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

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Theorem

Let $ABCD$ be a quadrilateral.

Then:

$ABCD$ is a parallelogram

if and only if:

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

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


Proof

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 $AD \parallel BC$.

$\Box$


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.

$\blacksquare$


Sources