Question

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer 1

Let ABCD be a quadrilateral circumscribing a circle centred at O such that it touches the circle at point P, Q, R, and S. Let us join the vertices of the quadrilateral ABCD to the centre of the circle.

Consider ΔOAP and ΔOAS,

AP = AS  ....(Tangents from the same point)

OP = OS  ....(Radii of the same circle)

OA = OA  ....(Common side)

ΔOAP ≅ ΔOAS  ...(SSS congruence criterion)

Therefore, A ↔ A, P ↔ S, O ↔ O

And thus, ∠POA = ∠AOS

∠1 = ∠8

Similarly,

∠2 = ∠3

∠4 = ∠5

∠6 = ∠7

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360º

(∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º

2∠1 + 2∠2 + 2∠5 + 2∠6 = 360º

2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º

(∠1 + ∠2) + (∠5 + ∠6) = 180º

∠AOB + ∠COD = 180º

Similarly, we can prove that ∠BOC + ∠DOA = 180º

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer 2

Let ABCD be a quadrilateral circumscribing a circle with centre at O.

Join OA, OB, OC and OD.

Here, Let ∠DAO = ∠BAO = ∠1  ...[∵ AB and AD are tangents]

Similarly, ∠ABO = ∠CBO = ∠2  ...[∵ BA and BC are tangents]

And Let ∠BCO = ∠DCO = ∠3  ...[∵ CB and CD are tangents]

And Let ∠CDO = ∠ADO = ∠4  ...[∵ DC and DA are tangents]

Sum of all angles at the centre is 360°.

Also, the sum of the angles in quadrilateral, ABCD = 360°

∴ 2(∠1 + ∠2 + ∠3 + ∠4) = 360°

⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°  ...(i)

Now, In ΔAOB, 

∠AOB = 180° – (∠1 + ∠2)   ...(ii)

In ΔCOD, ∠COD = 180° – (∠3 + ∠4)   ...(iii)

On adding equations (ii) and (iii), we get

∠AOB + ∠COD = 360° – (∠1 + ∠2 + ∠3 + ∠4)

= 360° – 180°

= 180°  ...[From equation (i)]

Since AB and CD subtend supplementary angles at O.

Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.