What does it mean for chords to be equidistant from the centre?

The distance from the centre to a chord is the length of the perpendicular from the centre to the chord. Two chords are equidistant from the centre when these two perpendiculars are equal.

Why are equal chords equidistant from the centre?

The perpendicular from the centre bisects each chord, so equal chords have equal half-chords. Together with the equal radii as hypotenuses, the two right triangles formed are congruent by hypotenuse-leg, so the perpendiculars are equal.

Is the converse true?

Yes. If two chords are the same distance from the centre, the right triangles have equal hypotenuses (the radii) and equal legs (the distances), so they are congruent and the half-chords, hence the chords, are equal.

Equal Chords and the Centre

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 $B$ to change the chord length ( $D$ follows), $C$ to move the second chord, $A$ for the radius or $O$ for the centre; the readout under the canvas shows the two chords and the two distances. Learn more about why it holds
Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene

Proof?AB and CD are equal chords. Drop the perpendiculars OM and ON from the centre: each meets its chord at the midpoint, so AM = AB/2 = CD/2 = CN. With the radii OA = OC, the right triangles OMA and ONC have equal hypotenuses and equal legs, so they are congruent (HL) and OM = ON. Drag B to change the chord length (D follows), C to move the second chord, A for the radius, O for the centre.Equal chords

DisplayLabelsMeasures

Drag B (chord length, D follows) or C round the circle, A for the radius, O for the centre

AB = 3.26,  CD = 3.26,  OM = 1.95,  ON = 1.95

Step 1 / 6

Step explanations

1Two equal chords

Given
AB and CD are two chords of the circle with centre O, and they are equal: AB = CD (one tick each). To prove: the two chords are the same distance from the centre, that is, the perpendiculars from O to the two chords are equal.

Drag B: both chords change length together. Learn more about the statement

Getting Started with the Visualizer

The Statement

The Perpendiculars from the Centre

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 7

Step through the proof, then drag

B

to change the length of the chords (

D

follows so the second chord stays equal), or

C

to move the second chord round the circle, and watch the two distances in the readout stay equal. 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?Equal chordsDisplayLabelsMeasuresDrag B (chord length, D follows) or C round the circle, A for the radius, O for the centre

AB = 3.26,  CD = 3.26,  OM = 1.95,  ON = 1.95

← Back▶ PlayNext →ResetNormal ▾Step 1 / 6

Step explanations

1Two equal chords

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

2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc

3Draggable points — B slides round the circle and sets the length of both chords, C carries the second chord round, A sets the radius, O moves the circle; the hollow points D, M and N follow

4Readout — the line under the figure: the two chords and the two distances OM and ON, 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?Equal chordsDisplayLabelsMeasuresDrag B (chord length, D follows) or C round the circle, A for the radius, O for the centre

AB = 3.26,  CD = 3.26,  OM = 1.95,  ON = 1.95

← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6

Step explanations

1Two equal chords

2The perpendiculars OM and ON

Construction
Drop the perpendicular from O to each chord: OM ⟂ AB with M on AB, and ON ⟂ CD with N on CD. The distance from the centre to a chord is the length of this perpendicular, so the claim is OM = ON.

3AM = CN

Perpendicular bisects the chord
The perpendicular from the centre bisects the chord, so M is the midpoint of AB and N the midpoint of CD: AM = AB/2 and CN = CD/2. Since AB = CD, AM = CN (two ticks).

4OA = OC

Equal radii
Join O to A and to C. Both are radii of the circle, so OA = OC (three ticks). They are the hypotenuses of the right triangles OMA and ONC.

5OMA ≅ ONC

HL congruence
The triangles OMA and ONC are right-angled at M and N, with equal hypotenuses OA = OC and equal legs AM = CN. By the hypotenuse-leg criterion they are congruent.

6OM = ON

Corresponding parts
The remaining legs of the congruent triangles are equal: OM = ON. Equal chords are equidistant from the centre. Read backwards (OM = ON and OA = OC give AM = CN by HL, so AB = CD), the same figure proves the converse: chords equidistant from the centre are equal. ∎

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1The two right triangles — OMA and ONC shaded, with the square marks at M and N

2The ticks — one tick for the equal chords, two for the half-chords AM = CN, three for the radii OA = OC

3The two distances — OM and ON, the perpendiculars from the centre, equal at the end

4Justification tags — Given, Construction, Perpendicular bisects the chord, Equal radii, HL congruence, Corresponding parts

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

The Statement

The distance from the centre of a circle to a chord is the length of the perpendicular from the centre to the chord. The theorem (Euclid III.14) says that equal chords are equidistant from the centre, and conversely chords equidistant from the centre are equal: for chords

AB

and

CD

with perpendiculars

OM

and

ON

AB = CD \iff OM = ON

It is the circle's version of a simple idea: a chord is determined, up to turning it about the centre, by how far it is from the centre. Longer chords are closer, the diameter passes through the centre itself.

