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
- 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
- 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.23 \ \text{(ii)}$