Opposite Sides Equal implies Parallelogram

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Theorem

Let $ABCD$ be a convex quadrilateral with $AB = CD$ and $BC = AD$.

Then $ABCD$ is a parallelogram.


Proof

Join $AC$.

\(\text {(1)}: \quad\) \(\ds AB\) \(=\) \(\ds CD\) Given
\(\text {(2)}: \quad\) \(\ds BC\) \(=\) \(\ds DA\) Given
\(\text {(3)}: \quad\) \(\ds AC\) \(=\) \(\ds CA\) Equality is Reflexive
\(\text {(4)}: \quad\) \(\ds \Delta ABC\) \(=\) \(\ds \Delta CDA\) SSS from $(1)$, $(2)$, and $(3)$
\(\text {(5)}: \quad\) \(\ds \angle BCA\) \(=\) \(\ds \angle DAC\) from $(4)$
\(\text {(6)}: \quad\) \(\ds \angle BAC\) \(=\) \(\ds \angle DCA\) from $(4)$
\(\text {(7)}: \quad\) \(\ds BC\) \(\parallel\) \(\ds DA\) Equal Alternate Angles implies Parallel Lines from $(5)$
\(\text {(8)}: \quad\) \(\ds BA\) \(\parallel\) \(\ds DC\) Equal Alternate Angles implies Parallel Lines from $(6)$

From $(7)$ and $(8)$, it follows by definition that $ABCD$ is a parallelogram.

$\blacksquare$