#### Question

Prove that the paralleogram circumscribing a circle, is a rhombus

#### Solution

Given: A parallelogram ABCD circumsribes a circle with centre O.

To prove: AB = BC = CD = AD

Proof: we know that the lengths of tangents drawn from an exterior point to a circle are equal.

∴ 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

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Solution Prove that the Paralleogram Circumscribing a Circle, is a Rhombus Concept: Circles Examples and Solutions.