The Nine Point Circle
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The diagram below contains a triangle \(\Delta ABC\) with circumcentre \(O\), centroid \(G\) and orthocentre \(H\). The Euler line through the points \(O\), \(G\) and \(H\) is coloured in black. The point \(N\) is the midpoint of the segment \(OH\). The Nine Point Circle Theorem says that the midpoints of each side, the altitudes of each vertex and the midpoints \(P\), \(Q\) and \(R\) of the segments \(AH\), \(BH\) and \(CH\) respectively all lie on the same circle, and that this circle has centre \(N\).
You can see this explicitly in the following picture. By moving the vertices \(A\), \(B\) and \(C\) you can see how the Euler line and the Nine Point Circle depend on these points.
An amazing property of the nine point circle is that the radius is half of the radius of the circumcircle. This formula has a very simple proof, as shown in the diagram below. Since \(N\) is the centre of the nine point circle and \(P\) is on the circle, then \(|NP|\) is the radius of the nine point circle, and similarly \(|OA|\) is the radius of the circumcircle. Recall that \(N\) is the midpoint of \(OH\), and so \(|HN| = \frac{1}{2} |OH|\). Similarly, \(P\) is the midpoint of \(AH\) and so \(|AP| = \frac{1}{2} |AH|\). Therefore we can apply Thales' theorem to show that \(NP\) is parallel to \(OA\) and that \(|NP| = \frac{1}{2} |OA|\).
