Rectangles, Rhombi, and Squares

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Learning goals

Prove the special diagonal properties of rectangles and rhombi using triangle congruence
Calculate lengths and angles by combining special properties with parallelogram properties
Choose sufficient classification tests and state their required hypotheses
Explain the square hierarchy and disprove unrestricted converses with counterexamples

Special members of one family

A rectangle is a parallelogram with four right angles. A rhombus is a parallelogram with four congruent sides. A square has both features, so every square is both a rectangle and a rhombus.

These definitions keep rectangles and rhombi within the parallelogram family. Their right-angle and equal-side conditions add structure without removing any parallelogram property.

Every parallelogram has congruent opposite sides, supplementary consecutive angles, and diagonals that bisect each other. The parallelogram properties lesson proves these facts; here we use them to investigate the extra conditions imposed by right angles or equal sides.

Throughout this lesson, quadrilaterals are convex and their names list vertices in boundary order. An intersection named $O$ lies on both diagonals.

Why a rectangle has congruent diagonals

A rectangle’s two diagonals have equal lengths. For example, a rectangle with adjacent sides $6$ cm and $8$ cm has a $10$ cm diagonal, by the Pythagorean theorem; the other diagonal crosses a right triangle with the same two leg lengths, so it also measures $10$ cm.

The common side and equal opposite sides give a proof that works without choosing side lengths.

The two diagonals belong to triangles with matching sides and included right angles.

A rectangle has congruent diagonals#

Let $ABCD$ be a rectangle. Compare triangles $DAB$ and $CBA$ .

Their sides $AD$ and $BC$ are congruent because the rectangle is a parallelogram. Side $AB$ is common, and the included angles at $A$ and $B$ are right angles. SAS gives

\triangle DAB\cong\triangle CBA.

The corresponding third sides are the diagonals, so

BD=AC.

Nothing in this argument depended on the particular side lengths $6$ and $8$ .

This adds a new fact to diagonal bisection. Bisection compares the two halves of each individual diagonal; congruent diagonals compare one whole diagonal with the other.

Check your understanding

Why can the rectangle proof compare triangles $DAB$ and $CBA$ by SAS?

A They share AB, opposite sides AD and BC are congruent, and their included angles are right B Their diagonals were already proved congruent before the proof began C All three angles of each triangle are right D Any two triangles sharing one side are congruent

Worked example 1 Using two descriptions of the diagonals

Problem. Rectangle $ABCD$ has diagonals meeting at $O$ . The lengths, in centimeters, are

\begin{gathered} AC=5x+4\\ BD=8x-11. \end{gathered}

Find $x$ , the length of $AC$ , and the length of $BO$ .

The expressions describe whole diagonals; O marks their intersection.

Solution. The whole diagonals are congruent, so equate their expressions.

\begin{aligned} 5x+4&=8x-11\\ 15&=3x\\ x&=5. \end{aligned}

Substitute into either whole-diagonal expression.

\begin{aligned} AC&=5(5)+4\\ AC&=29\text{ cm}. \end{aligned}

Since a rectangle is a parallelogram, $O$ bisects $BD$ . Therefore

BO=\frac{29}{2}\text{ cm}.

Check your understanding

A rectangle's diagonals meet at $O$ . If $AO$ is $11$ cm, how long is $BD$ ?

11

cm B

22

cm C

44

cm D It cannot be determined

What equal sides do to the diagonals

A rhombus’s diagonals cross at right angles and bisect its vertex angles. These are different statements: one describes the intersection, and the other describes how a diagonal splits an angle at an endpoint.

If two equal adjacent angles together make a straight angle, their common measure is $180^\circ\div2$ , or $90^\circ$ . We will find that situation inside a rhombus.

The side and midpoint facts are available before proving perpendicularity.

A rhombus has perpendicular diagonals#

Let the diagonals of rhombus $ABCD$ meet at $O$ . Since the rhombus is a parallelogram, $BO$ and $DO$ are congruent. Also, $AB$ and $AD$ are congruent by the rhombus definition, and $AO$ is common.

SSS gives

\triangle AOB\cong\triangle AOD.

Corresponding angles $AOB$ and $AOD$ are congruent. Since $B,O,D$ are collinear, these adjacent angles add to a straight angle. Each is $90^\circ$ , so the diagonals are perpendicular.

