If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.

Let a circle touch the sides AB, BC, CD and DA of parallelogram ABCD at P, Q, R and S respectively.

We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

Now, from point A, AP and AS are tangents to the circle.

So, AP = AS ………. (1)

Similarly, we also have

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

From figure,

⇒ AP + BP = AB, BQ + CQ = BC, CR + DR = CD and AS + DS = AD.

∴ AB + CD = AD + BC ……….(5)

As,

AB = CD and BC = AD [Opposite sides of a parallelogram]

Substituting values in equation 5, we get :

⇒ AB + AB = BC + BC

⇒ 2AB = 2 BC

⇒ AB = BC

∴ AB = BC = CD = DA.

Hence, proved that ABCD is a rhombus.