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Prove that the parallelogram circumscribing a circle is a rhombus.

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To prove: Parallelogram ABCD is a rhombus. 

ABCD is a parallelogram in which ‘O’ is the centre of the interior circle drawn in the parallelogram. 

Tangents AC and BD are drawn which intersect at ‘OP’.

Sides of the parallelogram AB. BC. CD and AD touches at the points P, Q, R and S respectively. OP, OS joined.

We have OP ⊥ AB, OS ⊥ AD.

In ⊥∆ OPB and ∆OSD, 

∠OPB = ∠OSD =90° 

OB = OD (tangent is bisected) 

OP = OS (Radii) 

∴ ∆OPB = ∆OSD 

∴ PB = SD ……………. (i) 

AP = AS ……………….. (ii) 

(∵ tangent of the external point) 

By combining eqn. (i) and eqn.(ii), 

AP + PB = AS + SC 

AB = AD 

Similarly, AB = BC = CD = DA 

∴ Parallelogram ABCD is a rhombus

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