What is the circumcentre of a triangle?

The point where the three perpendicular bisectors of the sides meet. It is equidistant from the three vertices, so it is the centre of the circle through them, the circumcircle.

Why do the three perpendicular bisectors meet at one point?

The crossing point of two of them is equidistant from all three vertices, by two side-angle-side congruences. A point equidistant from A and C lies on the perpendicular bisector of AC, so the third bisector passes through the same point.

Is the circumcentre always inside the triangle?

No. It is inside for an acute triangle, on the hypotenuse for a right triangle, and outside for an obtuse triangle.

The Perpendicular Bisectors Meet at the Circumcentre

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 midpoints, the bisectors and $O$ follow, and the readout under the canvas keeps the three distances from $O$ equal. 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?The perpendicular bisectors of AB and BC meet at O. At the midpoint P of AB the two right triangles OPA and OPB are congruent (SAS), so OA = OB; at the midpoint Q of BC, OB = OC in the same way. So OA = OC, the triangle OAC is isosceles, and its median OR is perpendicular to AC: the third perpendicular bisector passes through O too, and the circle about O through A passes through B and C. Drag any vertex.Circumcentre

DisplayLabelsMeasures

Drag A, B or C; the bisectors and O follow

OA = 3.09, OB = 3.09, OC = 3.09

Step 1 / 6

Step explanations

1Two perpendicular bisectors meet at O

Given
P and Q are the midpoints of AB and BC; the dashed lines are the perpendicular bisectors of those sides, and they meet at O. To prove: OA = OB = OC, and the perpendicular bisector of AC passes through O as well.

The midpoints and O are computed; drag a vertex and they follow. Learn more about the statement

Getting Started with the Visualizer

The Statement

The First Bisector

The Second Bisector

Three Equal Distances

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 perpendicular bisectors keep meeting at one point, with the circle about it still passing through all three corners. 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

Proof?CircumcentreDisplayLabelsMeasuresDrag A, B or C; the bisectors and O follow

OA = 3.09, OB = 3.09, OC = 3.09

← Back▶ PlayNext →ResetNormal ▾Step 1 / 6

Step explanations

1Two perpendicular bisectors meet at O

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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 midpoints P, Q, later R, and the crossing point O follow

4Readout — the line under the figure: the distances OA, OB, OC in the frame's units, 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?CircumcentreDisplayLabelsMeasuresDrag A, B or C; the bisectors and O follow

OA = 3.09, OB = 3.09, OC = 3.09

← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6

Step explanations

1Two perpendicular bisectors meet at O

2OA = OB

SAS congruence
Join O to A and B. The triangles OPA and OPB share the side OP, have PA = PB, and have right angles at P between those sides. So they are congruent, and OA = OB.

3OB = OC

SAS congruence
Join O to C. In the same way the triangles OQB and OQC share OQ, have QB = QC, and have right angles at Q: congruent, so OB = OC.

4OA = OB = OC

Equal to the same
OA = OB and OB = OC, so all three distances are equal (three ticks each): O is equidistant from the three vertices.

5The third perpendicular bisector passes through O

Isosceles median
OA = OC, so triangle OAC is isosceles with apex O. Its median OR, to the midpoint R of AC, is perpendicular to AC: the triangles ORA and ORC agree in three sides, so their angles at R are equal and form a straight angle. Hence OR is the perpendicular bisector of AC, and it passes through O.

6The circle about O through A passes through B and C

Definition of a circle
A circle is the set of points at one distance from its centre. OA = OB = OC, so the circle with centre O and radius OA passes through all three vertices: the circumcircle, with O its circumcentre. ∎

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1The circumcentre O — where the three perpendicular bisectors meet, equidistant from A, B, C

2The circumcircle — centre O, radius OA; it passes through all three vertices

3The third bisector OR — the median of the isosceles triangle OAC, perpendicular to AC

4Justification tags — Given, SAS congruence (twice), Equal to the same, Isosceles median, Definition of a circle

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

The Statement

The perpendicular bisector of a segment is the line through its midpoint at right angles to it. A triangle has three sides, so three perpendicular bisectors, and the theorem says they all pass through one point

O

, which is at the same distance from the three vertices:

OA = OB = OC

The circle with centre

O

and radius

OA

therefore passes through all three vertices; it is the circumscribed circle, or circumcircle, and

O

is the circumcentre. This is the fact behind drawing a circle through three given points, and it is how the centre of a circle is found from any three points on it, as on the chord bisector page.

The proof has the shape of every concurrence proof on these pages: take two of the three lines, find their crossing point, show it has a property, and show that the property forces the third line through it. In the player the midpoints and

O

are hollow, computed from the vertices. The readout under the canvas shows the three distances from

O

to the vertices, in the frame's own units, live while you drag.

