In Fig. 3, a quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC

From the figure,

AP = AS [Tangents from an external point to the same circle are equal]

DS = DR [Tangents from an external point to the same circle are equal]

CR = CQ [Tangents from an external point to the same circle are equal]

BQ = BP [Tangents from an external point to the same circle are equal]

⇒ AB + CD

= AP + BP + DR + CR

= AS + BQ + DS + CQ [AP = AS, DS = DR, CR = CQ, BQ = BP]

= AS + DS + BQ + CQ

= AD + BC

Hence Proved.