Prove that a rectangle circumscribing a circle is a square.

Solution:
As the length of tangents from an external point to a circle are equal
Thus,
$AP = AS$
$BP = BQ$
$DR = DS$
$CR = CQ$
Adding the above equations,
$AB + CD = BC + AD$
As $AB = CD$ & $BC = AD$ (opp. sides of rectangle)
$\Rightarrow AB = AD$
$\therefore ABCD$ is a square