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.
$\blacksquare$
Sources
- 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.24$