Given: A circle with center 'O'.

A parallelogram ABCD, circumscribing the given circle.

Let P, Q, R, S be the points of contact.

Required to prove: □ABCD is a rhombus.

Proof: AP = AS............(1)

[∵ tangents drawn from an external point to a circle are equal]

BP = ------------(2)

CR = -----------(3)

DR = ------------(4)

Adding (1), (2), (3) and (4) we get

AP + BP + CR + DR = AS + BQ + CQ + DS

(AP + ) + (CR + DR) = (AS + DS) + (BQ + CQ)

AB + DC = AD + BC

AB + AB = AD + AD

[∵ Opposite sides of a parallelogram are equal]

2AB = 2AD

AB =

Hence, AB = CD and AD = BC [∵ Oppsite sides of a parallelogram]

∴ AB = BC = CD = AD

Thus □ABCD is a rhombus (Q.E.D)