Question

Prove that a parallelogram circumscribing a circle is a rhombus.

Answer 1

Let ABCD be the rhombus circumscribing the circle with centre O, such that AB, BC, CD and DA touch the circle at points P, Q, R and S respectively. We know that the tangents drawn to a circle from an exterior point are equal in length.

∴ AP = AS  ...(i) [Tangents from A]

BP = BQ   ...(ii) [Tangents from B]

CR = CQ  ...(iii) [Tangents from C]

DR = DS  ...(iv) [Tangents from D]

∴ AB + CD = AP + BP + CR + DR

= AS + BQ + CQ + DS  ...[From (i), (ii), (iii), (iv)]

= (AS + DS) + (BQ + CQ)

= AD + BC

Hence, (AB + CD) = (AD + BC)

⇒ 2AB = 2AD   ...[∵ Opposite sides of a parallelogram are equal.]

⇒ AB = AD

∴ CD = AB = AD = BC

Hence, ABCD is a rhombus.