Prove that the rectangle circumscribing a circle is a square.

Hint: We first use the theorem of tangents’ length being the same from an outside point. We use that to find the sum of the opposite sides being equal. Then we use the rectangle properties to find the rectangle being a square.

Complete step by step solution:
We first try to draw a diagram for the rectangle circumscribing a circle.

In the figure, we got a circle with centre O where the rectangle ABCD touches the circle at points P, Q, R, S.
We know that the theorem of the tangent of a circle tells us that the length of the two tangents on a circle from an outside point will be equal.
For our given circle there are 4 outside points A, B, C, D which have two tangents for each of the points.
For point A, we have AS and AP which gives $ AS=AP $ .
Similarly, $ BP=BQ;CQ=CR;DR=DS $ .
We add these four equalities and get
 $
   AS+BQ+CQ+DS=AP+BP+CR+DR \\
  \Rightarrow AD+BC=AB+CD \;
 $
Now it's given that the quadrilateral is rectangle which means the opposite sides are equal.
So, $ AD=BC;AB=CD $ . This means
 $
   AD+BC=AB+CD \\
  \Rightarrow 2AD=2AB \\
  \Rightarrow AD=AB \;
 $
This proves the consecutive sides are equal also.
Therefore, the rectangle is square.
Thus proved, the rectangle circumscribing a circle is a square.

Note: We need to remember that the properties of $ AD+BC=AB+CD $ , the sum of the opposite sides being equal is not only for the rectangle. It can be applied for any quadrilateral.