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Illustration:

Consider a cyclic quadrilateral $$ABCD$$ and extend the side $$AB$$ to $$E$$.

Here, the angles $$\angle ABC$$ and $$\angle CBE$$ are a linear pair.

It is known that the sum of the linear angle is $$180^{\circ}$$.

So, $$\angle ABC$$ $$+$$ $$\angle CBE$$ $$=$$ $$180^{\circ}$$             $$……$$ $$\text{equation }(1)$$

Also, in a cyclic quadrilateral, the sum of the interior opposite angles is $$180^{\circ}$$.

So, $$\angle ABC$$ $$+$$ $$\angle ADC$$ $$=$$ $$180^{\circ}$$             $$……$$ $$\text{equation }(2)$$

From equation $$(1)$$ and $$(2)$$, we have:

$$\angle ABC$$ $$+$$ $$\angle CBE$$ $$=$$ $$\angle ABC$$ $$+$$ $$\angle ADC$$

Hence, $$\angle CBE$$ $$=$$ $$\angle ADC$$.

Similarly, all the other angles can be proved.

Based on this result, the following theorem is obtained.
Theorem: If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.

Explanation:

The theorem states that if any side of a cyclic quadrilateral is produced (extended), then the exterior angle formed on that side is the same as the interior opposite angle.

Here, in the figure, the side $$AB$$ is produced to $$X$$, the side $$BC$$ is produced to $$Y$$, the side $$CD$$ is produced to $$Z$$ and the side $$DA$$ is produced to $$W$$. Then:

$$\angle CBX = \angle CDA$$

$$\angle DCY = \angle DAB$$

$$\angle ADZ = \angle ABC$$

$$\angle BAW = \angle BCD$$
Example:
Find the unknown angle $$z$$ in the figure.

Solution:

By the exterior angle of a cyclic quadrilateral theorem, $$\text{Exterior angle}$$ $$=$$ $$\text{Interior opposite angle}$$.

Hence, $$\angle RSX$$ $$=$$ $$\angle PQR$$

Therefore, $$z$$ $$=$$ $$112^{\circ}$$.