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 Δ𝐴𝐵𝑄
∠𝐵𝑂𝑄 = ∠𝑂𝐴𝑄 + ∠𝑂𝑄𝐴………(𝑖) (Exterior angle)
In Δ𝐴𝑂𝑄
𝑂𝐴 = 𝑂𝑄 (radii of the same circle)
Thus, Δ𝐴𝑂𝑄 is an isosceles triangle.
So, ∠𝑂𝐴𝑄 = ∠𝑂𝑄𝐴………(𝑖𝑖)
Equate (𝑖) and (𝑖𝑖)
∠𝐵𝑂𝑄 = ∠𝑂𝐴𝑄 + ∠𝑂𝐴𝑄
From (𝑖) and (𝑖𝑖)
∠𝐵𝑂𝑄 = 2∠𝑂𝐴𝑄……(𝑖𝑖𝑖)
Similarly, ∠𝐵𝑂𝑃 = 2∠𝑂𝐴𝑃……(𝑖𝑣)
Adding (𝑖𝑖𝑖) and (𝑖𝑣)
∠𝐵𝑂𝑄 + ∠𝐵𝑂𝑃 = 2∠𝑂𝐴𝑃 + 2∠𝑂𝐴𝑄
∴ ∠𝐵𝑂𝑄 + ∠𝐵𝑂𝑃 = 2(∠𝑂𝐴𝑃 + ∠𝑂𝐴𝑄)
But ∠𝐵𝑂𝑄 + ∠𝐵𝑂𝑃 = ∠𝑃𝑂𝑄
and ∠𝑂𝐴𝑃 + ∠𝑂𝐴𝑄 = ∠𝑃𝐴𝑄
From the figure,
∠𝑃𝑂𝑄 = 2∠𝑃𝐴𝑄
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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