Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel

Theorem

Let $ABCD$ be a quadrilateral.

Then:

$ABCD$ is a parallelogram

if and only if:

$AB = CD$ and $AB \parallel CD$

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

Proof

Sufficient Condition

Let $ABCD$ be a parallelogram.

Then $AB \parallel CD$ by definition.

From Opposite Sides and Angles of Parallelogram are Equal it follows that $AB = CD$.

$\Box$

Necessary Condition

Let $AB = CD$ and $AB \parallel CD$.

Then from Lines Joining Equal and Parallel Straight Lines are Parallel, $AD \parallel BC$.

Thus we have $AB \parallel CD$ and $AD \parallel BC$, and so by definition $ABCD$ is a parallelogram.

$\blacksquare$

Sources

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