Correct option is: (B) a rhombus
Let sides AB, BC, CD and AD of parallelogram ABCD touches circle at points P, S, R an Q respectively.
\(\because\) Tangents drawn from a fixed outer point to a circle are of equal length.
\(\therefore\) AP = AQ ...(1) (Tangents to circle from outer point A).
BP = BS...(2) (Tangents to circle from outer point B).
CR = CS ...(3) (Tangents to circle from outer point C).
DR = DQ ...(4) (Tangents to circle from outer point D).
By adding equations (1), (2), (3) & (4) we get.
AP + BP + CR + DR = Aq + BS + CS + DQ.
= (AP + BP) + (CR + DR) = (AQ + DQ) + (BS + CS).
= AB + CD = AD + BC.
= 2 AB = 2BC (\(\because\) Opposite sides are equal in a parallelogram).
= AB = BC.
\(\therefore\) AB = BC = AD = CD.
\(\therefore\) Parallelogram ABCD is a rhombus.
