The three altitudes of a triangle intersect at a point called the orthocenter of the triangle. Suppose that the vertices of a \(\triangle ABC\) lie on a circle of radius 1 centered at 0. Show that the centroid, the orthocenter and the circumcenter of \(\triangle ABC\) are collinear. This line is called the Euler line of the triangle. Note that the circumcenter of \(\triangle ABC\) is just 0 by our assumption.