topic badge

9.05 Classifying quadrilaterals

Lesson

A quadrilateral is a four-sided 2D figure. By showing that quadrilaterals have specific properties, we can narrow it down to the most specific type of quadrilateral.

The hierarchy of parallelograms

We define a parallelogram as any quadrilateral whose opposite side lengths are parallel. In some instances, a parallelogram can also be a rhombus, rectangle or square, depending on what knowledge we know about the parallelogram.

Parallelogram

Properties of parallelograms:

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

Bisecting diagonals

We define a rhombus as a parallelogram whose all four sides are congruent.

Rhombus

Properties of a rhombus:

  • Four congruent sides
  • Diagonals are perpendicular
  • Each diagonal bisects a pair of opposite angles

We define a rectangle as a parallelogram whose interior angles are right angles.

Rectangle

Properties of rectangles:

  • Four right angles
  • Diagonals are congruent
  • Diagonals bisect each other

And lastly, we define a square as a parallelogram whose all four sides are congruent and and all interior angles are right angles.

Square

Properties of squares:

  • Four congruent sides and four congruent angles
  • A special case of both a rectangle and a rhombus
  • Diagonals are perpendicular

 

If we can prove that a parallelogram possesses one of these necessary conditions, then we can prove that it is a rhombus, rectangle or square respectively.

 

Exploration

Consider the following theorem.

If a parallelogram has one pair of consecutive sides that are congruent, then the parallelogram is a rhombus.

To begin proving the above theorem, we can first label the parallelogram by its set of vertices, $PQRS$PQRS.

Parallelogram $PQRS$PQRS


From the given information, we know that a pair of consecutive sides are congruent. So let's choose $\overline{PQ}$PQ and $\overline{SP}$SP. Ultimately we could have chosen another pair, but the proof would be the same. We write this as a geometric statement $\overline{PQ}\cong\overline{SP}$PQSP.

Parallelogram $PQRS$PQRS with congruent sides
 

We also know that the $PQRS$PQRS is a parallelogram, and one feature of the parallelogram is that opposite sides are congruent. We can state this as $\overline{PQ}\cong\overline{RS}$PQRS and $\overline{SP}\cong\overline{QR}$SPQR. Congruence is transitive, meaning that if two things are congruent to the same thing, they are also congruent. This means that $\overline{PQ}\cong\overline{QR}$PQQR and $\overline{SP}\cong\overline{RS}$SPRS and more generally all the sides are congruent. By definition, a rhombus is a parallelogram with all sides congruent. So $PQRS$PQRS is a rhombus and we are finished with the proof.

We can formalize the above steps into a two-column proof where each line contains a geometric statement in the left column and a corresponding reason in the right column.

 

Two-column proof

Given the parallelogram $PQRS$PQRS, and the fact that $\overline{PQ}\cong\overline{SP}$PQSP, prove that $PQRS$PQRS is a rhombus.

Statements Reasons
$PQRS$PQRS is  a parallelogram and $\overline{PQ}\cong\overline{SP}$PQSP Given

$\overline{PQ}\cong\overline{RS}$PQRS and $\overline{SP}\cong\overline{QR}$SPQR

If a quadrilateral is a parallelogram, then its opposite sides are congruent.

$\overline{PQ}\cong\overline{QR}\cong\overline{RS}\cong\overline{SP}$PQQRRSSP

Transitive property of congruence
$PQRS$PQRS is a rhombus Definition of a rhombus

The final line contains the statement that the quadrilateral is a rhombus, which is what we wanted to prove.

Practice questions

question 1

In order to show that a given parallelogram is a rectangle, which of the following must be proved?

  1. The parallelogram has four congruent sides.

    A

    The parallelogram has four right angles.

    B

    The parallelogram has opposite sides congruent.

    C

    The parallelogram has opposite sides parallel.

    D

    The parallelogram has opposite angles congruent.

    E

question 2

Given that $PQRS$PQRS is a parallelogram, and $\overline{PR}$PR and $\overline{QS}$QS are perpendicular, prove that $PQRS$PQRS is a rhombus.

 

The hierarchy of trapezoids

We define a trapezoid as any quadrilateral with one pair of opposite sides that are parallel.

