Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel

Theorem
Let $ABCD$ be a quadrilateral.

Then:
 * $ABCD$ is a parallelogram


 * either $AB = CD$ and $AD = BC$
 * or $AB \parallel CD$ and $AD \parallel BC$
 * or $AB \parallel CD$ and $AD \parallel BC$

where $AB \parallel CD$ denotes that $AB$ is parallel to $CD$.

Sufficient Condition
Let $ABCD$ be a parallelogram.

Then by definition:
 * $AB \parallel CD$ and $AD \parallel BC$

and from Opposite Sides and Angles of Parallelogram are Equal:
 * $AB = CD$ and $AD = BC$

From Conjunction implies Disjunction, it follows that:
 * either $AB = CD$ and $AD = BC$
 * or $AB \parallel CD$ and $AD \parallel BC$.

Necessary Condition
Let $ABCD$ be such that:


 * either $AB = CD$ and $AD = BC$
 * or $AB \parallel CD$ and $AD \parallel BC$.

First suppose that $AB = CD$ and $AD = BC$.

Then from Opposite Sides Equal implies Parallelogram, $ABCD$ is a parallelogram.

Now suppose that $AB \parallel CD$ and $AD \parallel BC$.

Then $ABCD$ is a parallelogram by definition.

Hence the result by Proof by Cases.