Circles: Properties of Chords and Angles
Angle Subtended by a Chord at the Centre
A chord connects two points on the circumference of a circle. When the endpoints of a chord are joined to the centre of the circle, an angle is formed at the centre. This angle is known as the angle subtended by the chord at the centre.
In the figure, the chord $\overline{AB}$ joins points A and B on the circle. The line segments (radii) $\overline{OA}$ and $\overline{OB}$ are drawn from the centre O to the endpoints of the chord. The angle formed at the centre, $\angle AOB$, is the angle subtended by the chord $\overline{AB}$ at the centre O.
The size of this central angle depends on the length of the chord. A longer chord subtends a larger angle at the centre. The longest chord (the diameter) subtends a straight angle ($180^\circ$) at the centre.
Theorem: Equal chords subtend equal angles at the centre.
This is a fundamental theorem relating chord lengths and central angles in a circle.
Statement:
Equal chords of a circle (or of congruent circles) subtend equal angles at the centre.
Given:
A circle with centre O.
$\overline{AB}$ and $\overline{CD}$ are two chords of the circle such that their lengths are equal ($AB = CD$).
To Prove:
The angles subtended by these chords at the centre are equal, i.e., $\angle AOB = \angle COD$.
Proof:
Consider $\triangle AOB$ and $\triangle COD$.
OA = OC
(Radii of the same circle)
OB = OD
(Radii of the same circle)
AB = CD
(Given)
Since all three corresponding sides are equal, by the SSS (Side-Side-Side) congruence criterion:
$\triangle AOB \cong \triangle COD$
As the triangles are congruent, their corresponding angles are equal by CPCT (Corresponding Parts of Congruent Triangles).
$\angle AOB = \angle COD$
(CPCT)
Hence, equal chords of a circle subtend equal angles at the centre.
Converse Theorem: Chords subtending equal angles at the centre are equal.
This is the converse of the previous theorem and is also a fundamental result.
Statement:
If the angles subtended by the chords of a circle (or of congruent circles) at the centre are equal, then the chords are equal.
Given:
A circle with centre O.
$\overline{AB}$ and $\overline{CD}$ are two chords of the circle such that the angles they subtend at the centre are equal ($\angle AOB = \angle COD$).
To Prove:
The lengths of the chords are equal, i.e., $AB = CD$.
Proof:
Consider $\triangle AOB$ and $\triangle COD$.
OA = OC
(Radii of the same circle)
$\angle AOB = \angle COD$
(Given)
OB = OD
(Radii of the same circle)
By the SAS (Side-Angle-Side) congruence criterion:
$\triangle AOB \cong \triangle COD$
By CPCT (Corresponding Parts of Congruent Triangles):
AB = CD
(CPCT)
Hence, if the angles subtended by chords at the centre are equal, then the chords are equal.
Equal Chords and Their Distances from the Centre (Theorem and Converse)
The relationship between the length of a chord and its position relative to the centre of the circle is described by important theorems concerning the distance from the centre. The distance of a chord from the centre is defined as the shortest distance from the centre to the chord, which is the length of the perpendicular segment from the centre to the chord.
In the figure, $\overline{OM}$ is perpendicular to chord $\overline{AB}$. The length OM is the distance of the chord $\overline{AB}$ from the centre O.
Recall the property: The perpendicular from the centre of a circle to a chord bisects the chord. This means if $\overline{OM} \perp \overline{AB}$, then M is the midpoint of $\overline{AB}$, so $AM = MB = \frac{1}{2} AB$.
Theorem: Equal chords are equidistant from the centre.
Statement:
Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).
Given:
A circle with centre O.
$\overline{AB}$ and $\overline{CD}$ are two equal chords of the circle, so $AB = CD$.
$\overline{OM}$ is the perpendicular from O to $\overline{AB}$ ($OM \perp AB$), and $\overline{ON}$ is the perpendicular from O to $\overline{CD}$ ($ON \perp CD$).
OM and ON are the distances of chords $\overline{AB}$ and $\overline{CD}$ from the centre O, respectively.
To Prove:
The distances are equal, i.e., $OM = ON$.
Construction:
Join $\overline{OA}$ and $\overline{OC}$. ($\overline{OA}$ and $\overline{OC}$ are radii of the circle).
Proof:
Since the perpendicular from the centre of a circle to a chord bisects the chord, M is the midpoint of $\overline{AB}$ and N is the midpoint of $\overline{CD}$.
