Triangle Centres & the Euler Line

Drag any vertex. Three of the four centres never leave the red line, and the fourth lands on it only when the triangle is isosceles.

Worked example — Triangle Centres & the Euler Line

Centroid G	(2.667, 1.667)
Circumcentre O	(3.000, 1.700)
Orthocentre H	(2.000, 1.600)
Incentre I	(2.491, 1.687)
HG : GO	2.000
Incentre off the line	0.0372

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Triangle Centres & the Euler Line

A triangle centre is a point picked out by the triangle itself rather than by the page it is drawn on: move or turn or scale the triangle and the point follows. Four of them are classical, and each is where one family of three lines happens to meet. The medians meet at the centroid, the perpendicular bisectors of the sides at the circumcentre, the altitudes at the orthocentre, and the angle bisectors at the incentre. That those three lines meet at all, in each of the four cases, is already a theorem.

What each symbol means

G: the centroid, the average of the three vertices, and the point the triangle would balance on
O: the circumcentre, equidistant from all three vertices, so the circle through them is centred there
H: the orthocentre, where the three altitudes meet; the only one of the four with no circle of its own
I: the incentre, equidistant from all three sides, and the only one that is always inside

Breaks when: On the equilateral triangle, which is the one case this page has nothing to say about. All four centres collapse to the same point, every line through it contains all of them, and the Euler line is not any particular line. The page prints "no line at all" there rather than choosing one.

Three constructions that share nothing, and one line 🖖

The centroid is where the three medians cross. The circumcentre is where the three perpendicular bisectors cross. The orthocentre is where the three altitudes cross. Those are three different families of line, and no one of the three definitions mentions either of the others.

Drag a vertex anywhere you like. The three points move, and they stay on one straight line.

The centroid is always between the other two, and always in the same place: HG is exactly twice GO. The HG : GO row reads 2.000 no matter how you drag. Euler worked it out in 1763 and the paper appeared in 1767; the line has carried his name ever since.

The incentre is invited only when two sides are equal 🖖

The fourth centre is where the angle bisectors cross. It is the one that is always inside the triangle, which is why a circle can be inscribed at it, and it is the one Euler's line does not pass through.

Not usually, anyway. The lower chart plots the incentre's distance from the line as the selected vertex slides sideways, and on the triangle this page opens on the curve touches zero five times. Each touch is a triangle with two equal sides: one where C sits above the midpoint of AB, two where CA has grown to match AB, and two where CB has.

That is the whole condition. An isosceles triangle has an axis of symmetry, every centre has to sit on it, and the Euler line is that axis.

At exactly 90° one centre lands on an edge and another on a corner 🖖

An acute triangle keeps all four centres inside itself. Push one angle past 90° and two of them leave together, the circumcentre and the orthocentre, and they leave on opposite sides.

The moment in between is worth stopping at. Press Right angle. The Circumcentre O row reads (2.000, 1.500), which is the midpoint of the hypotenuse from (4, 0) to (0, 3), and the Orthocentre H row reads (0.000, 0.000), which is the right-angle vertex itself. One on an edge, one on a corner, both on the boundary at the same instant.

The incentre never leaves. A circle that fits inside the triangle cannot have its centre anywhere else.

Problem solved in full

A line nobody drew 6 steps
A triangle has vertices A = (0, 0), B = (6, 0) and C = (2, 5). Find the centroid, the circumcentre and the orthocentre from their own definitions, show that the three are collinear, and find the ratio in which the centroid divides the segment between the other two.

Open this problem in the tool
1. The centroid first, because it is the one that needs no geometry. The medians meet at the average of the vertices, so G = ((0 + 6 + 2)/3, (0 + 0 + 5)/3) = (8/3, 5/3). The Centroid G row reads (2.667, 1.667).
  
  $G = \frac{A + B + C}{3} = (\frac{0 + 6 + 2}{3}, \frac{0 + 0 + 5}{3}) = (\frac{8}{3}, \frac{5}{3})$
2. The circumcentre is the point equidistant from all three vertices. Equidistant from A and B puts it on x = 3, the perpendicular bisector of a horizontal segment. Equidistant from A and C gives 3² + y² = (3 − 2)² + (y − 5)², and the y² cancels off both sides: 9 = 1 + 25 − 10y, so 10y = 17 and y = 1.7. O = (3, 1.7), and the Circumcentre O row agrees.
  
