Why does the perpendicular from the centre bisect a chord?

Joining the centre to the ends of the chord gives two right triangles with equal hypotenuses, the radii, and a common leg, the perpendicular. By the Pythagorean theorem both halves of the chord equal the square root of r² − OM², so they are equal.

Does the perpendicular bisector of a chord pass through the centre?

Yes. The perpendicular from the centre bisects the chord, so the centre lies on the perpendicular bisector. That is how the centre of a circle is found: draw two chords and intersect their perpendicular bisectors.

How long is a chord at distance d from the centre?

2√(r² − d²). Each half of the chord is a leg of a right triangle with hypotenuse r and other leg d.

The Perpendicular from the Centre Bisects a Chord

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
Slide $A$ or $B$ round the circle, or drag $O$ to move the circle; the foot $M$ follows, and the readout under the canvas keeps $AM$ and $MB$ 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?Join the centre O to the ends of the chord. OA and OB are radii, so the triangles OMA and OMB are right triangles with equal hypotenuses and the common leg OM. Pythagoras gives AM² = OA² − OM² and MB² = OB² − OM², the same number, so AM = MB. Slide A or B round the circle: the foot of the perpendicular stays at the midpoint.Perpendicular bisects the chord

DisplayLabelsMeasures

Drag A or B round the circle, O to move the circle

AM = 1.72, MB = 1.72  (OM = 1.57)

Step 1 / 6

Step explanations

1A chord and the perpendicular from the centre

Given
AB is a chord of the circle with centre O, and the perpendicular from O meets the chord at M. To prove: AM = MB, the perpendicular from the centre bisects the chord.

M is built from O, A and B; slide A or B and it follows. 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 slide the ends of the chord round the circle and watch the foot of the perpendicular stay at the midpoint. 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?Perpendicular bisects the chordDisplayLabelsMeasuresDrag A or B round the circle, O to move the circle

AM = 1.72, MB = 1.72  (OM = 1.57)

← Back▶ PlayNext →ResetNormal ▾Step 1 / 6

Step explanations

1A chord and the perpendicular from the centre

Given
AB is a chord of the circle with centre O, and the perpendicular from O meets the chord at M. To prove: AM = MB, the perpendicular from the centre bisects the chord.

1234567

1Proof — the theorem loaded in the player; the ? gives its idea in one line

2Display — Labels shows the point names

3Draggable points — A and B slide round the circle, O moves the circle; the hollow foot M follows

4Readout — the line under the figure: AM and MB in the frame's units, with OM, 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?Perpendicular bisects the chordDisplayLabelsMeasuresDrag A or B round the circle, O to move the circle

AM = 1.72, MB = 1.72  (OM = 1.57)

← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6

Step explanations

1A chord and the perpendicular from the centre

Given
AB is a chord of the circle with centre O, and the perpendicular from O meets the chord at M. To prove: AM = MB, the perpendicular from the centre bisects the chord.

2Join O to A and to B

Construction
OA and OB are radii of the same circle, so OA = OB = r (one tick each). With OM they make two triangles, OMA and OMB, which share the side OM.

3Two right triangles on the common leg OM

Right triangles
OM ⟂ AB, so the angles at M are right: OMA and OMB are right triangles with hypotenuses OA and OB, and OM is a leg of both.

4AM² = OA² − OM², MB² = OB² − OM²

Pythagorean theorem
In the right triangle OMA, OA² = OM² + AM², so AM² = OA² − OM². In the right triangle OMB, in the same way, MB² = OB² − OM².

5The two right-hand sides are the same

Equal radii
OA = OB = r, and OM is the same segment in both formulas. So AM² and MB² are both r² − OM²: AM² = MB².

6AM = MB

Positive roots
Lengths are positive, and positive numbers with equal squares are equal. So AM = MB (two ticks each): M is the midpoint of the chord. ∎

12345

1The two halves — AM and MB, two ticks each at the last step

2The two right triangles — OMA and OMB, equal hypotenuses OA and OB, common leg OM

3The right angle at M — the perpendicular from the centre, whose foot is the midpoint

4Justification tags — Given, Construction, Right triangles, Pythagorean theorem, Equal radii, Positive roots

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

The Statement

A chord is a segment joining two points of a circle. Drop the perpendicular from the centre

O

onto a chord

AB

and call its foot

M

. The theorem, Proposition 3 of Book III of Euclid's Elements, says that the perpendicular bisects the chord:

AM = MB

The converse holds too: the line from the centre to the midpoint of a chord is perpendicular to it. Together they say that the perpendicular bisector of any chord passes through the centre, which is how the centre of a circle is found.

The proof on this page is the same argument as on the two tangents page in a new setting: two right triangles with equal hypotenuses and a common leg, and the Pythagorean theorem applied in each. In the player

M

is hollow, since it is computed from

O

A

and

B

; the readout under the canvas shows

AM

and

MB

in the frame's own units, with

OM

beside them, live while you drag.

