Prove that the parallelogram circumscribing a circle is a rhombus.
Consider a parallel gram ABCD circumscribing a circle with center O.

To Prove: ABCD is a rhombus or AB = BC = CD = AD
We know, tangents from an external point to a circle are equal
AP = AS
BP = BQ
DR = DS
CR + CQ
Adding above equations, we get
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Now, In a parallelogram AB = CD and BC = AD
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Hence, Proved!
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