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Question

Prove that a parallelogram circumscribing a circle is a rhombus.

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Solution

Given ABCD be a parallelogram circumscribing a circle with centre O.
To Prove : ABCD is a rhombus.

We know that the tangents drawn to a circle from an exterior point are equal is length.
AP = AS, BP = BQ, CR = CQ and DR = DS.
AP+BP+CR+DR = AS+BQ+CQ+DS
(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)
AB+CD=AD+BC

or 2AB=2AD (since AB=DC and AD=BC of parallelogram ABCD)
AB=BC=DC=AD
Therefore, ABCD is a rhombus.


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