A historical research

^{1} 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 theorem**^{2}
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 dilatations**^{3}
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

δ_{C,r}(X) = r(X - C) + C

Let f = δ

_{G,-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
l

_{DC'}
and
l

_{HS} 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 = δ

_{H,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 theorem**^{4}
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