Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., $\angle CDE = \angle ABC$, where E is a point on the extension of side CD).

In cyclic quadrilateral ABCD:

$\angle ABC + \angle ADC = 180^\circ$ ...(i)

Since E lies on the extension of CD

$\angle ADC$ and $\angle ADE$ form a linear pair.

So, $\angle ADC + \angle ADE = 180^\circ$ ...(ii)

From (i) and (ii), we have

$\angle ABC + \angle ADC = \angle ADC + \angle ADE$

Subtracting $\angle ADC$ from both sides, we get

$\angle ABC = \angle ADE$

Therefore, the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Hence proved.