In the diagram, O is the centre of the circle - NSC Mathematics - Question 9 - 2023 - Paper 2

In triangle OAB and OCD, we have:

$\angle OAB = 2\bar{B}$ $\angle OCD = 2\bar{D}$

Since OA and OC are radii of the circle.

The total angle around point O is equal to $360^{ ext{o}}$. Thus:

$\angle OAB + \angle OCD = 360^{ ext{o}}$

Substituting the angle relationships into the equation gives:

$2\bar{B} + 2\bar{D} = 360^{ ext{o}}$

Dividing through by 2:

$\bar{B} + \bar{D} = 180^{ ext{o}}$

This proves that the opposite angles in a cyclic quadrilateral are supplementary.