The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc.
Step-by-step solution
To find: Show that ∠ACB = 2∠ADB
Idea: Split the angles with the line DC. Triangles DCA and DCB are isosceles (two sides are radii), and the exterior angle of a triangle equals the sum of the two opposite interior angles. Two cases arise, depending on whether the extension of DC meets the arc AFB.
- Case 1: DC extended meets arc AFB at E. In ΔDCB, CB = CD (radii), so ∠CBD = ∠CDB. ∠BCE is an exterior angle of ΔBCD, so ∠BCE = ∠CBD + ∠CDB = 2∠BDC.1 mark
- Similarly, in ΔDCA, CA = CD, so ∠CAD = ∠CDA, and the exterior angle ∠ACE = ∠CAD + ∠CDA = 2∠CDA.1 mark
- E lies inside ∠ACB and ∠ADB, so ∠ACB = ∠BCE + ∠ECA = 2(∠BDC + ∠CDA) = 2∠BDA. So ∠ACB = 2∠ADB.1 mark
- Case 2: DC extended meets the circle at E outside arc AFB. As before, ∠ACE = ∠ADC + ∠CAD = 2∠ADC (since CA = CD), and ∠BCE = ∠BDC + ∠CBD = 2∠BDC (since CB = CD).1 mark
- Now the parts are subtracted: ∠ACB = ∠ACE − ∠BCE and ∠ADB = ∠ADC − ∠BDC. So ∠ACB = 2(∠ADC − ∠BDC) = 2∠ADB.1 mark
- Consequence: ∠ADB = ½∠ACB for every point D on the circle outside the arc, so all such angles are equal: angles in the same segment are equal.
Answer to write in the exam
Given: arc AFB, centre C, D on the circle outside arc AFB. To prove: ∠ACB = 2∠ADB.
Join DC and produce it to E.
Case 1 (E on arc AFB):
CB = CD (radii) ⇒ ∠CBD = ∠CDB
∠BCE = ∠CBD + ∠CDB = 2∠BDC (exterior angle of ΔBCD)
CA = CD (radii) ⇒ ∠CAD = ∠CDA; ∠ACE = ∠CAD + ∠CDA = 2∠CDA (exterior angle of ΔADC)
∠ACB = ∠BCE + ∠ECA = 2(∠BDC + ∠CDA) = 2∠ADB
Case 2 (E outside arc AFB):
∠ACE = 2∠ADC, ∠BCE = 2∠BDC (as above)
∠ACB = ∠ACE − ∠BCE = 2(∠ADC − ∠BDC) = 2∠ADB
∴ ∠ACB = 2∠ADB in both cases.
Common mistakes that cost marks
- Proving only case 1. When the line DC misses the arc, the angles must be subtracted, not added.
- Writing ∠BCE = ∠CBD (forgetting the second interior angle). The exterior angle equals the sum of both opposite interior angles.
- Using a point D on the arc AFB itself; the theorem is about points outside the arc.
How this can come in the exam
O is the centre of a circle, A, B, C are on the circle and C lies on the major arc AB. If ∠ACB = 48°, then ∠AOB is
- 24°
- 48°
- 96°
- 132°
Show answer
(C) 96°
The angle at the centre is double the angle at the circle: ∠AOB = 2 × 48° = 96°.
A, B, C lie on a circle with centre O, C on the major arc AB, and ∠OAB = 25°. Find ∠AOB and ∠ACB.
Show answer
OA = OB (radii), so ∠OBA = ∠OAB = 25° (1 mark). ∠AOB = 180° − 25° − 25° = 130° (1 mark). ∠ACB = ½∠AOB = 65° (1 mark).Try one yourself
A, B, P are points on a circle with centre O, and P lies on the major arc AB. If ∠AOB = 2x + 30° and ∠APB = 55°, find x.
Show answer
∠AOB = 2∠APB = 110°, so 2x + 30 = 110 and x = 40°.
More questions like this
- The angle subtended by a diameter at any point on the circle is 90°.
- In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?
- Let A and B be two points on a circle with centre O.
- Find x in the figure.
- If a line segment AB joining two points A, B subtends equal angles at two other points C, D that lie on the same side of AB, then the four points lie on a circle.