Opening scene, frozen

The perpendicular bisectors of AB and BC, dashed, meeting at the hollow point O.

The First Bisector

Take the perpendicular bisector of

AB

and the perpendicular bisector of

BC

. They are not parallel, since

AB

and

BC

are not, so they cross at a point; call it

O

. Join

O

A

and to

B

.

At the midpoint

P

AB

there are now two right triangles,

OPA

and

OPB

. They share the side

OP

, they have

PA = PB

because

P

is the midpoint, and the angles between those sides are both right angles because the line is perpendicular to

AB

. Two sides and the included angle: the triangles are congruent by side-angle-side, and

OA = OB

Every point on the perpendicular bisector of

AB

is equidistant from

A

and

B

; this step proves that general fact for the point

O

The first bisector, frozen

OPA and OPB: OP shared, PA = PB, right angles at P; so OA = OB.

The Second Bisector

The same argument at the midpoint

Q

BC

. Join

O

C

. The right triangles

OQB

and

OQC

OQ

, have

QB = QC

, and have right angles at

Q

, so they are congruent by side-angle-side and

OB = OC

O

lies on the perpendicular bisector of

BC

, so it is equidistant from

B

and

C

. The player fills the second pair of triangles in two more colours.

The second bisector, frozen

OQB and OQC in the same way: OB = OC.

Three Equal Distances

OA = OB

from the first bisector and

OB = OC

from the second, so

OA = OB = OC

all three marked with three ticks. The point

O

is at one distance from all three vertices. That is already half the theorem: the two bisectors meet at a point equidistant from

A

B

and

C

. What remains is to show that the third perpendicular bisector cannot miss this point.

Three equal distances, frozen

OA = OB = OC, three ticks each.

The Third Bisector

Since

OA = OC

, the triangle

OAC

is isosceles with apex

O

. Let

R

be the midpoint of

AC

and draw the median

OR

. The triangles

ORA

and

ORC

have

OA = OC

RA = RC

and the common side

OR

, so they are congruent by side-side-side, and their angles at

R

are equal. Those two angles make up the straight angle along

AC

, so each is

90°

OR \perp AC

. This is the argument of the rhombus diagonals page, where the same isosceles median appeared.

So

OR

is a line through the midpoint of

AC

perpendicular to

AC

: it is the perpendicular bisector of

AC

, and it passes through

O

. All three perpendicular bisectors meet at

O

The third bisector, frozen

The isosceles median OR, perpendicular to AC: the third bisector passes through O.

Why It Holds

A circle is the set of all points at a fixed distance from its centre. Since

OA = OB = OC

, the circle with centre

O

and radius

OA

passes through

A

B

and

C

: the circumcircle, drawn at the last step. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given bisectors, side-angle-side twice, equality to the same thing, the isosceles median, and the definition of a circle.

The proof also shows that the circumcircle is the only circle through the three vertices. Any centre of such a circle is equidistant from

A

and

B

, so it lies on the perpendicular bisector of

AB

; likewise on that of

BC

; and two lines meet in at most one point. Three points that are not in a line determine exactly one circle.

The conclusion, frozen

The circumcircle about O through A, B and C.

Dragging the Figure

Drag a vertex and watch the circumcentre move. For a triangle with all angles acute,

O

is inside. Pull

A

down until the angle at

A

is a right angle and

O

lands on the midpoint of

BC

: the hypotenuse is a diameter, which is Thales' theorem read backwards. Pull

A

down further, making the angle obtuse, and

O

moves outside the triangle, beyond

BC

. The readout keeps the three distances equal throughout.

A drag that would flatten the triangle, or carry a vertex or the circumcentre out of the frame, is refused; a very flat triangle sends

O

far away, which is the geometry of the theorem, not a limit of it.

After dragging A, frozen

A pulled down and right: the angle at A is wider and O sits lower, nearer the base, still equidistant from the vertices.

Consequences

• One circle through three points. Three points not in a line lie on exactly one circle, with centre at the crossing of two perpendicular bisectors.
• Where the centre is. Inside an acute triangle, at the midpoint of the hypotenuse of a right triangle, outside an obtuse triangle.
• The circumradius. The Law of Sines gives

a / \sin A = 2R

, where

R = OA

is the circumradius; the circumcentre is also where the inscribed angle theorem's central angles live, with

\angle BOC = 2\angle A

.
• Equidistant from the vertices, not the sides. The incentre, where the angle bisectors meet, is equidistant from the sides instead; the two centres coincide only in an equilateral triangle.
• The Euler line. The circumcentre, the centroid and the orthocentre lie on one line, with the centroid a third of the way from the circumcentre to the orthocentre.