Question

A quadrilateral ABCD is drawn to circumscribe a circle (Figure given). Prove that AB + CD = AD + BC.

Answer 1

Since lengths of two tangents drawn from an external point of circle are equal,

Therefore,
AP = AS, BP = BQ and DR = DS
CR = CQ [Where P, Q, R and S are the points of contact]
Adding all these, we have
(AP + BP) + (CR + RD) = (BQ + CQ) + (DS + AS)
⇒ AB + CD = BC + DA
Hence proved.