Lesson 5 ~40 min Geometrical Figures C · Path +85 XP

Tests for Quadrilaterals and Solving Problems

A property runs one way: if it is a parallelogram, then the diagonals bisect. A test runs the other: if the diagonals bisect, then it IS a parallelogram. Tests are converses, and each one has to be earned.

Today's hook: You are told a quadrilateral has one pair of opposite sides that are both equal and parallel. Is that enough to conclude it is a parallelogram? It is. Change "the same pair" to "different pairs" and it is not, and an isosceles trapezium is the counterexample. Tests are precise for a reason.

0/5QUESTS

Think First

warm-up

You have proved that the diagonals of a parallelogram bisect each other. Now suppose you meet a quadrilateral whose diagonals bisect each other, and you know nothing else about it. Do you think it must be a parallelogram? Write down your answer, and then write down what would have to be true for your answer to be justified.

Record your answer in your workbook.

The Big Idea

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A property says: if the figure is of this kind, then this is true. A test says the reverse: if this is true, then the figure is of that kind. Each test is the converse of a property, and a converse is never free. It must be proved separately, and some do not hold at all.

$$\text{property: } P \Rightarrow Q \qquad\qquad \text{test: } Q \Rightarrow P$$

The difference matters in practice because they are used at opposite moments. You cite a property when you already know what the figure is and want a fact from it. You cite a test when you have some facts and want to establish what the figure is. Confusing them is the commonest structural error in geometry proofs.

$P \Rightarrow Q \quad \text{versus} \quad Q \Rightarrow P$

Same pair, both conditions

One pair equal AND parallel works. Split across two pairs, it fails.

A test is a converse

Proving a property never proves the corresponding test.

One counterexample kills it

To disprove a proposed test, one figure is enough.

What You'll Master

objectives

Know

The four tests for a parallelogram, and the tests for a rhombus and a rectangle
That a test is the converse of a property and requires its own proof

Understand

Why the equal and parallel sides in the third parallelogram test must be the SAME pair
Why a single counterexample is enough to reject a proposed test

Can Do

Prove a quadrilateral is a parallelogram, rhombus or rectangle by citing the appropriate test
Reject a proposed test by constructing a counterexample
Solve numerical and non-numerical Euclidean problems combining congruence, similarity and these tests

Words You Need

vocabulary

TestA set of conditions sufficient to conclude that a figure is of a particular kind. The converse of a property.

SufficientEnough on its own to force the conclusion.

NecessarySomething that must be true, but which may not be enough on its own.

CounterexampleA single figure satisfying the conditions but not being the claimed shape, which disproves a proposed test.

Isosceles trapeziumA trapezium whose two non-parallel sides are equal. The standard counterexample to a mis-stated parallelogram test.

Euclidean geometryGeometry built by deduction from definitions and previously proved theorems.

The Four Tests for a Parallelogram

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Any ONE of these is sufficient. You do not need more than one, and you should say which you are using.

Test	Condition
1	both pairs of opposite sides parallel (the definition)
2	both pairs of opposite sides equal
3	one pair of opposite sides both equal and parallel
4	the diagonals bisect each other

Test 3 is the one to read carefully. The two conditions must apply to the same pair of sides. Applied to different pairs the claim is false, and the counterexample is an isosceles trapezium: its two non-parallel sides are equal to each other, and the other pair is parallel, yet it is certainly not a parallelogram.

Test 2 is also a good illustration of a test being a converse. The property "opposite sides of a parallelogram are equal" was proved in Lesson 4; test 2 claims the reverse, and needs its own proof.

Proving a Test

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Here is test 4 proved, so that the shape of a test proof is visible.

Given: quadrilateral $ABCD$ whose diagonals $AC$ and $BD$ meet at $M$, with $AM = CM$ and $BM = DM$.
To prove: $ABCD$ is a parallelogram.

Statement	Reason
$AM = CM$, $BM = DM$	given
$\angle AMB = \angle CMD$	vertically opposite angles
$\triangle AMB \equiv \triangle CMD$	SAS
$\angle BAM = \angle DCM$	matching angles of congruent triangles
$AB \parallel DC$	equal alternate angles on the transversal $AC$

Repeating the argument on the other pair of triangles gives $AD \parallel BC$, so both pairs of opposite sides are parallel and the figure satisfies the definition. Notice the destination: a test proof always ends by establishing the definition.

Tests for Rhombuses and Rectangles

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These build on the parallelogram tests, which is why they are short.

A quadrilateral is a rhombus if: all four sides are equal (the definition); or it is a parallelogram with two adjacent sides equal; or it is a parallelogram whose diagonals are perpendicular.

A quadrilateral is a rectangle if: it is a parallelogram with one right angle (the definition); or it is a parallelogram whose diagonals are equal.

The pattern is worth noticing. Each test says parallelogram, plus one extra fact. So the efficient strategy for an unfamiliar figure is to establish the parallelogram first, using whichever of the four tests the given information suits, and then add the single condition that upgrades it.

Strategy

Ask what you have been given. Sides suggest test 2 or 3; a crossing point suggests test 4; parallel arrows suggest the definition.

Disproving a Proposed Test

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To show a set of conditions is NOT sufficient, you need exactly one figure: something that satisfies every condition and is not the claimed shape.

Claim: "a quadrilateral with two pairs of adjacent sides equal is a rhombus".

Counterexample: a kite. It has two pairs of adjacent equal sides by definition, and a kite with sides $3, 3, 5, 5$ is plainly not a rhombus. The claim fails.

