What is the Euler line?

The line through the circumcentre, the centroid and the orthocentre of a triangle. The centroid lies between the other two, a third of the way from the circumcentre to the orthocentre.

Why are the circumcentre, centroid and orthocentre collinear?

The segment from the orthocentre to the point opposite A on the circumcircle is bisected by the midpoint of BC, so the median AD and the segment HO are both medians of the triangle AHA′. They meet at its centroid, which is also the centroid G of ABC; so G lies on HO.

Is the incentre on the Euler line?

Only in an isosceles triangle, where the line is the axis of symmetry and every centre lies on it. In a general triangle the incentre is off the line.

The Euler Line

How to use

▶ 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
Drag $A$ , $B$ or $C$ ; the three centres and the construction follow, and the readout under the canvas keeps $HG / GO$ at $2$ and the angle $HGO$ at $180°$ . Learn more about why it holds
Labels toggles the point names. The Step explanations log keeps every step so far. Learn more about reading a scene

Proof?Let A′ be the point of the circumcircle opposite A. The angles at B and C in the semicircle are right, so A′B ∥ CH and A′C ∥ BH and BHCA′ is a parallelogram: its diagonals bisect each other, so the midpoint D of BC is also the midpoint of HA′. Then AD and HO are two medians of triangle AHA′, meeting at its centroid, two thirds of the way from A along AD: that point is G. So H, G, O are collinear and HG = 2 GO. Drag any vertex.Euler line

DisplayLabelsMeasures

Drag A, B or C; O, G, H and the construction follow

HG / GO = 2.00,  angle HGO = 180.0°

Step 1 / 6

Step explanations

1Three centres of one triangle

Given
O is the circumcentre of ABC, G its centroid and H its orthocentre, each found on its own page. To prove: H, G and O lie on one line, with HG = 2 GO.

O, G and H are computed as on their own pages; drag a vertex and they follow. Learn more about the statement

Getting Started with the Visualizer

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 7

Step through the proof, then drag the vertices and watch the three centres slide along one line, always in the same order and the same proportion. 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 hollow centres O, G, H and every construction point follow

4Readout — the line under the figure: the ratio HG / GO and the angle HGO, live while you drag

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 6 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 6 of 6

Proof?Euler lineDisplayLabelsMeasuresDrag A, B or C; O, G, H and the construction follow

HG / GO = 2.00,  angle HGO = 180.0°

← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6

Step explanations

1Three centres of one triangle

Given
O is the circumcentre of ABC, G its centroid and H its orthocentre, each found on its own page. To prove: H, G and O lie on one line, with HG = 2 GO.

2The point A′ opposite A on the circumcircle

Construction
Draw the circumcircle and the diameter from A through O to A′. Join A′ to B and to C, and draw the altitudes from B and from C, which pass through H.

3BHCA′ is a parallelogram

Thales + parallels
AA′ is a diameter, so the angles at B and C in the semicircle are right: A′B ⟂ AB and A′C ⟂ AC. The altitude from C is also perpendicular to AB, so A′B ∥ CH; the altitude from B is perpendicular to AC, so A′C ∥ BH. Both pairs of opposite sides parallel: BHCA′ is a parallelogram.

4D is the midpoint of HA′

Parallelogram diagonals
The diagonals of a parallelogram bisect each other. The diagonals of BHCA′ are BC and HA′, so they cross at the midpoint D of BC, and D is the midpoint of HA′ as well: HD = DA′.

5AD and HO are medians of triangle AHA′

Medians and centroid
In triangle AHA′, D is the midpoint of the side HA′, so AD is a median; and O is the midpoint of the side AA′, so HO is a median. The medians of a triangle meet at its centroid, which lies two thirds of the way from A along AD.

6G lies on HO, and HG = 2 GO

Same point
AD is also a median of the original triangle ABC, and the point two thirds of the way from A along it is the centroid G of ABC. So G is the centroid of AHA′ too: it lies on the median HO, two thirds of the way from H, which gives HG = 2 GO. The line through H, G and O is the Euler line. ∎

12345

1The Euler line — dashed, through H, G and O, with G a third of the way from O to H

2The parallelogram BHCA′ — two sides are altitudes, two are chords to the antipode A′

3The triangle AHA′ — with its medians AD and HO meeting at G

4Justification tags — Given, Construction, Thales + parallels, Parallelogram diagonals, Medians and centroid, Same point

5Replay — at the last step Play becomes Replay and Next is off

The Statement

A triangle has a circumcentre

O

, where the perpendicular bisectors meet; a centroid

G

, where the medians meet; and an orthocentre

H

, where the altitudes meet. In an equilateral triangle the three coincide; in any other triangle they are three different points, and Euler showed in 1765 that they always lie on one line, with the centroid between the other two and twice as close to the circumcentre:

H, G, O \text{ collinear}, \qquad HG = 2 \cdot GO

The line is the Euler line. The incentre is not on it, except in the equilateral case.

The proof builds a second triangle,

AHA'

, whose centroid turns out to be

G

and whose medians are the segments

AD

and

HO

; the centroid page then does the rest. In the player all three centres are hollow, computed from the vertices, as are the points of the construction. The readout under the canvas shows the ratio

HG / GO

and the angle

HGO

, which stay at

2

and

180°

while you drag.

Opening scene, frozen

The triangle with its three centres O, G, H, all hollow.