In the player

D

is built as

C

turned about

O

by the angle

AOB

, so the chord

CD

is always equal to

AB

; drag

B

to change both lengths at once, or

C

to move the second chord. The readout shows the two chords and the two distances.

Opening scene, frozen

Two equal chords AB and CD, one tick each.

The Perpendiculars from the Centre

Drop the perpendicular from

O

to each chord:

OM \perp AB

with

M

AB

, and

ON \perp CD

with

N

CD

. The square marks show the right angles. The claim to prove is now a statement about two segments:

OM = ON

Each is the shortest segment from

O

to its chord, which is why it is called the distance (the perpendicular is shortest).

The perpendiculars from the centre, frozen

OM ⟂ AB and ON ⟂ CD, with square marks at M and N.

Half of Each Chord

The perpendicular from the centre bisects the chord:

M

is the midpoint of

AB

and

N

the midpoint of

CD

. So

AM = \tfrac12 AB = \tfrac12 CD = CN

The two half-chords are equal (two ticks); they will be one pair of legs in a pair of right triangles.

Half of each chord, frozen

AM = CN, two ticks: the perpendicular bisects each chord.

The Radii

Join

O

A

and to

C

. Both are radii, so

OA = OC

(three ticks). In the right triangles

OMA

and

ONC

these radii are the hypotenuses, opposite the right angles at

M

and

N

; the half-chords

AM

and

CN

are one pair of legs, and the distances

OM

and

ON

are the other.

The radii, frozen

OA = OC, three ticks; the right triangles OMA and ONC shaded.

Hypotenuse and Leg

The triangles

OMA

and

ONC

are right-angled, with equal hypotenuses

OA = OC

and equal legs

AM = CN

. By the hypotenuse-leg criterion they are congruent:

\triangle OMA \cong \triangle ONC

(Without HL:

OM^2 = OA^2 - AM^2 = OC^2 - CN^2 = ON^2

by Pythagoras, which proves the same thing in one line.)

Hypotenuse and leg, frozen

Equal hypotenuses, equal legs: OMA ≅ ONC.

Why It Holds

Corresponding parts of congruent triangles are equal, so the remaining legs agree:

OM = ON

Equal chords are equidistant from the centre. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, a construction, the perpendicular bisecting the chord, equal radii, HL congruence and corresponding parts.

The converse uses the same picture read backwards: if

OM = ON

, the right triangles

OMA

and

ONC

have equal hypotenuses (the radii) and equal legs

OM = ON

, so they are congruent by HL,

AM = CN

, and doubling gives

AB = CD

The conclusion, frozen

OM = ON: equal chords are equidistant from the centre.

Dragging the Figure

Drag

B

round the circle: both chords change length together, since

D

is placed so that

CD = AB

, and the two distances in the readout change together and stay equal; a longer chord sits closer to the centre. Drag

C

to carry the second chord anywhere round the circle without changing its length. Drag

A

to change the radius or

O

to move the circle.

The player keeps the chords from passing through the centre (then

M

and

N

would coincide with

O

), keeps them apart from each other, and keeps the figure in the frame. A drag that would break one of these is refused.

After dragging B, frozen

B slid round the circle: both chords longer and both nearer the centre, still equally far.

Consequences

• Longer is nearer. The same right triangle gives

OM^2 = r^2 - (AB/2)^2

: of two unequal chords the longer is nearer the centre, and the diameter, the longest chord, passes through it (Euclid III.15).
• The locus of midpoints. The midpoints of all chords of a given length lie on a circle about

O

of radius

OM

; each such chord is tangent to that inner circle (tangent perpendicular to radius).
• Regular polygons. The sides of a regular polygon inscribed in a circle are equal chords, so they are all the same distance from the centre: that distance is the apothem, the radius of the inscribed circle.
• Congruent arcs. Equal chords cut off equal arcs and subtend equal angles at the centre; the inscribed angle theorem then makes them subtend equal angles at the circumference.
• Equal tangents. The two tangents from a point are proved equal by the same HL argument on the radii to the points of contact.