A special type of a trapezoid, called an isosceles trapezoid, occurs when the two legs of the trapezoid are congruent and hence the diagonals are congruent.

Isosceles trapezoid

 

There are a number of theorems that explain equivalent ways of identifying whether a trapezoid is isosceles. We call these special theorems criteria.

  1. If a trapezoid has congruent base angles, then it is an isosceles trapezoid.
Trapezoid with base angles congruent

 

  1. If a trapezoid has opposite angles that are supplementary, then it is an isosceles trapezoid.
Trapezoid with $\angle1$1 and $\angle3$3 are supplementary,
and $\angle2$2 and $\angle4$4 are supplementary

 

  1. If a trapezoid has congruent diagonals, then it is an isosceles trapezoid.

Trapezoid with congruent diagonals

 

Exploration

Consider the following theorem.

If a trapezoid has congruent base angles, then it is an isosceles trapezoid.

To begin proving the above theorem, we can first label the trapezoid by its set of vertices, $ABCD$ABCD. Let $\angle ADC$ADC and $\angle BCD$BCD be congruent.

Trapezoid $ABCD$ABCD

 

We next construct the pair of altitudes $\overline{AP}$AP and $\overline{BQ}$BQ, which form the pair of triangles $\triangle APD$APD and $\triangle BQC$BQC. If we can show that these triangles are congruent, then we can show that $\overline{AD}$AD is congruent to $\overline{BC}$BC, which will mean that $ABCD$ABCD is an isosceles trapezoid.

Trapezoid $ABCD$ABCD with altitudes drawn

 

Firstly we are given that $\angle ADC$ADC and $\angle BCD$BCD are congruent. Since altitudes create right angles with the base of a trapezoid, $\angle APD$APD and $\angle BQC$BQC are right angles. All right angles are congruent, so we have that $\angle APD$APD and $\angle BQC$BQC are congruent. The altitudes themselves are congruent, so $\overline{AP}$AP is congruent to $\overline{BQ}$BQ. This means that $\triangle APD\cong\triangle BQC$APDBQC are congruent by angle-angle-side congruence theorem.

It follows that the two sides $\overline{AD}$AD and $\overline{BC}$BC are congruent because they are corresponding parts of the two congruent triangles. Hence the trapezoid $ABCD$ABCD is isosceles, by definition.

 

We can formalize the above steps into a two-column proof where each line contains a geometric statement in the left column and a corresponding reason in the right column.

Statements Reasons

$ABCD$ABCD is a trapezoid, $\overline{AP}$AP and $\overline{BQ}$BQ are altitudes, and $\angle ADC\cong\angle BCD$ADCBCD

Given
$\angle APD$APD and $\angle BQC$BQC are right angles Definition of an altitude
$\angle APD\cong\angle BQC$APDBQC All right angles are congruent.
$\overline{AP}\cong\overline{BQ}$APBQ Altitudes of a trapezoid are congruent
$\triangle APD\cong\triangle BQC$APDBQC

Angle-angle-side congruence theorem

$\overline{AD}\cong\overline{BC}$ADBC Corresponding parts of congruent triangles are congruent (CPCTC)
$ABCD$ABCD is an isosceles trapezoid Definition of an isosceles trapezoid

 

Below is a summary of the criteria for a trapezoid to be an isosceles trapezoid.

Criteria for isosceles trapezoids

A trapezoid is an isosceles trapezoid if any of the following are true:

  • The base angles are congruent.
  • The opposite angles are supplementary.
  • The diagonals are congruent.

Practice Questions

Question 1

In order to show that a given trapezoid is isosceles, which of the following must be proved?

  1. The trapezoid has diagonals that bisect each other.

    A

    The trapezoid has diagonals that are perpendicular.

    B

    The trapezoid has diagonals that are congruent.

    C

    The trapezoid has diagonals that are perpendicular bisectors of each other.

    D

Question 2

Given that $ABCD$ABCD is a trapezoid with altitudes $\overline{AP}$AP and $\overline{BQ}$BQ and that $\angle ADC$ADC is congruent to $\angle BCD$BCD, prove that $ABCD$ABCD is an isosceles trapezoid.

What is Mathspace

About Mathspace