Parallelograms are Congruent if Two Adjacent Sides and Included Angle are respectively Equal

Theorem
Let $ABCD$ and $EFGH$ be parallelograms.

Then $ABCD$ and $EFGH$ are congruent if:
 * $2$ adjacent sides of $ABCD$ are equal to $2$ corresponding adjacent sides of $EFGH$
 * the angle between those $2$ adjacent sides on both $ABCD$ and $EFGH$ are equal.

Proof
let the $2$ adjacent sides of $ABCD$ be $AB$ and $BC$.

Let the $2$ corresponding adjacent sides of $EFGH$ be $EF$ and $FG$ such that $AB = EF$ and $BC = FG$.

Hence let $\angle ABC = \angle EFG$.

From Triangle Side-Angle-Side Equality we have that $\triangle ABC = \triangle EFG$.

From Quadrilateral is Parallelogram iff Both Pairs of Opposite Angles are Equal:
 * $\angle ABC = \angle ADC$
 * $\angle EFG = \angle EHG$

From Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel, $AB = CD$ and $EF = GF$.

Thus again by Triangle Side-Angle-Side Equality we have that $\triangle ADC = \triangle EHG$.

So:
 * $\triangle ABC + \triangle ADC = \triangle EFG + \triangle EHG$

and the result follows.