Properties of Parallelograms went from "parallelogram" to its properties. Now we go the other way: which properties are enough to prove that a quadrilateral is a parallelogram? Each test below is a converse of a property, and each one is a theorem.
- Both pairs of opposite sides are parallel (the definition).
- Both pairs of opposite sides are congruent.
- Both pairs of opposite angles are congruent.
- The diagonals bisect each other.
- One pair of opposite sides is both parallel and congruent.
Why it works
Here is the proof of the second test. Suppose $AB = CD$ and $BC = DA$. Draw diagonal $\overline{AC}$. Then $\triangle ABC \cong \triangle CDA$ by SSS, so $\angle BAC \cong \angle DCA$ by CPCTC. Those are alternate interior angles for $\overline{AB}$ and $\overline{DC}$ with transversal $\overline{AC}$, so by the converse of the Alternate Interior Angles Theorem, $\overline{AB} \parallel \overline{DC}$. The same argument with the other pair of angles gives $\overline{AD} \parallel \overline{BC}$.
Key idea
A quadrilateral is a parallelogram if both pairs of opposite sides are congruent, or both pairs of opposite angles are congruent, or its diagonals bisect each other, or one pair of opposite sides is both parallel and congruent.
Worked example: Solving so it works
The diagonals of $ABCD$ meet at $E$, with $AE = 3x - 2$, $EC = x + 6$, $BE = y + 5$ and $ED = 2y - 1$. Find $x$ and $y$ so that $ABCD$ is a parallelogram.
- If the diagonals bisect each other, it is a parallelogram. So set $AE = EC$ and $BE = ED$.
- $3x - 2 = x + 6$ gives $x = 4$. $y + 5 = 2y - 1$ gives $y = 6$.
Answer: $x = 4$ and $y = 6$
Worked example: A proof with coordinates
Is $A(-1, 1)$, $B(4, 2)$, $C(6, 6)$, $D(1, 5)$ a parallelogram?
- Test the diagonals: midpoint of $\overline{AC}$ is $\left(\frac{-1 + 6}{2}, \frac{1 + 6}{2}\right) = \left(\frac{5}{2}, \frac{7}{2}\right)$.
- Midpoint of $\overline{BD}$ is $\left(\frac{4 + 1}{2}, \frac{2 + 5}{2}\right) = \left(\frac{5}{2}, \frac{7}{2}\right)$.
- Same point, so the diagonals bisect each other.
Answer: Yes, $ABCD$ is a parallelogram.
Worked example: Not enough
A quadrilateral has one pair of parallel sides and the other pair congruent. Must it be a parallelogram?
- An isosceles trapezoid fits: its bases are parallel and its legs are congruent.
- But its legs are not parallel, so it is not a parallelogram.
Answer: No. The parallel pair and the congruent pair must be the same pair.
Common mistake
Mixing two half-tests. "One pair parallel" plus "the other pair congruent" is not a test; neither is "one pair of congruent sides" alone. Check each test on the list exactly as stated, or sketch a counterexample.