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The Nine-Point Circle

How to use
  1. ▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
  2. Drag AA, BB or CC; the nine points, the circle, OO and HH follow, and the readout under the canvas keeps the distances from NN equal and at half the circumradius. Learn more about why it holds
  3. Labels toggles the point names. The Step explanations log keeps every step so far. Learn more about reading a scene
Proof?Midsegments make EFQR a rectangle: EF and QR are parallel to BC, FQ and ER to the altitude AH, which is perpendicular to BC. Its diagonal EQ is a diameter of a circle through E, F, Q, R. The rectangle DEPQ has the same diagonal, so D and P are on that circle too; each foot of an altitude sees a diameter at a right angle, so the feet are on it as well. The centre is the midpoint of OH and the radius is half the circumradius. Drag any vertex.Nine-point circle
Display
Drag A, B or C; the nine points, the circle, O and H follow
ABCHaHbHcHDEFPQR
N to the nine points: 1.39 to 1.39; R / 2 = 1.39
Step 1 / 7

Step explanations

1Nine points in a triangle
Given
D, E, F are the midpoints of the sides; Ha, Hb, Hc the feet of the altitudes, which meet at the orthocentre H; and P, Q, R the midpoints of AH, BH, CH. To prove: all nine lie on one circle.
All nine points are computed from A, B, C; drag a vertex and they follow. Learn more about the statement

Getting Started with the Visualizer
The Statement
The First Rectangle
A Circle Through Four
The Second Rectangle
The Feet of the Altitudes
The Centre
Why It Holds
Dragging the Figure
Consequences



Getting Started with the Visualizer

DemoA run through the proof player
Step 0 of 8
Step through the proof, then drag the vertices and watch nine computed points stay on one circle, with its centre sliding along the Euler line. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.
Opening scene: the controls
1234567
1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names
3Draggable points — A, B and C move freely; the nine hollow points, N, O and H all follow
4Readout — the line under the figure: the smallest and largest distance from N to the nine points, and half the circumradius
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure
6Speed and step counter — how fast Play advances, and which of the 7 steps is on screen
7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it
Proof complete: step 7 of 7
12345
1The nine-point circle — through the three midpoints, the three feet and the three midpoints towards H
2The centre N — the midpoint of OH, where the diagonals of the two rectangles cross
3The two rectangles — EFQR and DEPQ, sharing the diagonal EQ
4Justification tags — Given, Midsegments, Rectangle diagonals, Midsegments, Converse of Thales, Parallelogram diagonals, Parallelogram sides
5Replay — at the last step Play becomes Replay and Next is off

The Statement

Every triangle has three natural sets of three points: the midpoints DD, EE, FF of its sides; the feet HaH_a, HbH_b, HcH_c of its altitudes; and the midpoints PP, QQ, RR of the segments from the vertices to the orthocentre HH. The theorem says that all nine lie on one circle, the nine-point circle, and more:

centre N=midpoint of OH,radius =12R\text{centre } N = \text{midpoint of } OH, \qquad \text{radius } = \tfrac{1}{2} R


where OO is the circumcentre and RR the circumradius. It was found early in the nineteenth century and is the last of the classical results about the centres of a triangle on these pages.

The proof is built from midsegments: they make two rectangles with a common diagonal, and a rectangle's four corners lie on the circle whose diameter is its diagonal. The feet join by the converse of Thales' theorem, and the centre and radius come from two parallelograms. In the player every point but AA, BB, CC is hollow, computed from the vertices. The readout under the canvas shows the smallest and largest distance from NN to the nine points, which agree while you drag, and half the circumradius beside them.
ABCHaHbHcHDEFPQR
Opening scene, frozen

The nine hollow points: the midpoints of the sides, the feet of the altitudes, and the midpoints of AH, BH, CH.


The First Rectangle

Take the four points EE, FF, QQ, RR. EFEF joins the midpoints of CACA and ABAB, so it is a midsegment of triangle ABCABC: parallel to BCBC and half of it. QRQR joins the midpoints of HBHB and HCHC, so it is a midsegment of triangle HBCHBC: also parallel to BCBC and half of it. So EFEF and QRQR are equal and parallel.

FQFQ joins the midpoints of ABAB and HBHB, a midsegment of triangle ABHABH, parallel to AHAH; and ERER is a midsegment of triangle ACHACH, also parallel to AHAH. Now AHAH is the altitude from AA, perpendicular to BCBC. So the sides of EFQREFQR are parallel in pairs to two perpendicular lines: it is a rectangle, filled in the player with its right angle marked at FF.
ABCHaHbHcHDEFPQR
The first rectangle, frozen

EFQR: two sides parallel to BC, two to the altitude AH.


A Circle Through Four

The diagonals of a rectangle are equal and bisect each other. So the diagonals EQEQ and FRFR cross at a point NN that is at the same distance from all four corners:

NE=NF=NQ=NRNE = NF = NQ = NR


The circle with centre NN and radius NENE passes through EE, FF, QQ and RR, and EQEQ is one of its diameters. The player draws the circle at this step; what follows is to show that the other five points lie on it as well.
ABCHaHbHcHDEFPQRN
A circle through four, frozen

The circle with diameter EQ, centred at N, through E, F, Q, R.


