How would you use given figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is $180^\circ$?

Given: $ABCD$ is a cyclic quadrilateral with centre $O$.

Since, $OA = OB = OC = OD$ [Radii of the same circle]

$\therefore$ Triangles $OAB$, $OBC$, $OCD$ and $ODA$ are isosceles, so their base angles are equal:

$$\begin{aligned} \angle OAB &= \angle OBA = p, \\ \angle OBC &= \angle OCB = q, \\ \angle OCD &= \angle ODC = u, \\ \angle ODA &= \angle OAD = v \end{aligned}$$

Now, the radius at each vertex divides that angle into two parts, so

$$\begin{aligned} \angle A &= p + v, \\ \angle B &= p + q, \\ \angle C &= q + u, \\ \angle D &= u + v \end{aligned}$$

Since the sum of the angles of a quadrilateral is $360^\circ$,

$$\begin{aligned} \angle A + \angle B + \angle C + \angle D &= 360^\circ \\ 2(p + q + u + v) &= 360^\circ \\ \therefore p + q + u + v &= 180^\circ \end{aligned}$$

Therefore,

$$\begin{aligned} \angle A + \angle C &= (p + v) + (q + u) = 180^\circ \\ \angle B + \angle D &= (p + q) + (u + v) = 180^\circ \end{aligned}$$

Hence, the opposite angles of a cyclic quadrilateral are supplementary.

Hence proved.