Conditions for a Quadrilateral to be a Parallelogram

The previous lesson laid out the properties of a parallelogram: opposite sides equal, opposite angles equal, diagonals bisecting each other, and so on. Now we ask the converse question: which of these properties is enough , by itself , to force a quadrilateral to be a parallelogram? The answer is surprising: each of them is.

Definitions

A quadrilateral is a parallelogram if both pairs of opposite sides are parallel. The conditions in this lesson are equivalent to this definition; any one of them suffices.

The five conditions

Condition 1. A quadrilateral is a parallelogram iff both pairs of opposite sides are parallel. (Definition.)

Condition 2. A quadrilateral is a parallelogram iff both pairs of opposite sides are equal.

Condition 3. A quadrilateral is a parallelogram iff one pair of opposite sides is both parallel AND equal.

Condition 4. A quadrilateral is a parallelogram iff both pairs of opposite angles are equal.

Condition 5. A quadrilateral is a parallelogram iff the diagonals bisect each other.

Proofs of the converses

We already know parallelogram $\Rightarrow$ each of these properties. The converses need proof.

Condition 2 (sides equal $\Rightarrow$ parallelogram). In quadrilateral $ABCD$ suppose $AB = CD$ and $BC = AD$. Draw diagonal $AC$. In $\triangle ABC$ and $\triangle CDA$: $AB = CD$ (given), $BC = DA$ (given), $AC = CA$ (common). By SSS, the triangles are congruent. So $\angle BAC = \angle DCA$ by CPCT. These are alternate angles for line $AC$ cutting $AB$ and $CD$, so $AB \parallel CD$ (converse of alternate-interior-angles theorem). Similarly $\angle BCA = \angle DAC$, so $BC \parallel AD$. Both pairs parallel: parallelogram.

Condition 3 (one pair parallel and equal $\Rightarrow$ parallelogram). Suppose $AB \parallel CD$ and $AB = CD$ in quadrilateral $ABCD$. Draw diagonal $AC$. By alternate-interior-angles (using $AB \parallel CD$), $\angle BAC = \angle DCA$. Also $AC = CA$ (common). And $AB = CD$ (given). By SAS, $\triangle ABC \cong \triangle CDA$. So $BC = DA$ (CPCT). Now we have both pairs of opposite sides equal, hence parallelogram (by Condition 2).

Condition 4 (opposite angles equal $\Rightarrow$ parallelogram). Suppose $\angle A = \angle C$ and $\angle B = \angle D$. Then $\angle A + \angle B + \angle C + \angle D = 360^\circ$ becomes $2(\angle A + \angle B) = 360^\circ$, so $\angle A + \angle B = 180^\circ$. By the converse of the co-interior-angles theorem (sum $180^\circ$ on the same side of transversal forces parallelism), $AD \parallel BC$. Similarly $AB \parallel CD$. Parallelogram.

Condition 5 (diagonals bisect each other $\Rightarrow$ parallelogram). In quadrilateral $ABCD$ with diagonals $AC$ and $BD$ meeting at $O$, suppose $AO = OC$ and $BO = OD$. Then in $\triangle AOB$ and $\triangle COD$: $AO = CO$ (given), $BO = DO$ (given), and $\angle AOB = \angle COD$ (vertically opposite). By SAS, the triangles are congruent. So $AB = CD$ by CPCT. Similarly $\triangle BOC \cong \triangle DOA$, giving $BC = AD$. Both pairs of opposite sides equal: parallelogram.

Which condition to use?

Given a figure, pick the condition that matches the information you have.

• Given sides equal: use Condition 2.
• Given one pair parallel and equal: use Condition 3 (often the slickest).
• Given angles equal: use Condition 4.
• Given diagonals bisect each other: use Condition 5.

Condition 3 is particularly common in proofs: in many configurations, you only have parallelism and equality of one pair, which already does the job.

A typical exam problem

Claim. In quadrilateral $ABCD$, $AB \parallel CD$ and $AB = CD$. Prove $ABCD$ is a parallelogram, and that $BC \parallel AD$.

By Condition 3, $ABCD$ is a parallelogram. By the definition of parallelogram, both pairs of opposite sides are parallel , in particular $BC \parallel AD$.

This is a short proof , most of the work was done by Condition 3.

Worked examples

Example 1. In quadrilateral $ABCD$, $AB = 5, CD = 5, AB \parallel CD$. Is $ABCD$ a parallelogram?

Yes, by Condition 3 (one pair both equal and parallel).

Example 2. In quadrilateral $ABCD$, the diagonals bisect each other at $O$. Prove $ABCD$ is a parallelogram.

By Condition 5, $ABCD$ is a parallelogram.

Example 3. In quadrilateral $ABCD$, opposite angles are equal. Prove $ABCD$ is a parallelogram.

By Condition 4, $ABCD$ is a parallelogram.

Example 4. Three points $A, B, C$ are given with $C$ outside line $AB$. Find a fourth point $D$ such that $ABCD$ is a parallelogram.

Use Condition 3: choose $D$ such that $CD = AB$ and $CD \parallel AB$. Plot $D$ accordingly.

Example 5. In $\triangle ABC$, let $D, E, F$ be the midpoints of $BC, CA, AB$. Prove that $AFDE$ is a parallelogram.

We need to show one pair of opposite sides is equal and parallel. $FD$ joins midpoints of $AB$ and $BC$, so by the midpoint theorem (lesson 5), $FD \parallel AC$ and $FD = \tfrac{1}{2} AC$. Since $E$ is the midpoint of $AC$, $AE = \tfrac{1}{2} AC$. So $FD = AE$ and $FD \parallel AE$. By Condition 3, $AFDE$ is a parallelogram.

Try it yourself

1. State the five conditions for a quadrilateral to be a parallelogram.
2. Prove that a quadrilateral with both pairs of opposite sides equal is a parallelogram.
3. Prove that a quadrilateral whose diagonals bisect each other is a parallelogram.
4. In quadrilateral $PQRS$, $PQ \parallel SR$ and $PQ = SR$. Prove $PQRS$ is a parallelogram.
5. In quadrilateral $ABCD$, opposite angles $\angle A = \angle C$ and $\angle B = \angle D$. Prove $ABCD$ is a parallelogram.
6. Why does "one pair of opposite sides equal" alone not suffice (without parallelism)?
7. Why does "one pair of opposite sides parallel" alone not suffice (without equality)?
8. In a quadrilateral, given $AB = CD$ and $AD = BC$ , what type is it (at least)?
9. In a quadrilateral, given $AB \parallel CD$ and $AD \parallel BC$ , what type is it?
10. Prove that the midpoints of the sides of any quadrilateral form a parallelogram (use midpoint theorem and Condition 3).

Pitfalls / Insight

• Single conditions matter. Don't try to verify all of them; one is enough.
• One pair of opposite sides "parallel and equal" is the most powerful condition. It is a single statement combining two pieces of data, and it works in many configurations.
• Diagonal bisection is often a clean route when diagonals are drawn in the figure.

Insight. Five conditions , each equivalent to "parallelogram" , give you flexibility. Whatever information the figure offers, one of the five usually fits. Spotting the right one is half the proof.