For angle bisection, consider a rhombus with a $116^\circ$ angle at $A$ . The two triangles on either side of $AC$ have matching side lengths, so they split that angle into two $58^\circ$ parts. The same matching argument proves the result for every rhombus.

Each rhombus diagonal bisects its endpoint angles#

In rhombus $ABCD$ , triangles $ABC$ and $ADC$ share $AC$ . The other two corresponding side pairs are congruent because all four sides of a rhombus are congruent. SSS gives

\triangle ABC\cong\triangle ADC.

Corresponding angles show that $AC$ bisects the angles at its own endpoints:

\begin{gathered} \angle BAC\cong\angle DAC\\ \angle BCA\cong\angle DCA. \end{gathered}

Likewise, SSS gives congruence of triangles $ABD$ and $CBD$ . Thus $BD$ bisects the angles at $B$ and $D$ . The $116^\circ$ example uses this side matching; that particular angle measure was not needed.

Worked example 2 Finding a diagonal and a side

Problem. Rhombus $ABCD$ has diagonals meeting at $O$ . The lengths $AO$ and $BO$ are $9$ cm and $12$ cm. Find $AC$ and $AB$ .

The half-diagonals form the legs of a right triangle.

Solution. The diagonals bisect each other, so the whole diagonal is twice $AO$ .

\begin{aligned} AC&=2(9)\\ AC&=18\text{ cm}. \end{aligned}

The diagonals are perpendicular, making triangle $AOB$ a right triangle with hypotenuse $AB$ .

\begin{aligned} AB^2&=9^2+12^2\\ AB^2&=225\\ AB&=15\text{ cm}. \end{aligned}

Check your understanding

In rhombus $PQRS$ , diagonal $PR$ splits the angle at $P$ into two equal parts. One part measures $34^\circ$ . What is the measure of the whole angle at $Q$ ?

34^\circ

68^\circ

112^\circ

146^\circ

Tests that identify special parallelograms

A property starts with a shape’s name and draws a conclusion. A sufficient test starts with evidence and guarantees the shape’s name. The previous lesson proves tests for establishing that a quadrilateral is a parallelogram; establish that fact before using the tests below.

A parallelogram with congruent diagonals must be a rectangle. For instance, if two consecutive angles are proved equal and their sum is $180^\circ$ , both are $90^\circ$ .

Congruent diagonals are given; right angles must be proved.

Congruent diagonals identify a rectangle among parallelograms#

Suppose $ABCD$ is a parallelogram with congruent diagonals. Compare triangles $DAB$ and $CBA$ again. Their third sides $DB$ and $CA$ are now given congruent, rather than obtained as a conclusion.

Opposite sides $AD$ and $BC$ are congruent, and $AB$ is common. SSS gives congruent triangles, so the consecutive angles $DAB$ and $CBA$ are congruent. They are also supplementary because $ABCD$ is a parallelogram.

Each of those angles is $90^\circ$ . Their opposite angles are congruent to them, giving four right angles, so $ABCD$ is a rectangle.

A parallelogram with perpendicular diagonals must be a rhombus. Perpendicularity supplies the included angles needed to compare two small triangles.

Midpoint information and the right angle provide an SAS comparison.

Two diagonal tests for a rhombus#

First suppose parallelogram $ABCD$ has perpendicular diagonals meeting at $O$ . Side $AO$ is common to triangles $AOB$ and $AOD$ . The sides $BO$ and $DO$ are congruent because parallelogram diagonals bisect each other, and the included angles at $O$ are right angles.

SAS gives congruent triangles, so $AB$ and $AD$ are congruent. Together with the two congruent opposite-side pairs of a parallelogram, this makes all four sides congruent. Thus $ABCD$ is a rhombus.

Now suppose instead that $ABCD$ is a parallelogram and $AC$ bisects angle $DAB$ . The bisector gives

\angle DAC\cong\angle BAC.

Since $AD$ is parallel to $BC$ , alternate interior angles give

\angle DAC\cong\angle ACB.

Thus triangle $ABC$ has congruent angles at $A$ and $C$ , so their opposite sides $BC$ and $AB$ are congruent. Equal adjacent sides, together with equal opposite sides, make the parallelogram a rhombus.

These tests require the whole angle at a vertex to be bisected. Splitting an unrelated angle somewhere on a side is different evidence.

Two simpler tests follow from earlier properties. A parallelogram with one right angle has four right angles, because consecutive angles are supplementary and opposite angles are congruent. A parallelogram with one pair of congruent adjacent sides has four congruent sides, because opposite sides are congruent.

