In the given figure, the parallelogram ABCD circumscribe a circle, touching circle at P, Q, R and S.

(a) Prove that: AB = BC

(b) What special name can be given to the parallelogram ABCD?

(a) We know that,

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

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

From figure,

⇒ AB + CD = (AP + BP) + (DR + CR)

⇒ AB + CD = (AS + BQ) + (DS + CQ)

⇒ AB + CD = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC …….(1)

We know that,

Opposite sides of parallelogram are equal.

∴ AB = CD and AD = BC

Substituting above values in equation (1), we get :

⇒ AB + AB = BC + BC

⇒ 2AB = 2BC

⇒ AB = BC.

Hence, proved that AB = BC.

(b) Since, AB = BC, AB = CD and AD = BC.

∴ AB = BC = CD = AD.

A parallelogram with all four sides equal is a rhombus.

Hence, ABCD is a rhombus.