Question

Prove that a parallelogram circumscribing a circle is a rhombus.

Answer 1

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.
$\therefore$ 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)
$\therefore$ AB+CD=AD+BC

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