Introduction
In a circle, if we draw a chord and join the two ends of the chord to the third point which is situated at the Centre or on the circle, then the angle made at the third point is the angle subtended by the chord of the Circle.

In the above figure, тИаPRQ and тИаPSQ are the angles subtended by the chord PQ on the points R and S in major and minor segments respectively. тИаPOQ is the angle subtended by the chord PQ at the Centre of the circle.
In a circle, if we draw two chords of equal lengths then the angles subtended by both the chords at the Centre of the circle are equal. We can prove this statement by theorem.
Explanation and Theorems Based on it
Theorem 1) Equal chords of a circle subtend equal angles at the Centre.

Given тАУ Chords PQ = RS
Prove that – тИаPOQ = тИаROS
Proof тАУ In тЦ│POQ and тЦ│ROS
PQ = RS (equal chords given)
OP = OR (Radii of the same circle)
OQ = OS (Radii of the same circle)
By the SSS rule, тЦ│POQ тЙЕ тЦ│ROS
So, corresponding parts of congruent triangles (CPCT) will be equal.
Therefore, тИаPOQ = тИаROS Hence Proved.
The converse of the above theorem is also true and here is its proof.
Converse of Theorem
Theorem 2) If the angles subtended by the chords of a circle at the Centre are equal, then the chords are equal.

Given – тИаPOQ = тИаROS
Prove that тАУ PQ = RS
Proof тАУ In тЦ│POQ and тЦ│ROS
OP = OR (Radii of the same circle)
тИаPOQ = тИаROS (equal angles given)
OQ = OS (Radii of the same circle)
By SAS rule, тЦ│POQ тЙЕ тЦ│ROS
So, by CPCT┬а┬а┬а┬а┬а┬а┬а┬а PQ = RS┬а┬а┬а┬а┬а┬а┬а┬а┬а┬а┬а┬а┬а┬а┬а┬а┬а┬а Hence Proved.
Perpendicular to a Chord from the Centre
In a circle, if we draw a perpendicular to a chord from the Centre then the perpendicular divides the chord into two equal parts. That means it bisects the chord. It can be proved by the theorem.
Theorem 3) The perpendicular drawn from the Centre of a circle to the chord bisects the chord.

Given тАУ PQ is a chord and OM тКе PQ
Prove that тАУ PM = MQ
Construction тАУ Joined Centre O to P and Q.
Proof тАУ In тЦ│POM and тЦ│QOM,
OP = OQ (radii of the same circle)
тИаOMP = тИаOMQ = 90┬░ (given OM тКе PQ)
OM = OM (common side)
By RHS rule, тЦ│POM тЙЕ тЦ│QOM
So, by CPCT, PM = MQ Hence Proved.
The Converse of the Above Theorem
Theorem 4) The line joining the Centre of the circle to the midpoint of the chord is perpendicular to the chord.

Given тАУ M is the midpoint of chord PQ, it means PM = MQ
Prove that тАУ OM тКе PQ
Construction – Joined Centre O to P and Q.
Proof – In тЦ│POM and тЦ│QOM,
OP = OQ (radii of the same circle)
PM = MQ (given)
OM = OM (common side)
By the SSS rule, тЦ│POM тЙЕ тЦ│QOM
So, by CPCT, тИаOMP = тИаOMQ ————–(1)
тИаOMP + тИаOMQ = 180┬░ (by linear pair of angles)
тИаOMP + тИаOMP = 180┬░ [from equation (1)]
2тИаOMP = 180┬░
тИаOMP = 180┬░/2 = 90┬░
Therefore, тИаOMP = тИаOMQ = 90┬░
So,┬а┬а┬а┬а┬а┬а┬а┬а┬а OM тКе┬аPQ ┬а┬а┬а┬а┬а┬а┬а┬а┬а┬аHence Proved.
Chords in a Circle and their Distances from the Centre
In a circle, if we draw chords of different lengths then we observe that the chord which is shorter in length is far from the Centre and the chord which is longer in length is near to the Centre.

In the figure, there are two chords PQ and RS drawn. Where chord RS > chord PQ. The distance from the Centre (perpendiculars drawn from the Centre) of both the chords are OM and ON respectively. We can clearly see that distance OM > ON. It shows that chord PQ is Smaller than RS so RS is nearer to the Centre than PQ.
If Chords are Equal тАУ If we draw two equal chords then the distance of both the chords from the Centre will be equal. This can be understood by the figure drawn below.

In the figure, there are two equal chords PQ and RS and their distance from the Centre is OM and ON respectively. If we measure both distances, we will find that both distances are equal. It means that equal chords have an equal distance from the Centre.
Theorem 5) Equal chords of a circle are equidistant from the Centre.

Given тАУ Chords AB = CD
Prove that тАУ OP = OQ
Construction тАУ Joined points A and D to Centre O.
Proof – тИ╡ OP тКе AB
тИ┤ AP = PB = ┬╜ AB (Perpendicular drawn from the Centre to a chord bisects the chord)
And OQ тКе CD
тИ┤ CQ = QD = ┬╜ CD (Perpendicular drawn from the Centre to a chord bisects the chord)
Here, AB = CD (given)
тИ┤ AP = QD ————(1)
Now, In тЦ│APO and тЦ│DQO,
AP = QD [from equation (1)]
OA = OD [Radii of the same circle]
тИаAPO = тИаDQO = 90┬░
By RHS rule, тЦ│APO тЙЕ тЦ│DQO
So, by CPCT, OP = OQ Hence Proved.
Note тАУ In the congruent circles, equal chords are equidistant from their corresponding Centres.
Converse of Theorem
Theorem 6) Chords equidistant from the Centre of a circle are equal in length.

Given тАУ OP = OQ
Prove that тАУ AB = CD
Construction тАУ Joined points A and D to Centre O.
Proof тАУ In тЦ│APO and тЦ│DQO,
OP = OQ (given)
OA = OD (Radii of the same circle)
тИаAPO = тИаDQO = 90┬░
By RHS rule, тЦ│APO тЙЕ тЦ│DQO
So, by CPCT, AP = QD
Multiplying both sides by 2
2AP = 2QD [тИ╡ AP = PB and CQ =QD]
Therefore, AB = CD Hence Proved.
Note тАУ Chords of congruent circles which are equidistant from the corresponding Centres are equal in length.
Example
Example тАУ In the given figure, the radius of a circle with Centre O is 6 cm. If OA тКе PQ, OB тКе RS, PQ тИе RS, PQ = 10 cm and RS = 8 cm. Then find AB.

Solution – тИ╡ OA тКе PQ and OB тКе RS
тИ┤ PA = AQ = ┬╜ PQ [perpendicular drawn from the Centre to a chord bisects the chord]
= ┬╜тип10 = 5 cm
RB = BS = RS [perpendicular drawn from the Centre to Chord bisects the chord]
= ┬╜тип8 = 4 cm
Radius OP = OR = 6 cm
In тЦ│PAO, by Pythagoras theorem,
OA2 = OP2 тАУ PA2
OA2 = (6)2 тАУ (5)2
OA2 = 36 тАУ 25 = 11
OA = тИЪ11 = 3.32 cm (Approximately)
Now, in тЦ│RBO, by Pythagoras theorem,
OB2 = OR2 тАУ RB2
OB2 = (6)2 тАУ (4)2
OB2 = 36 тАУ 16 = 20
OB = тИЪ20 = тИЪ(4тип5)
OB = 2тИЪ5 = 2тип2.2360 = 4.47 cm (Approximately)
Therefore, AB = OA + OB
AB = 3.32 + 4.47
AB = 7.79 cm Ans.
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