# Angle Subtended by a Chord at a Point

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We have already studied about circles and its parts in Class VI. Take a line segment PQ and a point R not on the line containing PQ. Join PR and QR (shown below). Then ∠ PRQ is called **the angle subtended by the line segment PQ at the point R**.

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For a given circle below, we have four points: P, Q, R and S. Let draw the chord PQ by joining P and Q.

What are angles POQ, PRQ and PSQ called in the above figure? ∠ POQ is **the angle subtended by the chord** PQ at the centre O, ∠ PRQ and ∠ PSQ are respectively **the angles subtended by PQ at points R and S on the major and minor arcs PQ**.

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Let us examine the relationship between the size of the chord and the angle subtended by it at the centre. Try to drag the labelled points on the circle and observe the chanage in angle and length of the chord.

What do you observe?

We can see that **the longer the chord is, the is the measure of the angle subtended by it at the centre**.

What will happen if you take two equal chords of a circle? Will the angles subtended at the centre be the same or not?

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Circle theorems are based on several key elements of circles, including **chords**, **angles**, radii, and arcs.

**Theorem 9.1**

**Equal Chords Subtend Equal Angles Theorem :** **Equal chords of a circle subtend equal angles at the centre.**

**Proof :**

**Given:** Two equal chords AB and EF of a circle with centre O (see above figure).

**To Prove:** ∠ AOB = ∠ EOF.

In triangles AOB and EOF,

OA = OE (as they are

OB = OF (as they are

AB = EF (

Therefore,

∆ AOB ≅ ∆ EOF by

This gives ∠ AOB = ∠

(Corresponding parts of congruent triangles i.e. CPCT)

**Remark :**

For convenience, the abbreviation CPCT will be used in place of ‘Corresponding parts of congruent triangles’, because we use this very frequently as you will see.

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Now if two chords of a circle subtend equal angles at the centre, what can you say about the chords? Are they equal or not? Let us examine this by the following **activity**:

Take a tracing paper and trace a circle on it. Cut it along the circle to get a disc. At its centre O, draw an angle AOB where A, B are points on the circle. Make another angle POQ at the centre equal to ∠AOB. Cut the disc along AB and PQ (see Fig. 9.5). You will get two segments ACB and PRQ of the circle. If you put one on the other, what do you observe? They cover each other, i.e., they are congruent. So AB = PQ.

Though you have seen it for this particular case, try it out for other equal angles too. The chords will all turn out to be equal because of the following theorem:

**Theorem 9.2**

**Equal Angles Give Equal Chords Theorem :** **If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.**

**Given:** ∠AOB = ∠EOF

**To Prove:** AB = EF

**Proof:** In the triangles △AOB and △ EOF:

(i) ∠AOB = ∠

(ii) OA =

(iii) OB =

Thus, by

Further by CPCT: AB = CD