The nine-point circle

There is one more remarkable theorem about the connection between certain interesting points in a given triangle. Using the notation from the previous two lectures, lets add a few more letters: K, L and M are the midpoints of the segments AH, BH and CH; AD, BE and CF are the three altitudes of $\Delta ABC$ (see Figure 1).

  

Figure 1: The nine-point circle

Theorem (Nine-point-circle theorem). The feet of the three altitudes (D, E and F) of any triangle, the midpoints of the three sides (A', B' and C'), and the midpoints of the segments from the three vertices to the orthocenter (K, L and M), all lie on the same circle, of radius $\frac{1}{2}R$ circle.

Proof. Since BC is a common side of the two triangles ABC and HBC, whose other sides are bisected, respectively, by C', B' and L, M, both the segments C'B' and B'M are parallel to BC (and half as long). Hence B'C'LM is a parallelogram. Since BC and AH are perpendicular, this perpendicular is a rectangle. Similarly, A'B'KL and C'A'MK are rectangles. Hence A'K, B'L and C'M are three diameters of the same circle. Since $\angle A'DK$ is a right angle, this (on A'K as a diameter) passes through D. Similarly, it passes through E and F.

Problem 1. The center of the nine-point circle lies on the Euler line. It is the midpoint of the segment connecting the orthocenter and circumcenter.

Proof. Since the three points K, L and M are diametrically opposite to A', B' and C' the two triangles are symmetric with respect to the center of the nine-point circle N. Then their orthocenters will be also symmetric. But since H is the orthocenter of $\Delta KLM$ and O is the orthocenter of $\Delta A'B'C'$ we have that N is the midpoint of HO.