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

Let us draw a quadrilateral ABCD circumscribe a circle with its centre O such a way that it touches the circle at points P, Q, R, and S. Now we get the following figure, after joining the vertices of ABCD, 

As we know, the lengths of tangents drawn from an external point to a circle are equal.

So considering the triangles OAP and OAS.

AP = AS (Tangents from the same point A)

OA = OA (Common side)

OP = OS (radius of the same circle)

So,  △OAP ≅ △OAS (by SSS congruency)

Also thus, ∠POA = ∠AOS

Which implies that ∠1 = ∠8

Similarly, other angles will be

∠4 = ∠5

∠2 = ∠3

∠6 = ∠7

Now by adding all these angles, we get

∠1+∠2+∠3 +∠4 +∠5+∠6+∠7+∠8 = 360°

Rearrange to get the required result, 

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

2∠1+2∠2+2∠5+2∠6 = 360°

Simplifying it further,

(∠1+∠2)+(∠5+∠6) = 180°

Thus, ∠AOB+∠COD = 180°

Similarly, we can prove for ∠BOC+∠DOA = 180°

Therefore, the opposite sides of any quadrilateral which is circumscribing a given circle will subtend supplementary angles at the centre of the circle.