AM = $\frac{1}{2}$ AB
(OM $\perp$ AB, so M is midpoint of AB) ... (1)
CN = $\frac{1}{2}$ CD
(ON $\perp$ CD, so N is midpoint of CD) ... (2)
We are given that the chords are equal:
AB = CD
(Given) ... (3)
From (1), (2), and (3), taking half of equal quantities results in equal quantities:
AM = CN
... (4)
Now, consider the right-angled triangles $\triangle OMA$ and $\triangle ONC$.
$\angle OMA = \angle ONC = 90^\circ$
(By construction, OM $\perp$ AB, ON $\perp$ CD)
OA = OC
(Radii of the same circle - these are the hypotenuses of the right triangles)
AM = CN
(From 4 - these are the corresponding sides)
By the RHS (Right angle - Hypotenuse - Side) congruence criterion:
$\triangle OMA \cong \triangle ONC$
By CPCT (Corresponding Parts of Congruent Triangles):
OM = ON
(CPCT)
Hence, equal chords of a circle are equidistant from the centre.
Converse Theorem: Chords equidistant from the centre are equal.
Statement:
Chords equidistant from the centre of a circle (or of congruent circles) are equal in length.
Given:
A circle with centre O.
$\overline{AB}$ and $\overline{CD}$ are two chords.
$\overline{OM}$ is the perpendicular from O to $\overline{AB}$ ($OM \perp AB$), and $\overline{ON}$ is the perpendicular from O to $\overline{CD}$ ($ON \perp CD$).
The chords are equidistant from the centre, so $OM = ON$.
To Prove:
The chords are equal in length, i.e., $AB = CD$.
Construction:
Join $\overline{OA}$ and $\overline{OC}$. ($\overline{OA}$ and $\overline{OC}$ are radii of the circle).
Proof:
Consider the right-angled triangles $\triangle OMA$ and $\triangle ONC$.
$\angle OMA = \angle ONC = 90^\circ$
(Given OM $\perp$ AB and ON $\perp$ CD)
OA = OC
(Radii of the same circle - Hypotenuse)
OM = ON
(Given - Side)
By the RHS (Right angle - Hypotenuse - Side) congruence criterion:
$\triangle OMA \cong \triangle ONC$
By CPCT (Corresponding Parts of Congruent Triangles):
AM = CN
... (1)
We know that the perpendicular from the centre to a chord bisects the chord.
AB = 2 $\times$ AM
(M is midpoint of AB)
CD = 2 $\times$ CN
(N is midpoint of CD)
From (1), we have $AM = CN$. Multiplying both sides by 2:
2 $\times$ AM = 2 $\times$ CN
Substituting the expressions for AB and CD:
AB = CD
Hence, chords equidistant from the centre are equal in length.
Angle Subtended by an Arc at the Centre and at any Point on the Remaining Part
An arc of a circle is a segment of its circumference. This arc can form angles at different locations relative to the circle. The two primary locations we consider are the centre of the circle and any point on the remaining part of the circle's circumference.
- The angle subtended by an arc at the centre is the angle formed by the two radii connecting the centre to the endpoints of the arc. For arc PQ, this is $\angle POQ$.
- The angle subtended by an arc at any point on the remaining part of the circle is the angle formed by the two chords connecting the endpoints of the arc to that point on the circumference. For arc PQ and point R on the remaining part of the circle, this is $\angle PRQ$.
The size of the angle subtended by an arc at the centre is directly related to the size of the angle subtended by the same arc at any point on the circumference.
Theorem: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
This is one of the most important theorems in circle geometry.
Statement:
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Given:
A circle with centre O.
Arc $\frown{PQ}$ is an arc of the circle.
$\angle POQ$ is the angle subtended by arc $\frown{PQ}$ at the centre O.
R is any point on the remaining part of the circle (the part of the circumference not containing arc $\frown{PQ}$). $\angle PRQ$ is the angle subtended by arc $\frown{PQ}$ at point R.
To Prove:
$\angle POQ = 2 \angle PRQ$.
Construction:
Join the line segment $\overline{RO}$ and extend it through O to a point M outside the circle.
Proof:
We will prove this theorem by considering three possible cases for the arc $\frown{PQ}$: when it is a minor arc, a semicircle, or a major arc.
Case 1: The arc $\frown{PQ}$ is a minor arc.
In this case, the point R lies on the major arc $\frown{PRQ}$.
In $\triangle POR$, $\overline{OP}$ and $\overline{OR}$ are radii of the same circle.
OP = OR
(Radii)
Therefore, $\triangle POR$ is an isosceles triangle. The angles opposite the equal sides are equal:
$\angle OPR = \angle ORP$
(Angles opposite equal sides in $\triangle POR$)
Consider the line segment $\overline{ROM}$ as a straight line. $\angle POM$ is an exterior angle of $\triangle POR$. By the Exterior Angle Property of a Triangle, an exterior angle is equal to the sum of the two opposite interior angles.
