Prove that the rectangle circumscribing a circle is a square.

Given: A rectangle $ABCD$ circumscribe a circle which touches the circle at $P,\ Q,\ R,\ S$.

To do: To prove that $ABCD$ is a square.

As tangents from external point are equal.

$AP = AS\ ..........\ ( 1)$

$PB = BQ\ ..........\ ( 2)$

$DR = DS\ ..........\ ( 3)$

$RC = QC\ ..........\ ( 4)$

Add (1), (2), (3) and (4)

$\Rightarrow AP+PB+DR+RC = AS +BQ+DS+QC$

$\Rightarrow AB +CD=AD+BC$

$\Rightarrow 2AB =2BC$

Adjacent sides are equal so $ABCD$ is a square.