Prove that a parallelogram circumscribing a circle is a rhombus.

ABCD is a parallelogram touching the circle at P, Q, R, S by sides AB, BC, CD, DA respectively.
We know that tangents drawn from the external point to a circle are equal.
$\therefore AP = AS$
quad --------(i)
$PB = BQ$
quad --------(ii)
$CR = CQ$
quad --------(iii)
$DR = DS$
quad ---------(iv)
Adding (i), (ii), (iii), (iv)
$(AP + PB) + (CR + DR) = (AS + DS) + (BQ + CQ)$
$AB + CD = AD + BC$
ABCD is a parallelogram
$\Rightarrow AB = CD, AD = BC$
$\Rightarrow 2AB = 2AD \Rightarrow AB = AD$
$\Rightarrow \text{ABCD is a rhombus.}$