Question:
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
Let ABCD be a parallelogram such that its sides touch a circle with centre O.
AP = AS [Tangents from an external point are equal]
BP = BQ
CR = CQ
DR = DS
Adding these equations
$A P+B P+C R+D R=A S+D S+B Q+C Q$
$\mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC}$
2AB = 2BC
AB = BC
AB = BC = CD = DA
ABCD is a rhombus.
Hence proved