Prove that, any rectangle is a cyclic quadrilateral.

To prove that any rectangle is a cyclic quadrilateral, we need to show that the four vertices of the rectangle lie on a single circle.

A rectangle is a quadrilateral with four right angles. Let's label the vertices of the rectangle as A, B, C, and D in a counterclockwise or clockwise order. Without loss of generality, let's assume that A is the top-left vertex, B is the top-right vertex, C is the bottom-right vertex, and D is the bottom-left vertex of the rectangle.

Now, let's consider the diagonals of the rectangle:

The diagonal AC connects the top-left vertex A to the bottom-right vertex C.
The diagonal BD connects the top-right vertex B to the bottom-left vertex D.
Since the opposite angles of a rectangle are congruent and equal to 90 degrees, angle A = angle C = 90 degrees, and angle B = angle D = 90 degrees.

In a cyclic quadrilateral, the opposite angles are supplementary, which means the sum of opposite angles is 180 degrees. Therefore, for a rectangle to be cyclic, we need to prove that the sum of opposite angles is 180 degrees.

Let's consider angle A and angle B:

angle A + angle B = 90 degrees + 90 degrees = 180 degrees.

Similarly, let's consider angle C and angle D:

angle C + angle D = 90 degrees + 90 degrees = 180 degrees.

Since the sum of opposite angles (A + B and C + D) in the rectangle is 180 degrees for both pairs, this implies that the four vertices A, B, C, and D lie on a single circle, making it a cyclic quadrilateral.

So, we have shown that any rectangle is indeed a cyclic quadrilateral because its vertices lie on a single circle, and the sum of opposite angles is 180 degrees.