Parallelogram with One Right Angle is Rectangle

Theorem

Let $\Box ABCD$ be a parallelogram.

Let one angle of $\Box ABCD$ be a right angle.

Then $\Box ABCD$ is a rectangle.

Proof

Without loss of generality, let $\angle ABC$ be a right angle.

Let $\angle CDA$ be opposite $\angle ABC$.

By Opposite Sides and Angles of Parallelogram are Equal:

$\angle CDA = \angle ABC$

By Sum of Internal Angles of Polygon:

the sum of the angles of $\Box ABCD$ is four right angles.

Hence: $\angle DAB + \angle BCD$ together equal two right angles.

By Opposite Sides and Angles of Parallelogram are Equal:

$\angle DAB = \angle BCD$

Hence, all four angles of $\Box ABCD$ are right angles.

By Opposite Sides and Angles of Parallelogram are Equal:

$AB = CD$

$AD = BC$

$\leadsto$

$\Box ABCD$ is a rectangle.

The result follows.

$\blacksquare$