If a parallelogram circumscribes a circle then prove that it must be a rhombus.

Explanation

Given a parallelogram, say ABCD, Let this parallelogram touch the circle at the point P, Q, R and S.
As AP and AS are tangents to the circle drawn from an external point A.
`AP=AS" "…(1)`
Similarly,`" "BP=BQ" "…(2)`
`CR=CQ" "...(3)`
`DR=DS" "...(4)`
Circles
Adding (1), (2), (3) and (4), we get
`implies" "AP+BP+CR+DR=AS+BQ+CQ+DS`
`implies" "(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)`
`implies" "AB+CD=AD+BC`
`implies" "AB+AB=AD+AD`
(CD=AB, DB=AD, opposite sides of a parallelogram)
`implies" "2AB=2AD`
`implies" "AB=AD`
Hence, ABCD is a rhombus. (`because` adjacent sides of a parallelogram are equal) Hence Proved.