Question

Prove that a parallelogram circumscribing a circle is a rhombus [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks

Since ABCD is a parallelogram,

AB = CD …(i)

BC = AD …(ii)

As the lengths of tangents drawn from an external point to a circle are equal

DR = DS

CR = CQ

BP = BQ

AP = AS

Adding all these equations, we obtain

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

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

CD + AB = AD + BC

On putting the values of equations (i) and (ii) in this equation, we obtain

2AB = 2BC

AB = BC …(iii)

Comparing equations (i), (ii), and (iii), we obtain

AB = BC = CD = DA

Hence, ABCD is a rhombus.