New YorkMathsSyllabus dot point

How do you prove a quadrilateral is a parallelogram, rectangle, rhombus, or square?

Prove theorems about parallelograms (opposite sides and angles congruent, diagonals bisect each other) and prove that a given quadrilateral is a parallelogram, rectangle, rhombus, or square using side, angle, and diagonal properties.

A NY Regents Geometry answer on quadrilateral proofs: the parallelogram properties, the ways to prove a parallelogram, and how the added conditions distinguish a rectangle, rhombus, and square.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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What this topic is asking
Parallelogram properties
Five ways to prove a parallelogram
Distinguishing the special quadrilaterals
Try this

What this topic is asking

The Regents Geometry exam (the Congruence, G-CO, and Expressing Geometric Properties with Equations, G-GPE, clusters) wants you to prove the properties of a parallelogram and to classify a quadrilateral as a parallelogram, rectangle, rhombus, or square by checking the right side, angle, and diagonal conditions. This is a frequent Part III or Part IV task, often as a coordinate proof.

Parallelogram properties

A parallelogram is a quadrilateral with both pairs of opposite sides parallel. From that single definition, several properties follow and are provable with congruent triangles (a diagonal splits the parallelogram into two congruent triangles):

Opposite sides are congruent.
Opposite angles are congruent.
Consecutive angles are supplementary.
The diagonals bisect each other.

These are properties you may cite once a figure is known to be a parallelogram, and they are also the targets of "prove the parallelogram has..." questions.

Five ways to prove a parallelogram

To prove a quadrilateral is a parallelogram, establish any one of these conditions:

On the coordinate plane, "parallel" is tested by equal slopes, "congruent" by equal distances (distance formula), and "bisecting diagonals" by a shared midpoint (midpoint formula).

Proving a parallelogram on the coordinate plane

Quadrilateral $PQRS$ has $P(0, 0)$ , $Q(4, 1)$ , $R(5, 4)$ , $S(1, 3)$ . Prove it is a parallelogram using one pair of sides both parallel and congruent.

step 1 Choose a pair of opposite sides

Take $\overline{PQ}$ and $\overline{SR}$ .

step 2 Compare their slopes

Slope $PQ = \frac{1 - 0}{4 - 0} = \frac{1}{4}$ . Slope $SR = \frac{4 - 3}{5 - 1} = \frac{1}{4}$ . Equal slopes, so $\overline{PQ} \parallel \overline{SR}$ .

step 3 Compare their lengths

$PQ = \sqrt{(4-0)^2 + (1-0)^2} = \sqrt{17}$ . $SR = \sqrt{(5-1)^2 + (4-3)^2} = \sqrt{17}$ . Equal lengths, so $\overline{PQ} \cong \overline{SR}$ .

step 4 Conclude

One pair of opposite sides is both parallel and congruent, so $PQRS$ is a parallelogram (condition 5).

Distinguishing the special quadrilaterals

Once a figure is a parallelogram, an extra condition upgrades it:

A rectangle is a parallelogram with a right angle, equivalently with congruent diagonals.
A rhombus is a parallelogram with four congruent sides, equivalently with perpendicular diagonals (or a diagonal that bisects the angles).
A square is both a rectangle and a rhombus: congruent and perpendicular diagonals, four equal sides, four right angles.

A clarifying point that decides many questions is which property is the distinguishing one. All parallelograms have bisecting diagonals, so that alone proves nothing special. Congruent diagonals point to a rectangle; perpendicular diagonals point to a rhombus; both together point to a square. On a coordinate proof, test a right angle with perpendicular slopes (product $-1$ ) and equal sides with the distance formula, then state explicitly which classification the evidence supports.

Try this

Q1. Name the extra diagonal condition that makes a parallelogram a rhombus. [1 credit]

Cue. Perpendicular diagonals.

Q2. On a coordinate grid, what do you compute to show two sides are parallel? [1 credit]

Cue. Their slopes; equal slopes mean parallel.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart I (multiple choice). Which property guarantees that a parallelogram is a rectangle? (1) the diagonals bisect each other (2) the diagonals are congruent (3) opposite sides are parallel (4) opposite angles are congruent

Show worked answer →

The correct answer is (2).

Every parallelogram already has bisecting diagonals, parallel opposite sides, and congruent opposite angles (choices 1, 3, 4 are true of all parallelograms), so none of those distinguishes a rectangle. A parallelogram is a rectangle exactly when its diagonals are congruent (equal length), which is equivalent to having right angles. Congruent diagonals is the defining extra condition for a rectangle.

Regents (style)6 marksPart IV (extended constructed response). Quadrilateral

ABCD

has vertices

A(0, 0)

B(4, 1)

C(6, 4)

, and

D(2, 3)

. Prove that

ABCD

is a parallelogram but not a rectangle, using coordinate geometry.

Show worked answer →

A 6-credit coordinate proof. Award credit for the parallelogram argument, the rectangle test, and a clear conclusion.

Parallelogram: show both pairs of opposite sides are parallel by equal slopes. Slope $AB = \frac{1 - 0}{4 - 0} = \frac{1}{4}$ ; slope $DC = \frac{4 - 3}{6 - 2} = \frac{1}{4}$ , so $AB \parallel DC$ . Slope $BC = \frac{4 - 1}{6 - 4} = \frac{3}{2}$ ; slope $AD = \frac{3 - 0}{2 - 0} = \frac{3}{2}$ , so $BC \parallel AD$ . Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram. Not a rectangle: a rectangle needs a right angle at a vertex, so test adjacent sides $\overline{AB}$ and $\overline{BC}$ . Their slopes are $\frac{1}{4}$ and $\frac{3}{2}$ , and the product $\frac{1}{4} \cdot \frac{3}{2} = \frac{3}{8} \neq -1$ , so the sides are not perpendicular and there is no right angle. (Equivalently, the diagonals have unequal lengths: $AC = \sqrt{6^2 + 4^2} = \sqrt{52}$ but $BD = \sqrt{(2 - 4)^2 + (3 - 1)^2} = \sqrt{8}$ .) Therefore $ABCD$ is a parallelogram but not a rectangle. The credits reward slopes for the parallel test, a perpendicularity or diagonal-length test for the right angle, and a clear stated conclusion.

Related dot points

Sources & how we know this

Regents Examination in Geometry — NYSED (2024)
New York State Next Generation Mathematics Learning Standards — NYSED (2017)