### Theory:

A four-sided figure inscribed in a circle such that all its vertices lie on the circumference of the circle is said to be a cyclic quadrilateral. Theorem: The sum of either pair of opposite angles of a cyclic quadrilateral is $$180^{\circ}$$.
Explanation: The theorem states that the sum of the interior opposite angles of a cyclic quadrilateral is $$180^{\circ}$$.

That is, $$\angle A + \angle C = 180^{\circ}$$.

And $$\angle B + \angle D = 180^{\circ}$$.

Proof of the theorem:

Consider a quadrilateral $$ABCD$$ whose vertices lie on the circumference of the circle with centre $$O$$.

Connect all the four vertices with centre $$O$$ to get four isosceles triangle $$AOB$$, $$BOC$$, $$COD$$ and $$DOA$$ where the sides $$OA$$, $$OB$$, $$OC$$ and $$OD$$ are the radii. The sum of the angles around the centre of a circle is $$360^{\circ}$$.

Also, the sum of all the interior angles of a triangle is $$180^{\circ}$$.

Hence:

$\begin{array}{l}\angle w\phantom{\rule{0.147em}{0ex}}+\angle w\phantom{\rule{0.147em}{0ex}}+\angle \mathit{AOD}\phantom{\rule{0.147em}{0ex}}=180\mathrm{°}\phantom{\rule{0.147em}{0ex}}\\ \angle x\phantom{\rule{0.147em}{0ex}}+\angle x\phantom{\rule{0.147em}{0ex}}+\angle \mathit{AOB}\phantom{\rule{0.147em}{0ex}}=180\mathrm{°}\\ \angle y\phantom{\rule{0.147em}{0ex}}+\angle y\phantom{\rule{0.147em}{0ex}}+\angle \mathit{BOC}\phantom{\rule{0.147em}{0ex}}=180\mathrm{°}\\ \angle z\phantom{\rule{0.147em}{0ex}}+\angle z\phantom{\rule{0.147em}{0ex}}+\angle \mathit{COD}\phantom{\rule{0.147em}{0ex}}=180\mathrm{°}\end{array}$

Adding all the equations we have:

$$2 (\angle w + \angle x + \angle y + \angle z) + \angle O = 4\times180^{\circ}$$

$$2 (\angle w + \angle x + \angle y + \angle z) + 360^{\circ} = 720^{\circ}$$

$$2 (\angle w + \angle x + \angle y + \angle z) = 720^{\circ} - 360^{\circ}$$

$$2 (\angle w + \angle x + \angle y + \angle z) = 360^{\circ}$$

$$\angle w + \angle x + \angle y + \angle z = 180^{\circ}$$

This is rewritten as, $$\angle A + \angle C = 180^{\circ}$$.

Similarly, $$\angle B + \angle D = 180^{\circ}$$.
Example:
Find the unknown angle $$x$$ in the given figure. Solution:

By the theorem, opposite angles of a cyclic quadrilateral are supplementary.

This implies:

$$\angle A + \angle C = 180^{\circ}$$

$$x + 115^{\circ} = 180^{\circ}$$

$$x = 180^{\circ} - 115^{\circ}$$

$$x = 65^{\circ}$$
Converse of Theorem: If the sum of a pair of opposite angles of a quadrilateral is $$180^{\circ}$$, the quadrilateral is cyclic.
Explanation: The theorem states that if the sum of the interior opposite angles of any quadrilateral is  $$180^{\circ}$$, then that quadrilateral is said to be cyclic. Here, in the figure, the sum of the $$\angle A$$ and $$\angle C$$ is $$180^{\circ}$$.

$$90^{\circ} + 90^{\circ} = 180^{\circ}$$

Thus, the quadrilateral $$ABCD$$ is cyclic.
Example:
Prove that a square inscribed in a circle is cyclic.

Proof:

Let $$ABCD$$ be the square inscribed in a circle. It is known that every angle of a square is $$90^{\circ}$$.

Here $$\angle A + \angle C = 90^{\circ} + 90^{\circ}$$

Implies, $$\angle A + \angle C$$ $$=$$ $$180^{\circ}$$

Similarly, $$\angle B + \angle D = 90^{\circ} + 90^{\circ}$$

Implies, $$\angle B + \angle D$$ $$=$$ $$180^{\circ}$$

According to the theorem, the pair of opposite angles of the square is supplementary.

Hence the square inscribed in a circle is cyclic.