Math Core

In the last lesson you started with a parallelogram and listed its properties. Now the question runs the other way: given a quadrilateral, how can you prove it is a parallelogram? Checking the definition directly (both pairs of sides parallel) is one way, but there are several shortcuts, and they are the converses of the properties you already know.

The parallelogram tests

Each property from the last lesson has a converse that is also true.

Ways to prove a quadrilateral is a parallelogram

A quadrilateral is a parallelogram if any one of these is true:

Both pairs of opposite sides are parallel. (This is 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 congruent and parallel.

Here is why test 2 works. Suppose $AB = CD$ and $BC = DA$ in quadrilateral $ABCD$ . Draw diagonal $\overline{AC}$ . Then $\triangle ABC \cong \triangle CDA$ by SSS, since $\overline{AC}$ is shared. By CPCTC, $\angle BAC \cong \angle DCA$ . Those are alternate interior angles for lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{DC}$ with transversal $\overleftrightarrow{AC}$ , so $\overline{AB} \parallel \overline{DC}$ . The same triangles give $\angle BCA \cong \angle DAC$ , so $\overline{BC} \parallel \overline{AD}$ . Both pairs of opposite sides are parallel, which is the definition.

Test 4 has a similar proof. If the diagonals meet at $E$ with $AE = CE$ and $BE = DE$ , then the vertical angles at $E$ give $\triangle ABE \cong \triangle CDE$ by SAS. CPCTC gives $\angle BAE \cong \angle DCE$ , alternate interior angles, so $\overline{AB} \parallel \overline{DC}$ . Doing the same with $\triangle ADE$ and $\triangle CBE$ gives $\overline{AD} \parallel \overline{BC}$ .

Watch out for half a test

Each test needs both pairs, except test 5, which needs both facts about the same pair. Mixing information from different pairs is not enough.

An isosceles trapezoid: DC is parallel to AB and legs AD and BC are congruent, but ABCD is not a parallelogram.

In the figure, $\overline{DC} \parallel \overline{AB}$ and $\overline{AD} \cong \overline{BC}$ . That is one pair parallel and the other pair congruent, and the shape is clearly not a parallelogram because $\overline{AD}$ and $\overline{BC}$ lean toward each other.

Common mistake

"One pair of sides parallel and the other pair congruent" does not prove a parallelogram, and neither does "one pair of opposite sides congruent" alone. For test 5, the pair you show is parallel must be the same pair you show is congruent.

Worked example: Is it enough?

Decide whether each set of facts about quadrilateral $ABCD$ proves it is a parallelogram.

$AB = CD$ and $BC = DA$ .
$\overline{AB} \parallel \overline{DC}$ and $AB = DC$ .
$\overline{AB} \parallel \overline{DC}$ and $AD = BC$ .
$\angle A \cong \angle C$ and $\angle B \cong \angle D$ .

Solutions.

Yes: both pairs of opposite sides are congruent (test 2).
Yes: one pair is both parallel and congruent (test 5).
No: the parallel pair and the congruent pair are different pairs. The isosceles trapezoid above is a counterexample.
Yes: both pairs of opposite angles are congruent (test 3).

Using algebra to make a parallelogram

A common question asks what values of the variables would make a quadrilateral a parallelogram. Pick a test, set the right parts equal, and solve.

Worked example: Opposite sides

In quadrilateral $ABCD$ , $AB = 2x + 3$ , $CD = x + 8$ , $BC = 4y - 1$ and $DA = y + 11$ . For what values of $x$ and $y$ is $ABCD$ a parallelogram?

Use test 2: make both pairs of opposite sides congruent.

\begin{aligned} 2x + 3 &= x + 8 & 4y - 1 &= y + 11 \\ x &= 5 & 3y &= 12 \\ & & y &= 4 \end{aligned}

With $x = 5$ and $y = 4$ , $AB = CD = 13$ and $BC = DA = 15$ , so $ABCD$ is a parallelogram.

Worked example: Diagonals

The diagonals of quadrilateral $PQRS$ meet at $E$ . $PE = x + 4$ , $RE = 3x - 6$ , $QE = 2y$ and $SE = y + 3$ . For what values of $x$ and $y$ is $PQRS$ a parallelogram?

Use test 4: $E$ must be the midpoint of both diagonals. So $x + 4 = 3x - 6$ , giving $x = 5$ , and $2y = y + 3$ , giving $y = 3$ . Then $PE = RE = 9$ and $QE = SE = 6$ .

Coordinate proofs

On the coordinate plane you can check the tests with three formulas:

Slope shows sides are parallel (equal slopes).
Distance shows sides are congruent (equal lengths).
Midpoint shows the diagonals bisect each other (same midpoint).

Worked example: Three ways to prove it

Prove that $A(-3, -1)$ , $B(2, 0)$ , $C(4, 4)$ , $D(-1, 3)$ are the vertices of a parallelogram.

Quadrilateral ABCD with its diagonals.Open in grapher →

Method 1: slopes (definition).

\begin{aligned} \text{slope of } \overline{AB} &= \frac{0 - (-1)}{2 - (-3)} = \frac{1}{5} & \text{slope of } \overline{DC} &= \frac{4 - 3}{4 - (-1)} = \frac{1}{5} \\[4pt] \text{slope of } \overline{BC} &= \frac{4 - 0}{4 - 2} = 2 & \text{slope of } \overline{AD} &= \frac{3 - (-1)}{-1 - (-3)} = 2 \end{aligned}

Both pairs of opposite sides have equal slopes, so both pairs are parallel.

Method 2: distances (test 2). $AB = \sqrt{5^2 + 1^2} = \sqrt{26}$ and $DC = \sqrt{5^2 + 1^2} = \sqrt{26}$ . $BC = \sqrt{2^2 + 4^2} = \sqrt{20}$ and $AD = \sqrt{2^2 + 4^2} = \sqrt{20}$ . Both pairs of opposite sides are congruent.

Method 3: midpoints (test 4). The midpoint of $\overline{AC}$ is $\left(\dfrac{-3 + 4}{2}, \dfrac{-1 + 4}{2}\right) = (0.5, 1.5)$ . The midpoint of $\overline{BD}$ is $\left(\dfrac{2 + (-1)}{2}, \dfrac{0 + 3}{2}\right) = (0.5, 1.5)$ . The diagonals share a midpoint, so they bisect each other.

Any one of these methods is a complete proof. Finish with a sentence that names the test, such as "Since both pairs of opposite sides are parallel, $ABCD$ is a parallelogram."

Tip

The midpoint method is usually fastest: two midpoint calculations and you're done. Use test 5 if a problem already hands you one pair of sides; then you need one slope comparison and one distance comparison, both for that same pair.

Practice

Practice 1

Which set of facts about quadrilateral $ABCD$ is not enough to prove it is a parallelogram?

Practice 2

In quadrilateral $ABCD$ , $\angle B \cong \angle D$ . Also $m\angle A = (5x - 20)^\circ$ and $m\angle C = (3x + 30)^\circ$ . For what value of $x$ is $ABCD$ a parallelogram?