Quadrilateral is Parallelogram iff Both Pairs of Opposite Angles are Equal

Theorem

Let $ABCD$ be a quadrilateral.

Then:

$ABCD$ is a parallelogram

if and only if:

$\angle ABC = \angle ADC$ and $\angle BAD = \angle BCD$.

Proof

Sufficient Condition

Let $ABCD$ be a parallelogram.

Then by Opposite Sides and Angles of Parallelogram are Equal:

$\angle ABC = \angle ADC$ and $\angle BAD = \angle BCD$.

$\Box$

Necessary Condition

Let $ABCD$ be such that:

$\angle ABC = \angle ADC$ and $\angle BAD = \angle BCD$.

This theorem requires a proof. In particular: Requires a diagram with a construction that allows the use of Equal Alternate Angles implies Parallel Lines, for example. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

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