**Theorem:** The angle subtended by the chord at the centre is double the angle subtended by it at any point on the circle.

**Given:** Circle PAQ with centre O, a chord PQ.

**To prove:** ∠*𝑃𝑂𝑄* = 2∠*𝑃𝐴𝑄*

**Construction: **Join *𝑂* and *𝐴* and extend the ray to *𝐵* where it touches the chord PQ.

### Proof:

In Δ*𝐴**𝐵**𝑄*

∠*𝐵𝑂𝑄* = ∠*𝑂𝐴𝑄* + ∠*𝑂𝑄𝐴*………(*𝑖*) **(Exterior angle)**

In Δ*𝐴**𝑂**𝑄*

*𝑂𝐴* = *𝑂𝑄* **(radii of the same circle)**

Thus, Δ*𝐴𝑂𝑄* is an isosceles triangle.

So, ∠*𝑂𝐴𝑄* = ∠*𝑂𝑄𝐴*………(*𝑖𝑖*)

Equate (*𝑖*) and (*𝑖𝑖*)

∠*𝐵𝑂𝑄* = ∠*𝑂𝐴𝑄* + ∠*𝑂𝐴𝑄*

From (*𝑖*) and (*𝑖**𝑖*)

∠*𝐵**𝑂**𝑄* = 2∠*𝑂**𝐴**𝑄*……(*𝑖**𝑖**𝑖*)

Similarly, ∠*𝐵**𝑂**𝑃* = 2∠*𝑂**𝐴**𝑃*……(*𝑖**𝑣*)

Adding (*𝑖**𝑖**𝑖*) and (*𝑖**𝑣*)

∠*𝐵𝑂𝑄* + ∠*𝐵𝑂𝑃* = 2∠*𝑂𝐴𝑃* + 2∠*𝑂𝐴𝑄*

∴ ∠*𝐵𝑂𝑄* + ∠*𝐵𝑂𝑃* = 2*(*∠*𝑂𝐴𝑃* + ∠*𝑂𝐴𝑄)*

But ∠*𝐵𝑂𝑄* + ∠*𝐵𝑂𝑃* = ∠*𝑃𝑂*𝑄

and ∠*𝑂𝐴𝑃* + ∠*𝑂𝐴𝑄* = ∠*𝑃 𝐴*𝑄

From the figure,

∠*𝑃**𝑂**𝑄* = 2∠*𝑃**𝐴**𝑄*

Therefore, the angle subtends by the chord at the centre (∠*𝑃𝑂𝑄*) equals twice the angle subtended at the circumference (∠*𝑃𝐴𝑄*).

### Example 1

Find the value of 𝑥, in terms of 𝑦 as shown in the figure below.

**Solution:**

∠*𝐴𝐶𝐵* = 2∠*𝐴𝐷𝐵* **(angle at the centre is twice the angle at the circumference)**

∴ *x = 2y*

### Example 2

In the diagram below, O is the centre of the circle through points L, M and N, if *∠MLN* = 74º and *∠MNL* = 39º, calculate *∠LON*.

**Solution:**

*∠LMN* = 180° – (74° + 39°) **(sum of angles in a triangle)**

*∠LMN* = 67°

Let’s redraw the diagram by joining radii \( \scriptsize \overline{OL}\) and \( \scriptsize \overline{ON}\) to centre O.

*∠LON* = 2 x 67° **(angle subtended at the centre is twice the angle at the circumference)**

*∠LON* = 134°

### Example 3

In the diagram below, O is the centre of the circle PQRS. *∠*PQR = 84°.

Find**(i) **reflex *∠POR***(ii)** *∠PSR*

**Solution:**

**(i) **reflex *∠POR*

Remember: A reflex angle is greater than 180° while an obtuse angle is greater than 90° but less than 180°

Obtuse *∠POR* = 2 x 84° **(angle subtended at the centre is 2 x the angle at the circumference)**

Obtuse *∠POR* = 168°

Reflex *∠POR* = 360° – 168° **(sum of angles at a point = 360°)**

Reflex *∠POR* = 192°

**(ii)*** ∠PSR*

*∠PSR* = \( \frac{1}{2} \scriptsize \: \times \: reflex \: \angle POR \)

**(angle subtended at the centre is 2 x the angle at the circumference)**

*∠PSR* = \( \frac{1}{2} \scriptsize \: \times \: 192^o \)

*∠PSR* = 96°

### Example 3

Below is part of a spider web with a hole in it. Points P, Q, and R all lie on the circumference of a circle, with centre C.*∠PRC* = 25°

Calculate the angle, *a*, of *ΔPQR*.

**Solution:**

Consider *ΔPCR*

\( \scriptsize \overline{CP} = \overline{CR} \) **(radii of the same circle)**

Thus, *ΔPCR* is an isosceles triangle.

*∠RPC* = *∠PRC* = 25° **(base angles of isosceles ΔPCR)**

*∠PCR *+ *∠RPC* + *∠PCR* = 180° **(sum of angles in** *ΔPCR*)

*∠PCR *+ *25°* + *25°* = 180°

*∠PCR* *= * 180° – 50°

*∠PCR* = 130°

But *∠PCR* = 2 x *∠PQR* **(angle subtended at the centre is 2 x the angle at the circumference)**

∴ *∠PQR* = \( \frac{1}{2} \scriptsize \: \times \: \angle PCR \)

*∠PQR* = \( \frac{1}{2} \scriptsize \: \times \: 130\)

*∠PQR* = 65°

a = *∠PQR* = 65°

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