$\angle POM = \angle OPR + \angle ORP$
(Exterior angle of $\triangle POR$)
Substitute $\angle OPR = \angle ORP$:
$\angle POM = \angle ORP + \angle ORP = 2 \angle ORP$
... (1)
Similarly, in $\triangle QOR$, $\overline{OQ}$ and $\overline{OR}$ are radii.
OQ = OR
(Radii)
Therefore, $\triangle QOR$ is an isosceles triangle. The angles opposite the equal sides are equal:
$\angle OQR = \angle ORQ$
(Angles opposite equal sides in $\triangle QOR$)
Consider the exterior angle $\angle QOM$ of $\triangle QOR$.
$\angle QOM = \angle OQR + \angle ORQ$
(Exterior angle of $\triangle QOR$)
Substitute $\angle OQR = \angle ORQ$:
$\angle QOM = \angle ORQ + \angle ORQ = 2 \angle ORQ$
... (2)
Add equation (1) and equation (2):
$\angle POM + \angle QOM = 2 \angle ORP + 2 \angle ORQ$
From the figure, $\angle POM + \angle QOM = \angle POQ$ (the angle subtended by arc PQ at the centre O). Also, $\angle ORP + \angle ORQ = \angle PRQ$ (the angle subtended by arc PQ at point R on the circumference).
$\angle POQ = 2 (\angle ORP + \angle ORQ)$
$\angle POQ = 2 \angle PRQ$
... (Proved for Case 1)
Case 2: The arc $\frown{PQ}$ is a semicircle.
In this case, the chord $\overline{PQ}$ is a diameter, and the centre O lies on $\overline{PQ}$. The angle subtended by the semicircle at the centre is a straight angle, $\angle POQ = 180^\circ$. The point R lies on the other semicircle.
The same logic using exterior angles of $\triangle POR$ and $\triangle QOR$ applies as in Case 1.
$\angle POM = 2 \angle ORP$
$\angle QOM = 2 \angle ORQ$
Adding these equations:
$\angle POM + \angle QOM = 2 (\angle ORP + \angle ORQ)$
In this case, $\angle POM + \angle QOM$ forms the straight angle $\angle POQ$, which is $180^\circ$. Also, $\angle ORP + \angle ORQ = \angle PRQ$.
$\angle POQ = 2 \angle PRQ$
180^\circ = 2 \angle PRQ
This means $\angle PRQ = 90^\circ$. This confirms the well-known result that the angle in a semicircle is a right angle. The theorem $\angle POQ = 2 \angle PRQ$ holds as $180^\circ = 2 \times 90^\circ$.
Case 3: The arc $\frown{PQ}$ is a major arc.
In this case, the point R lies on the minor arc $\frown{PRQ}$. The angle subtended at the centre by the major arc $\frown{PQ}$ is the reflex angle $\angle POQ$.
Using the same exterior angle property of $\triangle POR$ and $\triangle QOR$ as in Case 1, extended line segment ROM is used.
$\angle POM = 2 \angle ORP$
$\angle QOM = 2 \angle ORQ$
Adding these equations:
$\angle POM + \angle QOM = 2 \angle ORP + 2 \angle ORQ$
The sum of angles $\angle POM$ and $\angle QOM$ around point O, when measured from the side of the major arc, gives the reflex angle $\angle POQ$.
Reflex $\angle POQ = \angle POM + \angle QOM$
And $\angle ORP + \angle ORQ = \angle PRQ$.
Reflex $\angle POQ = 2 (\angle ORP + \angle ORQ)
Reflex $\angle POQ = 2 \angle PRQ$
... (Proved for Case 3)
In all three cases, the theorem holds: the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
This theorem is often abbreviated as "Angle at the centre is double the angle at the circumference (subtended by the same arc)".
Angles in the Same Segment are Equal
A chord divides a circle into two segments: a minor segment and a major segment. The angles subtended by a chord (or its corresponding arc) at various points on the circumference within the same segment have a special relationship.
In the figure, the chord $\overline{AB}$ defines two segments. The angles $\angle APB$ and $\angle AQB$ are both subtended by the arc $\frown{AB}$ (the minor arc in this illustration) at points P and Q, which lie in the major segment.
Theorem: Angles in the same segment of a circle are equal.
Statement:
Angles subtended by the same arc in the same segment of a circle are equal.
Given:
A circle with centre O.
Points P and Q are any two distinct points on the same segment of the circle formed by chord $\overline{AB}$.
(This means P and Q lie on the same arc formed by A and B, not including the arc between A and B that defines the segment containing P and Q).
