In the above figure, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA.

The figure shows that the tangents drawn from the exterior point to a circle are equal in length.
As DR and DS are tangents from exterior point D so, DR = DS —- (1)
As AP and AS are tangents from exterior point A so, AP = AS —- (2)
As BP and BQ are tangents from exterior point B so, BP = BQ —- (3)
As CR and CQ are tangents from exterior point C so, CR = CQ —- (4)
Adding the equation 1, 2, 3 & 4, we get
DR + AP + BP + CR = DS + AS + BQ + CQ
(DR + CR) + (AP + BP) = (DS + AS) + (BQ + CQ)
CD + AB = DA + BC
AB + CD = BC + DA
Hence proved.