Problem #WSP-5297

Problem

Let \(ABC\) be a non-isosceles triangle. The point \(G\) is the point of intersection of the medians \(AE\), \(BF\), \(CD\). The point \(H\) is the point of intersection of all heights. The point \(I\) is the center of the circumscribed circle of \(ABC\), or the point of intersection of all perpendicular bisectors to the segments \(AB\), \(BC\), \(AC\).
Prove that points \(I,G,H\) lie on one line and that the ratio \(IG:GH = 1:2\). The line that all of \(I\), \(G\) and \(H\) lie on is called the Euler line of triangle \(ABC\).