Worked example 3 Putting classification evidence in order

Problem. Convex quadrilateral $JKLM$ has diagonals $JL$ and $KM$ meeting at $O$ . You know

\begin{gathered} JO=OL\\ KO=OM\\ JL=KM\\ JL\perp KM. \end{gathered}

Prove that $JKLM$ is a square. Explain which evidence establishes the parallelogram first.

The marks record only the supplied diagonal facts.

Solution. The first two equalities say that the diagonals bisect each other. The converse parallelogram test therefore makes $JKLM$ a parallelogram.

Its congruent diagonals now prove that it is a rectangle.

Its perpendicular diagonals now prove that it is a rhombus.

A quadrilateral that is both a rectangle and a rhombus has four right angles and four congruent sides. It is a square.

Check your understanding

You know only that $ABCD$ is a parallelogram. Which additional fact guarantees a rhombus but does not by itself guarantee a square?

A Its diagonals are congruent B Its diagonals are perpendicular C Its diagonals bisect each other D It has one right angle

Why the hypothesis cannot disappear

Congruent diagonals alone do not identify a rectangle among all quadrilaterals. Perpendicular diagonals alone do not identify a rhombus either. The following construction satisfies even both diagonal conditions and still fails both conclusions.

All four lengths and the right angle are given; no midpoint claim is assumed.

The four rays from $O$ form right angles, with the given lengths shown. The polygon $ABCD$ is convex. Its diagonals have lengths

\begin{aligned} AC&=4+2\\ AC&=6\\ BD&=3+3\\ BD&=6. \end{aligned}

They are congruent and perpendicular. But $O$ does not bisect $AC$ , so this is not a parallelogram; therefore it is neither a rectangle nor a rhombus. This uses the necessary diagonal-bisection property, not any trapezoid or kite theorem.

Even a diagonal that bisects an endpoint angle is insufficient without the parallelogram hypothesis. In this same figure, right triangles $AOB$ and $AOD$ are congruent by SAS, using the equal $3$ -unit legs and their common $AO$ . Thus $AC$ bisects angle $DAB$ , yet the shape is not a rhombus.

Squares combine the two conditions

Within parallelograms, the two special diagonal properties work independently: congruent diagonals give a rectangle, and perpendicular diagonals give a rhombus. A square belongs to both families, so its diagonals are congruent, perpendicular, bisect each other, and bisect the vertex angles.

For a square, each vertex angle is $90^\circ$ , so each part cut by a diagonal is $45^\circ$ . A rotated square keeps all these properties.

Worked example 4 What can the given facts guarantee?

Problem. In each case below the quadrilateral is a parallelogram. State the strongest classification guaranteed by the givens, and decide whether a square is guaranteed.

Case A. The whole diagonals are congruent.

Case B. The whole diagonals are perpendicular and have different lengths.

Case C. One angle is right, and a diagonal bisects that whole angle.

Solution. Case A guarantees a rectangle. A rectangle with unequal adjacent sides satisfies the evidence, so a square is not guaranteed.

Case B guarantees a rhombus. Its unequal diagonals rule out a rectangle and therefore rule out a square.

Case C gives a rectangle by the right-angle test and a rhombus by the angle-bisector test. It guarantees a square.

Check your understanding

A convex quadrilateral has congruent, perpendicular diagonals. Which further fact guarantees that it is a square?

A One diagonal is longer than the other B The diagonals meet inside the quadrilateral C The diagonals bisect each other D One pair of adjacent sides has different lengths

Predict which conditions will make the diagonals equal or perpendicular. Try each control separately, then combine the right-angle and equal-side conditions. Use the earlier proofs to justify what appears.

Change one condition and watch the diagonals

The shape always stays a parallelogram

Scroll sideways to see the whole diagram.

Angle at A Adjacent side ratio Try a case

Classification within parallelograms: general parallelogram. Diagonal AC: about 5.59 units. Diagonal BD: about 4.33 units. Angle AOB: about 106.8°.

The controls always produce a parallelogram. Change angle A or the adjacent-side ratio to compare the rectangle, rhombus, and square cases.

Rectangles, Rhombi, and Squares

Learning goals

Special members of one family

Why a rectangle has congruent diagonals

What equal sides do to the diagonals

Tests that identify special parallelograms

Why the hypothesis cannot disappear

Squares combine the two conditions

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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