To Prove:
$\angle APB = \angle AQB$.
Proof:
Consider the arc $\frown{AB}$ (this is the arc that subtends the angles $\angle APB$ and $\angle AQB$). Let's assume it's the minor arc for the proof, but the proof holds for the major arc as well.
The angle subtended by the arc $\frown{AB}$ at the centre is $\angle AOB$.
According to the theorem "The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle" (Section I3):
For the angle subtended at point P (which is on the remaining part of the circle with respect to arc $\frown{AB}$):
$\angle AOB = 2 \angle APB$
(Angle at centre is double the angle at circumference)
This implies:
$\angle APB = \frac{1}{2} \angle AOB$
... (1)
Similarly, for the angle subtended by the same arc $\frown{AB}$ at point Q (which is also on the remaining part of the circle with respect to arc $\frown{AB}$):
$\angle AOB = 2 \angle AQB$
(Angle at centre is double the angle at circumference)
This implies:
$\angle AQB = \frac{1}{2} \angle AOB$
... (2)
From equations (1) and (2), since both $\angle APB$ and $\angle AQB$ are equal to half of the same angle $\angle AOB$, they must be equal to each other:
$\angle APB = \angle AQB$
Hence, angles in the same segment of a circle are equal.
If $\frown{AB}$ was a major arc, $\angle AOB$ would be the reflex angle. The angles $\angle APB$ and $\angle AQB$ would be subtended at points P and Q on the minor arc. The theorem $\text{Reflex } \angle AOB = 2 \angle APB$ (and $2 \angle AQB$) would still hold, leading to the same conclusion.
Angle in a Semicircle
A special case of the angle subtended by an arc is when the arc is a semicircle. A semicircle is formed by a diameter of the circle.
In the figure, $\overline{AB}$ is a diameter passing through the centre O. The arc $\frown{APB}$ (or the arc below $\overline{AB}$) is a semicircle. An angle formed by joining the endpoints of the diameter to any point P on the circumference is called the angle in a semicircle ($\angle APB$).
Theorem: The angle in a semicircle is a right angle.
This is a direct application of the theorem relating the angle at the centre and the angle at the circumference.
Statement:
The angle in a semicircle is a right angle ($90^\circ$).
Given:
A circle with centre O.
$\overline{AB}$ is a diameter of the circle.
P is any point on the circumference of the circle.
$\angle APB$ is the angle subtended by the diameter $\overline{AB}$ (or the semicircle arc $\frown{APB}$) at the point P on the circumference.
To Prove:
$\angle APB = 90^\circ$.
Proof:
The diameter $\overline{AB}$ is a straight line segment passing through the centre O. It forms an arc (a semicircle) that subtends an angle at the centre.
The angle subtended by the diameter $\overline{AB}$ at the centre O is the straight angle $\angle AOB$.
$\angle AOB = 180^\circ$
(Angle on a straight line)
The angle subtended by the same arc $\frown{APB}$ (the semicircle) at point P on the remaining part of the circle is $\angle APB$.
Using the theorem "The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle" (Section I3):
$\angle AOB = 2 \angle APB$
Substitute the value of $\angle AOB$:
180^\circ = 2 \angle APB
Now, solve for $\angle APB$:
$\angle APB = \frac{180^\circ}{2}$
$\angle APB = 90^\circ$
Hence, the angle in a semicircle is a right angle.
Converse Theorem: Chord subtending a right angle at the circumference is a diameter.
This is the converse of the theorem above, providing a condition for a chord to be a diameter.
Statement:
If an angle subtended by a chord at a point on the circle is $90^\circ$, then the chord is a diameter of the circle.
Given:
A circle with centre O.
$\overline{AB}$ is a chord.
P is a point on the circumference such that $\angle APB = 90^\circ$.
To Prove:
$\overline{AB}$ is a diameter of the circle.
Proof:
The chord $\overline{AB}$ subtends the angle $\angle APB$ at point P on the circumference.
The angle subtended by the same arc $\frown{APB}$ (or $\frown{AQB}$, depending on which side P is) at the centre O is $\angle AOB$.
Using the theorem "The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle":
$\angle AOB = 2 \angle APB$
We are given that $\angle APB = 90^\circ$. Substituting this value:
$\angle AOB = 2 \times 90^\circ$
$\angle AOB = 180^\circ$
An angle of $180^\circ$ at the centre means that the points A, O, and B are collinear, and O lies on the line segment $\overline{AB}$.
A chord that passes through the centre of the circle is defined as a diameter.
Therefore, $\overline{AB}$ is a diameter of the circle.
Hence proved.