Let ABCD be a cyclic quadrilateral in which the side AB is extended up to point P. We need to prove that the exterior angle \(\angle\)CBP is equal to the interior opposite angle \(\angle\)ADC. Proof: On the extended side AP, BC stands at point B. \(\therefore \angle\)CBP + \(\angle\)ABC = 180\(^\circ\) Again, since ABCD is a cyclic quadrilateral, \(\therefore \angle\)ADC + \(\angle\)ABC = 180\(^\circ\) \(\therefore \angle\)CBP + \(\angle\)ABC = \(\angle\)ADC + \(\angle\)ABC That is, \(\angle\)CBP = \(\angle\)ADC (Proved)