Given: A quadrilateral ABCD circumscribes a circle with centre O. Let P, Q, R, S be the points of contact of sides AB, BC, CD, DA respectively with the circle.

To Prove: $\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle AOD = 180^\circ$

Construction: Join OA, OB, OC, OD, OP, OQ, OR, OS.

Proof:

Since tangents drawn from an external point to a circle are equal (Theorem 10.2):

$$AP = AS, \quad BP = BQ, \quad CQ = CR, \quad DR = DS \quad \ldots \text{(i)}$$

In $\Delta AOP$ and $\Delta AOS$:

$$OP = OS \text{ (radii)}, \quad AP = AS \text{ (from i)}, \quad OA = OA \text{ (common)}$$ $$\therefore \Delta AOP \cong \Delta AOS \text{ (by SSS)}$$ $$\text{Hence, } \angle AOP = \angle AOS \quad \ldots \text{(ii)}$$

Similarly,

$$\Delta BOP \cong \Delta BOQ \text{ giving } \angle BOP = \angle BOQ \quad \ldots \text{(iii)}$$ $$\Delta COQ \cong \Delta COR \text{ giving } \angle COQ = \angle COR \quad \ldots \text{(iv)}$$ $$\Delta DOR \cong \Delta DOS \text{ giving } \angle DOR = \angle DOS \quad \ldots \text{(v)}$$

Now,

$$\angle AOB = \angle AOP + \angle BOP$$

And

$$\angle COD = \angle COR + \angle DOR = \angle COQ + \angle DOS \text{ (using iv, v)}$$

Adding:

$$\angle AOB + \angle COD = \angle AOP + \angle BOP + \angle COQ + \angle DOS$$ $$= \angle AOS + \angle BOQ + \angle COQ + \angle DOS \text{ (using ii, iii)}$$ $$= (\angle AOS + \angle DOS) + (\angle BOQ + \angle COQ)$$ $$= \angle AOD + \angle BOC \quad \ldots \text{(vi)}$$

Since angles around point O sum to $360^\circ$:

$$\angle AOB + \angle BOC + \angle COD + \angle AOD = 360^\circ$$

From (vi), $\angle AOB + \angle COD = \angle AOD + \angle BOC$

Let $\angle AOB + \angle COD = x$ and $\angle AOD + \angle BOC = y$

Then $x + y = 360^\circ$ and $x = y$

$$\therefore 2x = 360^\circ, \text{ giving } x = 180^\circ$$

Therefore, $\angle AOB + \angle COD = 180^\circ$ and $\angle AOD + \angle BOC = 180^\circ$

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre.