**Solution:**

**ABCD is a parallelogram,**

∴ AB = CD ... (i)

∴ BC = AD ... (ii)

From the figure, we observe that,

DR = DS (Tangents to the circle at D)

CR = CQ (Tangents to the circle at C)

BP = BQ (Tangents to the circle at B)

AP = AS (Tangents to the circle at A)

Adding all these,

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

⇒ (DR + CR) + (BP + AP) =

(DS + AS) + (CQ + BQ)

⇒ CD + AB = AD + BC ... (iii)

Putting the value of (i) and (ii) in equation (iii) we get,

2AB = 2BC

⇒ AB = BC ... (iv)

By Comparing equations (i), (ii), and (iv) we get,

AB = BC = CD = DA

**∴ ABCD is a rhombus.**