The Nine-Point Circle: One Circle Through Nine Points of Every Triangle

The nine points, one trio at a time

Take any triangle ABC. Three separate constructions each give you three points, and all nine end up on one circle.

• The three side midpoints — M_a, M_b, M_c, the middles of BC, CA, AB. They are the vertices of the medial triangle.
• The three feet of the altitudes — drop a perpendicular from each vertex onto the opposite side and mark where it lands. These are the vertices of the orthic triangle. Note the wording: the perpendicular is dropped onto the line containing the opposite side. In an obtuse triangle two of the three feet land outside the side itself, on its extension, and the theorem is unaffected.
• The three Euler points — the three altitudes meet at a single point, the orthocenter H. Mark the midpoint of AH, of BH and of CH. These are the vertices of the Euler triangle.

Because all nine are concyclic, the nine-point circle is simultaneously the circumcircle of the medial triangle, the circumcircle of the orthic triangle and the circumcircle of the Euler triangle. That is the cleanest way to hold the result in your head: three different triangles inscribed in one triangle, all sharing a circumcircle.

Two facts come free from the medial-triangle description. The medial triangle is similar to ABC with ratio 1/2, so its circumradius — the nine-point radius — is R/2. And three points already determine a circle, so the nine-point circle is pinned by the three midpoints alone; the interesting content of the theorem is that the other six points then land on it too.

Why they land on one circle

Dragging a triangle around and watching the circle keep up is convincing, but it is an illustration, not a proof. Here is the proof, in two moves.

Move 1: two half-scale copies of the circumcircle turn out to be the same circle. Consider the homothety (uniform scaling) h₁ centered at H with ratio 1/2. It sends each vertex to the midpoint of the segment from H to that vertex — that is, A, B, C go to the three Euler points. So the three Euler points lie on the image of the circumcircle: a circle of radius R/2 centered at h₁(O), which is the midpoint of OH.

Now consider the homothety h₂ centered at the centroid G with ratio −1/2. It sends A, B, C to the midpoints of the opposite sides (that is exactly the statement that the medians cut each other 2:1). So the three side midpoints lie on a circle of radius R/2 centered at h₂(O). Put O at the origin; then the vector to the centroid is G = H/3, and h₂(O) = G + (G − O)/2 = H/2 — the midpoint of OH again. Same center, same radius, so it is the same circle, and six of the nine points are already on it.

Move 2: Thales puts the altitude feet on it. Let A′ be the point of the circumcircle diametrically opposite A. Then BHCA′ is a parallelogram (both BH and A′C are perpendicular to AC, and both CH and A′B are perpendicular to AB), so its diagonals bisect each other and the midpoint of HA′ is M_a, the midpoint of BC. Apply h₁ again: it sends the antipodal pair A, A′ to E_a and M_a, so E_aM_a is a diameter of the nine-point circle. Finally, the foot F_a sees that diameter at a right angle: F_aM_a runs along line BC, F_aE_a runs along the altitude from A, and the altitude is perpendicular to BC. By the converse of Thales' theorem, F_a lies on the circle with diameter E_aM_a. The same argument covers F_b and F_c. That is all nine.

The argument never assumed the triangle was acute. It uses only signed relations between lines, which is why the obtuse case — feet outside the sides, orthocenter outside the triangle — needs no separate treatment.

Where the center sits: the Euler line

Euler proved in 1765 that the circumcenter O, the centroid G and the orthocenter H are always collinear, with OH = 3 OG. The nine-point center N joins them: it is the midpoint of OH. Putting O at 0 and H at 1 on that line gives G at 1/3 and N at 1/2, so the four centers divide the Euler line in the ratio

OG : GN : NH = 2 : 1 : 3.

In vector form with O as origin, H = A + B + C and N = (A + B + C)/2. The length of the Euler line has a clean closed form,

OH² = 9R² − (a² + b² + c²),

so the whole Euler line collapses to a point exactly when a² + b² + c² = 9R², which happens only for the equilateral triangle. There O = G = N = H, and the nine-point circle becomes concentric with the circumcircle at half its radius — which is also the incircle, since r = R/2 for an equilateral triangle. A useful sanity check on the obtuse case: H falls outside the triangle, O falls outside on the other side, and N, sitting halfway between them, can also land outside the triangle while its circle still passes through all nine points.

A worked triangle: sides 13, 14, 15

The 13–14–15 triangle is the standard test case because everything comes out rational. Place B = (0, 0), C = (14, 0), A = (5, 12). Then AB = 13, CA = 15, BC = 14, the semiperimeter is s = 21 and Heron gives area K = √(21·7·6·8) = 84.

From those: R = abc/4K = (13·14·15)/336 = 65/8 = 8.125, and r = K/s = 4. The centers are O = (7, 33/8), H = (5, 15/4), G = (19/3, 4), I = (6, 4), and

N = midpoint of OH = (6, 63/16),   nine-point radius = R/2 = 65/16 = 4.0625.

The nine-point circle is therefore (x − 6)² + (y − 63/16)² = (65/16)². Here are the nine points, each of which is exactly 65/16 from N:

• Side midpoints: (7, 0), (19/2, 6), (5/2, 6)
• Altitude feet: (5, 0), (224/25, 168/25), (350/169, 840/169)
• Euler points: (5, 63/8), (5/2, 15/8), (19/2, 15/8)

The three ugly-looking feet are the point of the exercise: 350/169 and 840/169 carry a denominator of 169 = 13² and look like they have nothing to do with the tidy midpoints, yet (350/169 − 6)² + (840/169 − 63/16)² works out to exactly (65/16)². The Euler-line check also lands: OH = √265/8 ≈ 2.0349, with OG ≈ 0.6783, GN ≈ 0.3391 and NH ≈ 1.0174 — the 2 : 1 : 3 split.

