Question

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

1. Prove that: AB = BC 
2. What special name can be given to the parallelogram ABCD?

Answer 1

a. We know that if two tangents are drawn to a circle from an exterior point, the tangents are equal in 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 a parallelogram are equal.

∴ AB = CD and AD = BC

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

⇒ AB + AB = BC + BC

⇒ 2AB = 2BC

⇒ AB = BC.

Hence, it is 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.