Given: ABCD is a parallelogram circumscribing a circle with centre O. Let the circle touch sides AB, BC, CD, and DA at points P, Q, R, and S respectively.

To prove: ABCD is a rhombus.

Proof:

Since ABCD is a parallelogram, opposite sides are equal:

$$AB = CD \text{ and } BC = AD \quad \ldots \text{(1)}$$

By Theorem 10.2, tangents drawn from an external point to a circle are equal in length.

From vertex A:

$$AP = AS \quad \ldots \text{(2)}$$

From vertex B:

$$BP = BQ \quad \ldots \text{(3)}$$

From vertex C:

$$CQ = CR \quad \ldots \text{(4)}$$

From vertex D:

$$DR = DS \quad \ldots \text{(5)}$$

Now,

$$AB + CD = (AP + BP) + (CR + DR) = AP + BP + CR + DR \quad \ldots \text{(6)}$$

And,

$$AD + BC = (AS + DS) + (BQ + CQ) = AS + DS + BQ + CQ \quad \ldots \text{(7)}$$

Using equations (2), (3), (4), (5) in equation (7):

$$AD + BC = AP + DR + BP + CR = AP + BP + CR + DR \quad \ldots \text{(8)}$$

From (6) and (8):

$$AB + CD = AD + BC$$

Using (1):

$$AB + AB = BC + BC, \text{ which gives } AB = BC$$

Since $AB = CD$, $BC = AD$ (parallelogram property), and $AB = BC$:

∴ All four sides are equal. Hence, ABCD is a rhombus.