Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
First draw a quadrilateral ABCD which will circumscribe a circle with its centre O in a way that it touches the circle at point P, Q, R, and S. Now, after joining the vertices of ABCD we get the following figure:
Now, consider the triangles OAP and OAS,
AP = AS (They are the tangents from the same point A)
OA = OA (It is the common side)
OP = OS (They are the radii of the circle)
So, by SSS congruency △OAP ≅ △OAS
So, ∠POA = ∠AOS
Which implies that ∠1 = ∠8
Similarly, other angles will be,
∠4 = ∠5
∠2 = ∠3
∠6 = ∠7
Now by adding these angles we get,
∠1+∠2+∠3 +∠4 +∠5+∠6+∠7+∠8 = 360°
Now by rearranging,
(∠1+∠8)+(∠2+∠3)+(∠4+∠5)+(∠6+∠7) = 360°
2∠1+2∠2+2∠5+2∠6 = 360°
Taking 2 as common and solving we get,
(∠1+∠2)+(∠5+∠6) = 180°
Thus, ∠AOB+∠COD = 180°
Similarly, it can be proved that ∠BOC+∠DOA = 180°