Given: AB and CD are two chords of a circle. When extended, BA and DC intersect at point P. To Prove: \(\angle\)PCB = \(\angle\)PAD Proof: \(\angle\)BCD = \(\angle\)BAD [Angles subtended by the same arc in a circle] Also, \(\angle\)PCB = 180° − \(\angle\)BCD And \(\angle\)PAD = 180° − \(\angle\)BAD \(\because \angle\)BCD = \(\angle\)BAD Therefore, 180° − \(\angle\)BCD = 180° − \(\angle\)BAD \(\therefore \angle\)PCB = \(\angle\)PAD [Proved]