Rectangle is Parallelogram

Theorem

Let $ABCD$ be a rectangle.

Then $ABCD$ is a parallelogram.

Proof

Let $ABCD$ be a rectangle.

Seeking a contradiction, assume $ABCD$ is not a parallelogram.

Without loss of generality let line segments $AD$ and $BC$ not be parallel.

Then extend $AD$ and $BC$ into two infinite straight lines.

By the Parallel Postulate, the lines will eventually meet at one side or the other.

Let their point of intersection be $E$.

There are two possibilities:

$ABE$ is a triangle

$CDE$ is a triangle.

Without loss of generality, let $ABE$ be a triangle.

If $ABE$ is a triangle, then by Sum of Angles of Triangle equals Two Right Angles, $\angle ABE + \angle BEA + \angle EAB$ add to two right angles.

But by the definition of a rectangle, $\angle ABE + \angle EAB$ add to two right angles.

As $\angle BEA$ is not a zero angle (else $ABD$ would not be a triangle), this cannot be so.

We are forced to conclude that that $ABCD$ is not a rectangle, a contradiction.

$\blacksquare$