Opening scene, frozen

The chord AB and the perpendicular OM from the centre, with its foot M.

Joining the Centre

The construction joins the centre to the two ends of the chord.

OA

and

OB

are radii, so they are equal,

OA = OB = r

; the player marks them with one tick each. The figure is now the isosceles triangle

OAB

with the perpendicular

OM

dropped from its apex onto its base.

The perpendicular cuts the isosceles triangle into two triangles,

OMA

and

OMB

, which share the side

OM

. The proof shows that they are mirror images of each other, and in particular that their bases

AM

and

MB

are equal.

The construction, frozen

OA and OB joined, marked equal: the isosceles triangle OAB cut by OM.

Two Right Triangles

OM

is perpendicular to

AB

, so both angles at

M

are right angles. The triangles

OMA

and

OMB

are therefore right triangles:

OMA

has its right angle at

M

, legs

OM

and

AM

, and hypotenuse

OA

;

OMB

has its right angle at

M

, legs

OM

and

MB

, and hypotenuse

OB

. The player fills them in two colours.

The two triangles have equal hypotenuses, both radii, and share the leg

OM

. Two right triangles with equal hypotenuses and one pair of equal legs are congruent, and that is the shortest way to finish; the next step gets the same result with the Pythagorean theorem, which keeps every step on a page of its own.

The right triangles, frozen

OMA and OMB, right-angled at M, on the common leg OM.

Pythagoras Twice

In the right triangle

OMA

the Pythagorean theorem gives

OA^2 = OM^2 + AM^2

, so

AM^2 = OA^2 - OM^2

In the right triangle

OMB

it gives

OB^2 = OM^2 + MB^2

, so

MB^2 = OB^2 - OM^2

Two formulas of the same shape, one for each half of the chord. What remains is to compare their right-hand sides.

Pythagoras twice, frozen

AM² = OA² − OM² in the orange triangle, MB² = OB² − OM² in the blue one.

The Same Numbers

OA

and

OB

are both radii, so

OA^2 = OB^2 = r^2

; and

OM

is the same segment in both formulas. So

AM^2 = r^2 - OM^2 = MB^2

The squares of the two halves are equal. The formula also gives the length of the chord outright: each half is

\sqrt{r^2 - d^2}

, where

d = OM

is the distance from the centre to the chord, so the chord has length

2\sqrt{r^2 - d^2}

. The nearer a chord is to the centre, the longer it is, and a chord through the centre, a diameter, is the longest of all.

The same numbers, frozen

OA = OB = r and OM shared: both differences are r² − OM².

Why It Holds

From

AM^2 = MB^2

AM = MB

is one step: lengths are positive, and two positive numbers with equal squares are equal. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given chord, a construction, the right angles, the Pythagorean theorem, equal radii, and the positive square root.

Seen differently, the theorem is a fact about isosceles triangles:

OAB

is isosceles with apex

O

, and the perpendicular from the apex to the base is the axis of symmetry of the triangle, so it hits the base at its midpoint. The isosceles base angles page builds its triangle from exactly this picture, with the apex on the perpendicular bisector of the base. The converse of this page's theorem, that the line from

O

to the midpoint is perpendicular, follows from the same symmetry: triangles

OMA

and

OMB

then have three pairs of equal sides, so the angles at

M

are equal, and two equal angles on a straight line are right angles.

The conclusion, frozen

AM = MB, two ticks each.

Dragging the Figure

Slide

A

B

round the circle: the chord tilts and changes length, the foot

M

moves with it, and the readout keeps

AM = MB

to the last digit. Bring

A

and

B

close together and the chord is short and far from the centre, with

OM

nearly a radius; pull them apart and the chord grows,

OM

shrinks, and as the chord approaches a diameter

M

approaches

O

. Drag

O

to move the circle.

A drag that would take the chord through the centre, or shrink it to a point, is refused: with

M = O

the two triangles collapse into the two halves of a diameter, and the theorem is still true but there is nothing left to draw.

After dragging B, frozen

B slid down the circle: the chord is tilted and longer, M nearer the centre, still the midpoint.

Consequences

• Finding the centre. The perpendicular bisector of every chord passes through the centre, so the centre is where the perpendicular bisectors of two chords meet (Euclid III.1). The same construction draws the circle through three given points.
• Equal chords. Chords at the same distance from the centre are equal, and equal chords are equally distant from the centre, since each half is

\sqrt{r^2 - d^2}

(Euclid III.14).
• Chord length. A chord at distance

d

from the centre has length

2\sqrt{r^2 - d^2}

; the distance

d

is the length of the perpendicular, which is the shortest segment from the centre to the chord's line, as the perpendicular is shortest page shows.
• The chord of contact. For two tangents from an outside point

P

, the line

OP

is perpendicular to the chord

AB

joining the points of contact and bisects it, by this theorem applied to the chord

AB

.
• Arcs. The perpendicular from the centre also bisects the arc cut off by the chord, since it is the axis of symmetry of the whole figure.