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

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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

Without loss of generality 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 Congruence:

$\triangle ABC = \triangle EFG$

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 Congruence:

$\triangle ADC = \triangle EHG$

So:

$\triangle ABC + \triangle ADC = \triangle EFG + \triangle EHG$

and the result follows.

$\blacksquare$


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