Rectangle is Parallelogram

Theorem

Let $ABCD$ be a rectangle.

Then $ABCD$ is a parallelogram.

Proof

Let $ABCD$ be a rectangle.

Aiming for a contradiction, suppose $ABCD$ is not a parallelogram.

Without loss of generality, let line segments $AB$ and $CD$ not be parallel.

By Euclid's Second Postulate, let us produce $AB$ and $CD$ into two infinite straight lines.

By the Parallel Postulate, the $AD$ and $BC$ will eventually meet at one side or the other.

Let their point of intersection be $E$.

There are two possibilities:

$ADE$ is a triangle

$CBE$ is a triangle.

Without loss of generality, let $\triangle ADE$ be a triangle:

By Sum of Angles of Triangle equals Two Right Angles, $\angle ADE + \angle DEA + \angle EAD$ add to two right angles.

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

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

Hence $ABCD$ is not a rectangle.

But by hypothesis, $ABCD$ is a rectangle.

This is a contradiction.

Hence by Proof by Contradiction, $AB$ and $CD$ are parallel.

The same argument can be used to demonstrate that $AD$ and $BC$ are also parallel.

Hence the result by definition of parallelogram.

$\blacksquare$