The Second Rectangle

Now the four points DD, EE, PP, QQ. DEDE is a midsegment of triangle CABCAB, parallel to ABAB; PQPQ is a midsegment of triangle HABHAB, also parallel to ABAB. DQDQ is a midsegment of triangle BCHBCH, parallel to CHCH; EPEP is a midsegment of triangle ACHACH, also parallel to CHCH. And CHCH is the altitude from CC, perpendicular to ABAB. So DEPQDEPQ is a rectangle too.

Its diagonals are EQEQ and DPDP. But EQEQ is the same segment as before, the diameter of the circle just drawn; the diagonals of this rectangle bisect each other at the midpoint of EQEQ, which is NN, and ND=NP=NEND = NP = NE. So DD and PP lie on the same circle, and DPDP is another diameter of it. Six points so far.
ABCHaHbHcHDEFPQRN
The second rectangle, frozen

DEPQ, with the same diagonal EQ: D and P join the circle.


The Feet of the Altitudes

The foot HaH_a lies on the side BCBC, which also contains DD, and on the altitude AHAH, which also contains PP. The altitude is perpendicular to the side, so

∠DHaP=90°\angle D H_a P = 90°


A point that sees a segment at a right angle lies on the circle with that segment as diameter; this is the converse of Thales' theorem, and it follows from the median to the hypotenuse being half the hypotenuse, as the rectangle page noted. DPDP is a diameter of the circle, so HaH_a lies on it.

The same argument puts HbH_b on the circle, since it sees the diameter EQEQ at a right angle (EE on CACA, QQ on the altitude BHBH), and HcH_c, which sees the diameter FRFR at a right angle. That makes nine points on one circle, and the theorem's first claim is proved.
ABCHaHbHcHDEFPQRN
The feet of the altitudes, frozen

Right angles at the feet, each facing a diameter: nine points on the circle.


The Centre

Where is NN? Bring in the circumcentre OO. The segment ODOD is perpendicular to BCBC, because OO lies on the perpendicular bisector of BCBC; so is the altitude AHAH. Two lines perpendicular to BCBC are parallel, so OD∥AHOD \parallel AH, and in particular OD∥PHOD \parallel PH. The Euler line page showed AH=2⋅ODAH = 2 \cdot OD, so OD=12AH=PHOD = \tfrac{1}{2} AH = PH.

ODHPODHP now has a pair of opposite sides equal and parallel, so it is a parallelogram, and its diagonals bisect each other. Its diagonals are DPDP and OHOH. The midpoint of DPDP is NN, so NN is also the midpoint of OHOH: the nine-point centre is halfway between the circumcentre and the orthocentre, on the Euler line.
ABCHaHbHcHDEFPQRNO
The centre, frozen

ODHP is a parallelogram: N is the midpoint of OH, two ticks each side.


Why It Holds

The same facts give the radius. OD=APOD = AP, since both equal 12AH\tfrac{1}{2} AH, and OD∥APOD \parallel AP. So OAPDOAPD is a parallelogram and its opposite sides are equal: DP=OADP = OA. But DPDP is a diameter of the nine-point circle and OAOA is the circumradius RR, so the nine-point circle has diameter RR, that is, radius 12R\tfrac{1}{2} R. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given points, midsegments twice, rectangle diagonals, the converse of Thales, and the two parallelograms.

Everything was assembled from earlier pages: the midsegment theorem, the rectangle and parallelogram pages, Thales, the Euler line. The nine-point circle is where those results meet.
ABCHaHbHcHDEFPQRNO
The conclusion, frozen

DP = OA = R, three ticks each: the radius is R / 2.


Dragging the Figure

Drag a vertex and watch the nine points move while the circle through them simply changes size and position; the readout keeps the smallest and largest distance from NN equal, and equal to half the circumradius. Watch NN too: it stays halfway between OO and HH, so as the triangle leans the whole Euler line swings and NN swings with it.

The player keeps the triangle acute, so that the feet of the altitudes lie on the sides and the two rectangles are readable, and it keeps OO and HH apart; the theorem holds for every triangle. Make the triangle nearly equilateral and the nine points pair up, each midpoint of a side approaching the foot of an altitude.
ABCHaHbHcHDEFPQRNO
After dragging A, frozen

A pulled right: the nine points have moved and the circle with them, N still halfway along OH.


Consequences

• Three circumcircles in one. The nine-point circle is the circumcircle of the medial triangle DEFDEF, of the orthic triangle HaHbHcH_aH_bH_c, and of the triangle PQRPQR.
• The centroid. The centroid GG lies on the Euler line a third of the way from OO to HH, and NN halfway; shrinking the circumcircle by half about GG, with a half turn, gives the nine-point circle.
• Feuerbach's theorem. The nine-point circle touches the inscribed circle and the three excircles; the point of contact with the incircle is the Feuerbach point.
• An orthocentric system. The four triangles formed by AA, BB, CC, HH taken three at a time share the same nine-point circle, since the four points are each the orthocentre of the other three.
• Right triangles. In a right triangle two of the feet coincide with the right-angle vertex, and the circle still passes through the remaining points, its centre at the midpoint of the median to the hypotenuse.