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

Quadrilateral ABCD is circumscribing a circle with centre O, touching at point P, Q, R and S.
join the points P, Q, R and S from the centre O.
In ΔOAP and ΔOAS,
OP = OS [radill of same circle]
AP = AS [Tangents drawn from point A]
AO = AO [Common]
ΔOPA ≅ ΔOCA [SSS Congruency rule]
Hence, ∠POA = ∠SOA or ∠1 = ∠8
Similarly, ∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7
Sum of all angles at point O is 360°. Therefore
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360°
⇒ 2∠1 + 2∠2 + 2∠5 + 2∠6 = 360
⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180°
⇒ ∠AOB + ∠COD = 180°
Similarly, we can prove ∠BOC + ∠DOA = 180°
Hence, the opposite sides subtend supplementary angles at the centre.