Feuerbach's theorem: the circle that touches four others

In 1822 Karl Wilhelm Feuerbach, then 22, published Eigenschaften einiger merkwürdigen Punkte des geradlinigen Dreiecks and proved something far stronger than concyclicity: the nine-point circle is tangent to the incircle and to each of the three excircles. Four tangencies, for every triangle. The contact point with the incircle is the Feuerbach point, catalogued as X(11) in Kimberling's Encyclopedia of Triangle Centers.

Tangency to the incircle is internal (the incircle sits inside the nine-point circle), so the distance between their centers must equal the difference of the radii:

|NI| = R/2 − r.

For the 13–14–15 triangle that predicts |NI| = 65/16 − 4 = 1/16 = 0.0625, and the coordinates above give N = (6, 63/16), I = (6, 4) = (6, 64/16) — a vertical separation of exactly 1/16. The two circles are a hair apart and touch at a single point.

Note what the formula requires: R/2 ≥ r, i.e. R ≥ 2r. That is Euler's inequality, which follows from Euler's own 1765 identity OI² = R(R − 2r); since a squared length cannot be negative, R ≥ 2r, with equality exactly when O = I, i.e. for the equilateral triangle. (Check: OI² = 8.125 × 0.125 = 1.015625 for our triangle, and |OI| from the coordinates is √1.015625 ≈ 1.0078.) Tangency to each excircle is external, so |NI_A| = R/2 + r_A, and with r_A = 12 that is 16.0625.

When nine points are not nine points

The circle is always there; the nine labels are not always nine places. It is worth knowing which cases collapse, because a demonstration that only ever shows a generic scalene triangle hides them.

• Equilateral: each altitude foot is the corresponding side midpoint, so the nine labels cover only 6 distinct points. And since r = R/2 here, the nine-point circle is the incircle — Feuerbach's tangency degenerates into coincidence.
• Isosceles (but not equilateral): the altitude from the apex lands on the midpoint of the base, so exactly one foot merges with one midpoint: 8 distinct points.
• Right-angled at C: the two legs are altitudes, so H = C, two of the feet collapse onto C, and two Euler points fall onto side midpoints. Only 5 distinct points survive: the three side midpoints, the foot of the altitude from C, and C itself — and only 4 if that right triangle is isosceles too, because then the foot from C is itself the midpoint of the hypotenuse. (The midpoint of the hypotenuse is the circumcenter, which is the classical Thales result in disguise.)
• Obtuse and scalene: all nine stay distinct, but two altitude feet, the orthocenter, and two Euler points all lie outside the triangle. The circle does not care. (An obtuse triangle that is also isosceles falls under the isosceles case above and has 8 — being obtuse does not by itself keep the nine apart.)

There is also a striking symmetry in the definition. Take the four points A, B, C, H: each one is the orthocenter of the triangle formed by the other three, which makes them an orthocentric system. The four triangles ABC, ABH, BCH, CAH therefore all have the same nine-point circle — one circle doing duty for four triangles, all four of which share the circumradius R as well. Relax the orthocentric condition to four arbitrary points, no three collinear, and the four nine-point circles no longer coincide but still meet at a common point: the Poncelet point of the quadruple.

Who found it, and what to call it

The attribution is genuinely tangled, and the circle carries different names in different countries.

• Euler, 1765. In Solutio facilis problematum quorundam geometricorum difficillimorum Euler established the Euler line and, in the same circle of ideas, that the medial and orthic triangles share a circumcircle — the six-point version. This is why French texts still say le cercle d'Euler. Historians differ on how explicitly Euler stated the six-point circle itself.
• Bevan and Butterworth, 1804–1805. Benjamin Bevan set a problem in the Ladies' Diary whose solution, given by John Butterworth, contains the fact that the circle through the feet of the altitudes has radius half the circumradius. J. S. Mackay's History of the Nine-Point Circle (Proc. Edinburgh Math. Soc. 11, 1892) is the standard source for this strand.
• Brianchon and Poncelet, 1821. Charles Brianchon and Jean-Victor Poncelet, in a joint paper in Gergonne's Annales de Mathématiques Pures et Appliquées vol. 11 (1820–21), gave what is generally taken as the first statement and proof of the full nine-point result, Euler points included.
• Feuerbach, 1822. Feuerbach's pamphlet proved the tangency theorem above. German texts call the circle der Feuerbachkreis in his honour, even though the tangency, not the nine points, is his distinctive contribution.
• Terquem, 1842. Olry Terquem is credited with the name that stuck in English and French — cercle des neuf points, the nine-point circle — and with a proof covering all nine.

So “Euler circle”, “Feuerbach circle” and “nine-point circle” all name the same object, and none of the three names points at the person who first wrote down all nine points. If you want a name that is unambiguous and cannot be wrong, nine-point circle is it: it describes the object rather than claiming a discoverer.

The classical circles attached to a triangle, with radii worked out for the 13–14–15 triangle (semiperimeter s = 21, area K = 84)

Circle	Center	Radius	13–14–15 triangle
Circumcircle	circumcenter O = X(3)	R = abc / 4K	R = 65/8 = 8.125
Nine-point circle	N = X(5), the midpoint of OH	R/2	65/16 = 4.0625
Incircle	incenter I = X(1)	r = K / s	4
Excircle opposite A	excenter I<sub>A</sub>	r<sub>A</sub> = K / (s &minus; a)	12 (the other two: 14 and 10.5)
Spieker circle	Spieker center X(10)	r/2 &mdash; it is the incircle of the medial triangle	2