The Antipode of A

The construction starts with the circumcircle, centre

O

through

A

B

C

. Draw the diameter from

A

through

O

and call its other end

A'

, the point of the circle opposite

A

; by construction

O

is the midpoint of

AA'

. Join

A'

B

and to

C

.

Also draw the two altitudes from

B

and from

C

; they pass through

H

, which is what the orthocentre is. These two altitudes and the two new segments

A'B

A'C

will turn out to make a parallelogram, and that parallelogram is the key to the proof.

The antipode of A, frozen

The circumcircle, the diameter AA′, the segments A′B and A′C, and the altitudes from B and C.

A Parallelogram

AA'

is a diameter, so by Thales' theorem the angles at

B

and at

C

in the semicircle are right angles:

A'B \perp AB

and

A'C \perp AC

. The altitude from

C

is also perpendicular to

AB

, and two lines perpendicular to the same line are parallel, so

A'B \parallel CH

. In the same way the altitude from

B

is perpendicular to

AC

, so

A'C \parallel BH

.

The quadrilateral

BHCA'

has both pairs of opposite sides parallel, so it is a parallelogram. The player fills it and marks the two pairs of parallels with chevrons. It is a strange-looking parallelogram, with

H

inside the triangle and

A'

outside, but the sides are parallel in pairs and that is all a parallelogram needs.

The parallelogram, frozen

Right angles at B and C in the semicircle: A′B ∥ CH and A′C ∥ BH, so BHCA′ is a parallelogram.

The Midpoint D

The diagonals of a parallelogram bisect each other. The diagonals of

BHCA'

are

BC

and

HA'

, so they cross at the midpoint of

BC

, which is

D

, and

D

is also the midpoint of

HA'

HD = DA'

The player draws

HA'

and marks its two halves. This is the fact the whole proof turns on: the midpoint of the side

BC

of the original triangle is the midpoint of the segment from the orthocentre to the antipode of

A

. (It also shows, by the midsegment theorem in triangle

AHA'

, that

OD = \tfrac{1}{2} AH

; that is a consequence, not a step.)

The midpoint D, frozen

The diagonals BC and HA′ bisect each other at D.

The Medians of AHA′

Now look at the triangle

AHA'

, filled in the player.

D

is the midpoint of its side

HA'

, so

AD

is a median of this triangle.

O

is the midpoint of its side

AA'

, so

HO

is a median of this triangle too.

The medians of a triangle meet at its centroid, which lies two thirds of the way from each vertex along its median. So the two medians

AD

and

HO

of triangle

AHA'

meet at a point two thirds of the way from

A

along

AD

, and two thirds of the way from

H

along

HO

The medians of AHA′, frozen

AD and HO are medians of the filled triangle AHA′; O is the midpoint of AA′.

Why It Holds

The segment

AD

joins the vertex

A

of the original triangle to the midpoint of the opposite side, so it is a median of

ABC

as well, and the point two thirds of the way from

A

along it is the centroid

G

ABC

. That is the same point found in the previous step. So

G

is the centroid of

AHA'

too, and it lies on the median

HO

of that triangle, two thirds of the way from

H

G \in HO, \qquad HG = \tfrac{2}{3} HO, \qquad GO = \tfrac{1}{3} HO, \qquad HG = 2 \cdot GO

The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the three given centres, a construction, Thales with parallels, the parallelogram's diagonals, the medians of the second triangle, and the identification of the two centroids.

The argument is a chain of earlier pages: Thales, the parallelogram pages, the centroid. Each was proved on its own; here they are only assembled.

The conclusion, frozen

The dashed Euler line through H, G and O.

Dragging the Figure

Drag a vertex and watch the three centres move:

H

fastest,

O

slowest,

G

in between, and always on one line with

G

a third of the way from

O

H

. The readout keeps the ratio at

2

and the angle at

180°

. Make the triangle nearly isosceles and the three points crowd together on the axis of symmetry; make it equilateral and they coincide, which is why the player refuses triangles with

H

and

O

too close to tell apart.

The player also keeps the triangle acute, so that

H

stays inside and the picture of the parallelogram is readable, and it keeps the circumcircle within the frame. The theorem itself holds for every triangle; for an obtuse one

H

and

O

are both outside, on opposite sides, with

G

still between them.

After dragging A, frozen

A pulled further right: the centres spread out along the line, G still a third of the way from O to H.

Consequences

• AH = 2 OD. In triangle

AHA'

OD

is a midsegment, so

OD \parallel AH

and

AH = 2 \cdot OD

: the distance from a vertex to the orthocentre is twice the distance from the circumcentre to the opposite side.
• The nine-point circle. The midpoint of

HO

is the centre of a circle through the three midpoints of the sides, the three feet of the altitudes and the three midpoints of

AH

BH

CH

; its radius is half the circumradius, and its centre is also on the Euler line.
• Special triangles. In a right triangle

H

is the right-angle vertex and

O

the midpoint of the hypotenuse, so the Euler line is the median to the hypotenuse. In an isosceles triangle it is the axis of symmetry. In an equilateral triangle the three centres coincide and there is no line.
• With vectors. Taking

O

as origin,

G = \tfrac{1}{3}(A + B + C)

and

H = A + B + C

, so

H = 3G

: the three points are on one line through the origin, with

OH = 3 \cdot OG

.
• The incentre. It lies on the Euler line only for isosceles triangles, where everything lies on the axis.