**Solution:**

**Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S. **

Join the vertices of the quadrilateral ABCD to the center of the circle.

In ΔOAP and ΔOAS,

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

OP = OS (**Radii of the circle**)

OA = OA (**Common side**)

ΔOAP ≅ ΔOAS (**SSS congruence condition**)

∴ ∠POA = ∠AOS

⇒∠1 = ∠8

Similarly we get,

∠2 = ∠3

∠4 = ∠5

∠6 = ∠7

Adding all these angles,

∠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.**