  $3^{2} + y^{2} = (3 - 2)^{2} + (y - 5)^{2} \Rightarrow 9 = 26 - 10 y \Rightarrow y = 1.7, O = (3, 1.7)$
3. The orthocentre needs two altitudes and nothing from step 2. The altitude from A is perpendicular to BC, whose direction is C − B = (−4, 5), so every point on it satisfies −4x + 5y = 0. The altitude from B is perpendicular to AC, direction (2, 5), and passes through (6, 0), so 2(x − 6) + 5y = 0. Substituting y = 4x/5 into the second gives 2x − 12 + 4x = 0, so x = 2 and y = 1.6. H = (2, 1.6).
  
  $- 4 x + 5 y = 0, 2 (x - 6) + 5 y = 0 \Rightarrow 6 x = 12 \Rightarrow H = (2, 1.6)$
4. Now the claim. From O the step to G is (8/3 − 3, 5/3 − 17/10) = (−1/3, −1/30), and the step to H is (2 − 3, 1.6 − 1.7) = (−1, −1/10). Two vectors are parallel when their cross product vanishes: (−1/3)(−1/10) − (−1/30)(−1) = 1/30 − 1/30 = 0. It vanishes exactly, in fractions, with no rounding anywhere.
  
  $(G - O) = (- \frac{1}{3}, - \frac{1}{30}), (H - O) = (- 1, - \frac{1}{10}), (- \frac{1}{3}) (- \frac{1}{10}) - (- \frac{1}{30}) (- 1) = 0$
5. The second vector is three times the first, so H is three times as far from O as G is, and G lies between them. That makes HG two parts to GO's one. In lengths, GO = √(1/9 + 1/900) = √101/30 = 0.334996 and HG = √(4/9 + 1/225) = √101/15 = 0.669992. The HG : GO row reads 2.000, and the √101 cancels rather than being rounded away.
  
  $GO = \frac{101}{30}, H G = \frac{101}{15}, \frac{H G}{GO} = 2$
6. The fourth centre does not join in. With a = |BC| = √41, b = |CA| = √29 and c = |AB| = 6, the incentre is (aA + bB + cC)/(a + b + c) = (2.491, 1.687). The Incentre off the line row reads 0.0372 — small, and not zero. Drag C to x = 3 and it becomes 0.0000, because the triangle is then isosceles.
  
  $I = \frac{a A + b B + c C}{a + b + c}, a = 41, b = 29, c = 6 \Rightarrow I = (2.491, 1.687)$
Answer

Exactly 2, and exactly 2 for every triangle you can draw. The centroid is not merely somewhere on the segment from the orthocentre to the circumcentre; it is two thirds of the way along it, always. What the tool cannot show you is why, because dragging a vertex is evidence and not a proof. The shortest proof takes the circumcentre as the origin, so that |A| = |B| = |C| = R, and looks at the single point P = A + B + C. Then P − A = B + C, and (B + C)·(C − B) = |C|² − |B|² = 0, so P sits on the altitude from A. The same two lines put it on the other two altitudes, which makes P the orthocentre. The centroid is P/3, one third of the way from the origin to P, and the origin was the circumcentre.

$H - O = 3 (G - O) ⟹ \frac{H G}{GO} = 2$

References (3)

The paper the line is named after, in which Euler works out all four centres in coordinates and then notices that three of them are collinear: L. Euler, "Solutio facilis problematum quorundam geometricorum difficillimorum." Novi Commentarii academiae scientiarum Petropolitanae 11 (for 1765), printed 1767, pp. 103–123. Euler Archive E325; reprinted in Opera Omnia I.26, pp. 139–157.
The nine-point circle, the fifth point this page draws and the one with the longest list of claimants: K. W. Feuerbach, Eigenschaften einiger merkwürdigen Punkte des geradlinigen Dreiecks. Wiessner, Nürnberg, 1822.
A modern textbook treatment of the four centres and the line, at the level this page is pitched: H. S. M. Coxeter and S. L. Greitzer, Geometry Revisited. Mathematical Association of America, New Mathematical Library 19, 1967, chapter 1.

By the 42Math team · Reviewed against published sources. Corrections welcome.