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

Let ABCD be the quadrilateral circumscribing a circle at the center O such that it touches the circle at the point P, Q, R, S. Let join the vertices of the quadrilateral ABCD to the center of the circle.
In ΔOAP and ΔOAS
AP = AS (Tangents from to same point A)
PO = OS (Radii of the same circle)
OA = OA (Common side)
so, ΔOAP = ΔOAS (SSS congruence criterion)
∠POA = ∠AOS (CPCT)
∠1 = ∠8
Similarly ∠2 = ∠3
∠4 = ∠5
∠6 = ∠7
∠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°
⇒ (∠1) + (∠2) + (∠5) + (∠6) = 180°
∠AOD + ∠COD = 180°
Similarly we can prove ∠BOC + ∠DOA = 180°
Hence proved.