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 $AC$.

\(\text {(1)}: \quad\)

\(\ds AB\)

\(=\)

\(\ds CD\)

by hypothesis

\(\text {(2)}: \quad\)

\(\ds BC\)

\(=\)

\(\ds DA\)

by hypothesis

\(\text {(3)}: \quad\)

\(\ds AC\)

\(=\)

\(\ds CA\)

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.

$\blacksquare$

Sources

• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): parallelogram