A historical research1 revealed that in the early 19th century, the nine-point circle theorem was independently discovered by English, French, German, and Swiss mathematicians! Sometimes the circle is called the "Euler circle," giving credit to Euler, or the "Feuerbach circle," giving credit to Karl Feuerbach, who have also noted that the circle is tangent to the incircle (at the Feuerbach point) and excircles.

Triangle ABC, orthocenter H, and the nine points.
```                                    C
/|\
/  | \
/    |  \
/      |   \
R/        |C"  \
/  \       |     \
/      \     |      \Q
B'  /          \   |    _, \
*              \ |_,-'    * A'
/               _,-H         \
/             _,-'   | \        \
/           _,-'       |   \       \
/       A"_,-'           |     \      \
/       _,-'               |       \B"   \
/     _,-'                   |         \    \
/   _,-'                       |           \   \
/ _,-'                           |             \  \
/,-'                               |               \ \
A-------------------------*---------+-----------------B
C'        S
```
The altitudes are AQ, BR, and CS. The sides BC, AC, AB have midpoints A', B', C'. Points A", B", C" are midpoints of AH, BH, CH. The centroid G and circumcenter D are not shown. The nine-point circle theorem claims that the nine points lie on a circle centered at N, the midpoint of D and H, on the Euler line.

Proof by the eight point circle theorem2

Note that quadrilaterals ABCH, ABHC, and AHBC all have perpendicular diagonals. Thus the eight point circle theorem applies. The sides of the quadrilaterals are perpendicular in a way that two pairs of points overlap, making them six point circles instead. The eight point circle theorem states that the circle center is located at the centroid of a quadrilateral. Since ABCH, ABHC, and AHBC all share the same centroid N, all six point circles share the same center. Since the circles share points on the circumference as well as the center, the circles are equivalent. One circle covers all nine points.

Furthermore, it can be shown that N lies on the Euler line. By the theorem of Snapper we know
-2(D - G) = H - G
-2D = H - 3G
D = -(1/2)H + (3/2)G
and Since N is the centroid of A, B, C, and H we know
N = (1/4)(A + B + C + H)
N = (3/4)(1/3)(A + B + C) + (1/4)H
N = (3/4)G + (1/4)H
N = (1/2)[(3/2)G - (1/2)H] + (1/2)H
N = (1/2)D + (1/2)H
So N is the midpoint of D and H.
QED

Proof by central dilatations3

This proof doesn't rely on the eight point circle theorem, which relies on the theorem of Thales. A central dilatation is a type of mapping that maps an images into a new image as if it was mapped by a lens: it can enlargen, shrink, and/or invert an image. The general equation of a central dilatation with center of dilatation C and a non-zero dilatation coefficient r is
δC,r(X) = r(X - C) + C
Let f = δG,-1/2. This function maps ABC to A'B'C', due to where the centroid G is located in a circle (as explained in the node centroid). It also maps the circumcircle O of ABC into the circumcircle O' of A'B'C'. If the nine-point circle exists, it must be O' because a triangle can only have one circumcircle (as explained in the node circumcenter), and the nine-point circle certainly circumscribes triangle A'B'C'. Hence N = f(D). Along with D = -(1/2)H + (3/2)G derived from the theorem of Snapper above, it can be shown that N is the midpoint of D and H:
N = f(D)
N = (-1/2)(D - G) + G
N = (-1/2)D + (1.5)G
N = (1/2)D - D + (1.5)G
N = (1/2)D - [-(1/2)H + (3/2)G] + (1.5)G
N = (1/2)D + (1/2)H - (3/2)G + (1.5)G
N = (1/2)D + (1/2)H
Since N is equidistant from D and H, N is also equidistant from lines lDC' and lHS by congruent triangles. Hence ΔC'NS is an isosceles triangle, and the distance to C' and S from N is equal. Similarly, Q, R, and S also lie on circle O'.

Consider dilitation g = δH,1/2. This maps points A, B, C to A", B", C". Hence A", B", C" lies on the circle centered at g(D). Then
g(D) = (1/2)(D - H) + H = (1/2)(D + H) = N
This circle has the same center as O' and the same radius (half the radius of the circumcircle of ΔABC). Hence they are the same circle. Points A", B", C" also lie on O'.
QED

Corollary: Hamlton's theorem4

In 1861, W. R. Hamilton (1805-1865) noticed that triangles ABC, ABH, AHC, HBC all have the same nine-point circle. Since the nine-point circle is half the width of the circumcircle, it follows that the four triangles have a circumcircle of same radius (but different centers).
Sources
1. Mackay, J. S. "History of the Nine-Point Circle." Proc. Edinburgh Math. Soc. 11, 19-61, 1892.
2. http://www.cut-the-knot.com/Curriculum/Geometry/SixPointCircle.html
3. "Vectors and Transformations in Plane Geometry" by Philippe Tondeur, Publish or Perish, Inc. 1993.
4. http://www.cut-the-knot.com/triangle/EulerLine.html