Prove that a parallelogram circumscribing a circle is a rhombus.

Given: ABCD is a ||gm.

To Prove: ABCD is a rhombus.

Theorem Used:

The length of two tangents drawn from an external point are equal.

Explanation:

Let ABCD is a ||gm and its sides touches the circle with centre O.

As A is external point and AS and AP are tangents,

By theorem stated above,

AP = AS …. (1)

Similarly, for point B,

BP = BQ …. (2)

For point C,

CR = CQ …. (3)

For point D,

DR = DS …. (4)

Add 1,2,3 and 4 to get

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

⇒ (AP + BP) + (CR + DR) = (AS + DS) +(BQ + CQ)

⇒ AB + CD = AD + BC

As ABCD is a ||gm,

⇒ AB = CD

AD = BC

⇒ 2 AB = 2AD

⇒ AB = AD

⇒ AB = AD = BC = CD