Learnify Academy is a tuition centre in Bahrain. Classes are for students in Bahrain only.Tuition classes in Bahrain only

Cyclic quadrilaterals · 5 marks

If two opposite angles of a quadrilateral add up to 180°, then the vertices of the quadrilateral lie on a circle, i.e., they are concyclic.

Answer: Take the circle through A, B, D. If C were outside it, CD would meet the circle at E and ∠BED (an exterior angle of ΔBEC) would be greater than ∠BCD; if C were inside, DC produced would meet it at E and ∠BCD would be greater than ∠BED. But ∠BCD = ∠BED in both cases (each is 180° − ∠BAD). So C lies on the circle: ABCD is cyclic.

Step-by-step solution

Given: ABCD is a quadrilateral; ∠BAD + ∠BCD = 180°
To find: Show that A, B, C, D lie on a circle

Idea: This is the converse of “opposite angles of a cyclic quadrilateral add up to 180°”. Suppose C is not on the circle through A, B, D, make a cyclic quadrilateral ABED with a point E on that circle, and get a contradiction from the exterior angle of a triangle. Both cases, C outside and C inside, are needed.

ABDCEC outsideABDCEC inside
  1. A, D, B are not collinear, so there is a circle through them. Suppose it does not pass through C. Then C is either outside or inside this circle.½ mark
  2. Case 1, C outside: let CD meet the circle at E. ABED is a cyclic quadrilateral, so ∠BAD + ∠BED = 180°. Also ∠BAD + ∠BCD = 180° (given). Hence ∠BED = ∠BCD.1 mark
  3. But ∠BED is an exterior angle of ΔBEC, so ∠BED > ∠BCE = ∠BCD. This contradicts ∠BED = ∠BCD.1 mark
  4. Case 2, C inside: produce DC to meet the circle at E. ABED is cyclic, so again ∠BAD + ∠BED = 180°, and with the given condition ∠BED = ∠BCD.1 mark
  5. Now D, C, E are on one line with C between D and E, so ∠BCD is an exterior angle of ΔBCE: ∠BCD > ∠BEC = ∠BED. This again contradicts ∠BED = ∠BCD.1 mark
  6. Both cases are impossible, so C lies on the circle through A, B, D: A, B, C, D are concyclic.½ mark
Assuming C is off the circle through A, B, D leads in both cases (outside and inside) to ∠BED = ∠BCD together with a strict inequality between them, which is impossible. So A, B, C, D are concyclic.

Answer to write in the exam

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

Draw the circle through A, B, D. Suppose C is not on it.

Case 1, C outside: CD meets the circle at E.

∠BAD + ∠BED = 180° (ABED cyclic) and ∠BAD + ∠BCD = 180° (given) ⇒ ∠BED = ∠BCD

∠BED > ∠BCE = ∠BCD (exterior angle of ΔBEC), a contradiction.

Case 2, C inside: DC produced meets the circle at E.

∠BED = ∠BCD (as above)

∠BCD > ∠BEC = ∠BED (exterior angle of ΔBCE), a contradiction.

∴ C lies on the circle through A, B, D; ABCD is cyclic.

Common mistakes that cost marks

  • Proving only the case where C is outside. The case where C is inside needs DC to be produced beyond C, and the exterior angle is then at C, not at E.
  • Assuming the circle passes through all four points at the start; that is what has to be proved.
  • Thinking both pairs of opposite angles must be checked. If one pair adds to 180°, the other pair does too (the four angles total 360°).

How this can come in the exam

Assertion–Reason (1 mark)

Assertion (A): Every rectangle is a cyclic quadrilateral.
Reason (R): If a pair of opposite angles of a quadrilateral add up to 180°, the quadrilateral is cyclic.

  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Show answer

(A) Both A and R are true, and R is the correct explanation of A.
In a rectangle opposite angles are 90° + 90° = 180°, so by R it is cyclic. Both are true and R explains A.

Short answer (3 marks)

ABCD is a trapezium with AB ∥ DC whose non-parallel sides AD and BC are equal. Show that ABCD is cyclic.

Show answerSince AD and BC are the equal non-parallel sides, ABCD is an isosceles trapezium, so its base angles are equal: ∠A = ∠B (1 mark). AB ∥ DC, so ∠A + ∠D = 180° (co-interior angles) (1 mark). Hence ∠B + ∠D = 180°: a pair of opposite angles is supplementary, so ABCD is cyclic (1 mark).

Try one yourself

In quadrilateral PQRS, ∠P = (5k)° and ∠R = (4k)°, and PQRS is known to be cyclic. Find k, ∠P and ∠R.

Show answer

5k + 4k = 180, so k = 20, ∠P = 100° and ∠R = 80°.

More questions like this

All Circles questions · All maths questions

← A cyclic quadrilateral has angles measuring ∠A = 80°, ∠B = 110°, ∠C = In a circle, a chord is 5 cm away from the centre. If the radius of th →