Math Core

Parallelograms show up everywhere: in floor tiles, in scissor lifts, in the way a rectangle leans when you push on its corner. They are defined by parallel sides, but that one fact forces a whole list of other properties. Once you know those properties, you can find missing sides, angles and diagonal lengths with a little algebra.

Definition

Parallelogram

A parallelogram is a quadrilateral with both pairs of opposite sides parallel.

In parallelogram $ABCD$ , the vertices go in order around the figure, so $\overline{AB} \parallel \overline{DC}$ and $\overline{AD} \parallel \overline{BC}$ . Opposite vertices are $A$ and $C$ , and $B$ and $D$ .

Parallelogram ABCD with diagonals AC and BD meeting at E.

Opposite sides and opposite angles

Draw diagonal $\overline{AC}$ . It cuts the parallelogram into two triangles, $\triangle ABC$ and $\triangle CDA$ , and you can prove they are congruent.

$\overline{AB} \parallel \overline{DC}$ , and $\overline{AC}$ is a transversal, so the alternate interior angles $\angle BAC$ and $\angle DCA$ are congruent.
$\overline{AD} \parallel \overline{BC}$ , and $\overline{AC}$ is again a transversal, so $\angle DAC \cong \angle BCA$ .
$\overline{AC} \cong \overline{CA}$ by the reflexive property.

By ASA, $\triangle ABC \cong \triangle CDA$ . By CPCTC, $\overline{AB} \cong \overline{CD}$ , $\overline{BC} \cong \overline{DA}$ , and $\angle B \cong \angle D$ . Drawing the other diagonal gives $\angle A \cong \angle C$ the same way.

Consecutive angles, like $\angle A$ and $\angle B$ , are same-side interior angles for the parallel lines $\overleftrightarrow{AD}$ and $\overleftrightarrow{BC}$ with transversal $\overleftrightarrow{AB}$ . So they are supplementary.

The diagonals

Now look at the two diagonals, which meet at $E$ . In $\triangle ABE$ and $\triangle CDE$ :

$\angle BAE \cong \angle DCE$ and $\angle ABE \cong \angle CDE$ (alternate interior angles, since $\overline{AB} \parallel \overline{DC}$ ),
$\overline{AB} \cong \overline{CD}$ (opposite sides, just proved).

By ASA, $\triangle ABE \cong \triangle CDE$ , so $\overline{AE} \cong \overline{CE}$ and $\overline{BE} \cong \overline{DE}$ . Each diagonal cuts the other in half: $E$ is the midpoint of both.

Properties of every parallelogram

If a quadrilateral is a parallelogram, then:

Its opposite sides are congruent.
Its opposite angles are congruent.
Its consecutive angles are supplementary.
Its diagonals bisect each other.

Worked example: Finding all four angles

In parallelogram $PQRS$ , $m\angle P = 68^\circ$ . Find the other three angles.

Opposite angles are congruent, so $m\angle R = 68^\circ$ . Consecutive angles are supplementary, so $m\angle Q = 180^\circ - 68^\circ = 112^\circ$ , and $m\angle S = 112^\circ$ too. Check: $68 + 112 + 68 + 112 = 360$ , the angle sum of any quadrilateral.

Worked example: Sides and angles with algebra

In parallelogram $ABCD$ , $AB = 3x + 4$ and $CD = 5x - 10$ . Also $m\angle A = (4y + 6)^\circ$ and $m\angle B = (6y - 26)^\circ$ . Find $AB$ , $m\angle A$ and $m\angle B$ .

$\overline{AB}$ and $\overline{CD}$ are opposite sides, so they are congruent:

3x + 4 = 5x - 10 \quad\Rightarrow\quad 14 = 2x \quad\Rightarrow\quad x = 7.

So $AB = 3(7) + 4 = 25$ . (Check: $CD = 5(7) - 10 = 25$ .)

$\angle A$ and $\angle B$ are consecutive, so they are supplementary:

(4y + 6) + (6y - 26) = 180 \quad\Rightarrow\quad 10y - 20 = 180 \quad\Rightarrow\quad y = 20.

So $m\angle A = 86^\circ$ and $m\angle B = 94^\circ$ .

Worked example: Using the diagonals

The diagonals of parallelogram $ABCD$ meet at $E$ . $AE = 2x + 1$ , $CE = 4x - 9$ , $BE = y + 3$ and $DE = 3y - 7$ . Find $AC$ and $BD$ .

The diagonals bisect each other, so $AE = CE$ and $BE = DE$ :

\begin{aligned} 2x + 1 &= 4x - 9 & y + 3 &= 3y - 7 \\ 10 &= 2x & 10 &= 2y \\ x &= 5 & y &= 5 \end{aligned}

Then $AE = 11$ , so $AC = 2 \cdot 11 = 22$ . And $BE = 8$ , so $BD = 2 \cdot 8 = 16$ .

Notice that $AC \ne BD$ in that example. That's normal.

Common mistake

The diagonals of a parallelogram bisect each other, but they are usually not congruent and usually not perpendicular. Don't set $AC = BD$ or assume a right angle at $E$ unless you know the parallelogram is a special kind (a rectangle, a rhombus or a square, which come later in this unit). Likewise, consecutive angles are supplementary, not congruent.

Parallelograms on the coordinate plane

Because the diagonals share a midpoint, you can find a missing vertex with the midpoint formula. Another way is to "copy the slide": moving from $A$ to $B$ is the same move as going from $D$ to $C$ , because those sides are parallel and congruent.

Worked example: Finding the fourth vertex

Three vertices of parallelogram $ABCD$ are $A(-2, 1)$ , $B(3, 2)$ and $C(5, 6)$ . Find $D$ .

From $B$ to $A$ you move $5$ left and $1$ down. Since $\overline{BA}$ and $\overline{CD}$ are parallel and congruent, the move from $C$ to $D$ is the same: $D = (5 - 5,\ 6 - 1) = (0, 5)$ .

Check with the diagonals. The midpoint of $\overline{AC}$ is $\left(\dfrac{-2 + 5}{2}, \dfrac{1 + 6}{2}\right) = (1.5, 3.5)$ . The midpoint of $\overline{BD}$ is $\left(\dfrac{3 + 0}{2}, \dfrac{2 + 5}{2}\right) = (1.5, 3.5)$ . They match, so the diagonals bisect each other.

Parallelogram ABCD. Both diagonals pass through their shared midpoint (1.5, 3.5).Open in grapher →

Tip

Before you finish a parallelogram problem, check that the four angles add to $360^\circ$ and that each pair of opposite sides really came out equal. A single substitution catches most algebra slips.