# Opposite Sides Equal implies Parallelogram

## 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.

$\blacksquare$