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Cyclic quadrilaterals · 4 marks

The sum of two opposite angles of a cyclic quadrilateral is 180°.

Answer: ∠BAD is half the angle that arc BCD subtends at the centre O, and ∠BCD is half the angle that arc BAD subtends at O. These two angles at O make a full turn, 360°, so ∠BAD + ∠BCD = ½ × 360° = 180°.

Step-by-step solution

Given: A, B, C, D lie on a circle with centre O (ABCD is a cyclic quadrilateral)
To find: Show that ∠BAD + ∠BCD = 180° (and likewise ∠ABC + ∠ADC = 180°)

Idea: Each angle of the quadrilateral is an angle at the circle subtended by an arc, so it is half the angle that arc subtends at the centre. Opposite vertices see the two complementary arcs BCD and BAD, whose angles at the centre fill the whole 360°.

ABCDO
  1. A lies on the circle outside arc BCD. So ∠BAD = ½ × (angle subtended by arc BCD at O). Moving from OB to OD through C sweeps the reflex ∠BOD, so ∠BAD = ½ reflex ∠BOD.1 mark
  2. C lies outside arc BAD. Moving from OB to OD through A sweeps the ordinary ∠BOD, so ∠BCD = ½∠BOD.1 mark
  3. Adding: ∠BAD + ∠BCD = ½(reflex ∠BOD + ∠BOD) = ½ × (complete angle at O).1 mark
  4. A complete turn at O is 360°, so ∠BAD + ∠BCD = ½ × 360° = 180°. In the same way ∠ABC + ∠ADC = 180°.1 mark
∠BAD = ½ reflex ∠BOD and ∠BCD = ½ ∠BOD, so ∠BAD + ∠BCD = ½ × 360° = 180°. Opposite angles of a cyclic quadrilateral are supplementary.

Check: Example: a rectangle is cyclic, and its opposite angles are 90° + 90° = 180° ✓.

Answer to write in the exam

Given: ABCD is cyclic with centre O. To prove: ∠BAD + ∠BCD = 180°.

Join OB, OD.

∠BAD = ½ reflex ∠BOD (arc BCD; angle at the centre is double the angle at the circle)

∠BCD = ½ ∠BOD (arc BAD)

∠BAD + ∠BCD = ½(reflex ∠BOD + ∠BOD) = ½ × 360°

∴ ∠BAD + ∠BCD = 180°. Similarly ∠ABC + ∠ADC = 180°.

Common mistakes that cost marks

  • Using the same angle ∠BOD for both A and C. The vertex A sees the arc through C (the reflex angle), and C sees the arc through A (the ordinary angle).
  • Saying opposite angles of a cyclic quadrilateral are equal. They are supplementary (add to 180°).
  • Applying the result to any quadrilateral. It holds only when all four vertices lie on one circle.

How this can come in the exam

MCQ (1 mark)

In a cyclic quadrilateral PQRS, ∠Q = 3∠S. Then ∠S is

  1. 30°
  2. 45°
  3. 60°
  4. 135°
Show answer

(B) 45°
∠Q + ∠S = 180°, so 4∠S = 180° and ∠S = 45°.

Short answer (2 marks)

ABCD is a cyclic quadrilateral in which ∠A − ∠C = 40°. Find ∠A and ∠C.

Show answer∠A + ∠C = 180° (opposite angles of a cyclic quadrilateral) and ∠A − ∠C = 40° (1 mark). Adding, 2∠A = 220°, so ∠A = 110° and ∠C = 70° (1 mark).

Try one yourself

The angles of a cyclic quadrilateral ABCD, taken in order, are ∠A = 4y, ∠B = 3y + 15°, ∠C = 2y + 30° and ∠D. Find y and ∠D.

Show answer

∠A + ∠C = 180°: 6y + 30 = 180, so y = 25°. ∠B = 90°, so ∠D = 180° − 90° = 90°.

More questions like this

All Circles questions · All maths questions

← If a line segment AB joining two points A, B subtends equal angles at A cyclic quadrilateral has angles measuring ∠A = 80°, ∠B = 110°, ∠C = →