Given: A circle with centre ‘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 = BQ ……. (2)
CR = CQ ……. (3)
DR = DS ……. (4)
Adding (1), (2), (3) and (4) we get
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + DC = AD + BC
AB + AB = AD + AD
[∵ Opposite sides of a parallelogram are equal]
2AB = 2AD AB = AD
Hence, AB = CD and AD = BC [∵ Opposite sides of a parallelogram]
∴ AB = BC = CD = AD
Thus □ ABCD is a rhombus (Q.E.D.)