Claim: "a quadrilateral whose diagonals are perpendicular is a rhombus".

Counterexample: again a kite, whose diagonals meet at right angles without the sides being equal. The correct test needs the figure to be a parallelogram first.

One counterexample is enough, and it must be described precisely: give the figure a name and, where useful, actual measurements, so that a reader can check it satisfies every condition.

Common Pitfalls

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Citing a property where a test is needed: "the diagonals bisect each other, and the diagonals of a parallelogram bisect each other, so it is a parallelogram".

Fix: that reasoning runs the implication backwards. Cite the TEST, "a quadrilateral whose diagonals bisect each other is a parallelogram", which is a separately proved result.

Applying test 3 with the equal pair and the parallel pair being different pairs.

Fix: the two conditions must describe the SAME pair of sides. An isosceles trapezium satisfies the split version and is not a parallelogram.

Trying to disprove a proposed test by general argument when one figure would do.

Fix: a single counterexample is a complete disproof. Name the figure and give measurements so it can be checked.

Watch Me Solve It · 3 examples

Watch Me Solve It · Choosing and citing a test

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PROBLEM

In quadrilateral $PQRS$, $PQ = SR$ and $PQ \parallel SR$. Prove that $PQRS$ is a parallelogram.

1

Identify what is given, and about which pair

$PQ = SR \text{ and } PQ \parallel SR$

Both conditions describe the SAME pair of sides, which is what the test requires.
2

Name the test

$\text{one pair of opposite sides equal and parallel}$

Test 3. Nothing further need be established.
3

State the conclusion

$PQRS \text{ is a parallelogram}$
4

Note the trap avoided

$\text{if the parallel pair were } QR \parallel PS \text{ instead}$

Then the conditions would describe different pairs and an isosceles trapezium would satisfy them, so no conclusion could be drawn.

AnswerA parallelogram, by the one-pair-equal-and-parallel test

Watch Me Solve It · Upgrading to a rhombus

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PROBLEM

$ABCD$ has diagonals that bisect each other at $M$, and $AB = BC$. Prove that $ABCD$ is a rhombus.

1

Establish the parallelogram first

$\text{diagonals bisect each other} \Rightarrow ABCD \text{ is a parallelogram (test 4)}$

Every rhombus test is "parallelogram plus one extra fact", so the parallelogram comes first.
2

Bring in the extra fact

$AB = BC \quad \text{(given)}$

These are two ADJACENT sides, which is what the rhombus test asks for.
3

Cite the rhombus test

$\text{a parallelogram with two adjacent sides equal is a rhombus}$
4

Conclude

$ABCD \text{ is a rhombus}$

The parallelogram property that opposite sides are equal then forces all four sides equal.

AnswerA rhombus

Watch Me Solve It · A numerical Euclidean problem

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PROBLEM

In $\triangle ABC$, $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. Given $AD = 6$, $DB = 4$ and $AE = 9$, find $EC$, and state whether $DECB$ is a parallelogram.

1

Prove the similarity

$\angle A \text{ common}, \quad \angle ADE = \angle ABC \text{ (corresponding angles)}$

$\triangle ADE \sim \triangle ABC \quad \text{(AA)}$

The standard parallel-line figure from Lesson 3.
2

Set up the ratio

$\frac{AD}{AB} = \frac{AE}{AC}$

$\frac{6}{10} = \frac{9}{AC}$

$AB = AD + DB = 10$. Using $AD$ rather than $AB$ here is the commonest slip.
3

Solve

$AC = \frac{10 \times 9}{6} = 15 \Rightarrow EC = 15 - 9 = 6$
4

Answer the second part

$DE \parallel BC \text{ but } DE \neq BC$

$DECB$ has one pair of parallel sides only, and the similarity ratio $\tfrac{6}{10}$ means $DE$ is shorter than $BC$. So it is a trapezium, not a parallelogram.

Answer$EC = 6$; $DECB$ is a trapezium, not a parallelogram

Brain Trainer · 4 problems

Brain Trainer · Test, property or neither

4 problems

Four quick problems. Work each one, then reveal the answer.

1 A quadrilateral has both pairs of opposite sides equal. Is it a parallelogram?

This is parallelogram test 2, a separately proved result.Yes
2 A quadrilateral has perpendicular diagonals. Is it a rhombus?

A kite is a counterexample; the figure must be a parallelogram first.No
3 A parallelogram has equal diagonals. What is it?

That is the rectangle test.A rectangle
4 One pair of sides is equal and a DIFFERENT pair is parallel. Parallelogram?

An isosceles trapezium satisfies this.No

Complete in your workbook.

Multiple Choice · 5 questions

MC1

Property or test

+10 XP

"The diagonals of a parallelogram bisect each other" is a property. The corresponding TEST states:

MC2

The same pair

+10 XP

Which is sufficient to prove a quadrilateral is a parallelogram?

MC3

Disproof

+10 XP

To show that a proposed test is NOT sufficient, you must:

MC4

Upgrading a parallelogram

+10 XP

A parallelogram whose diagonals are equal must be:

MC5

Choosing the efficient route

+10 XP

A quadrilateral's diagonals bisect each other and are perpendicular. The most efficient proof that it is a rhombus is to:

Short Answer · 3 questions

Apply the tests

+15 XP

SHORT ANSWER

For each quadrilateral, state what type it must be and name the test you used. If no conclusion follows, give a counterexample.
(a) $ABCD$ with $AB = DC$ and $AD = BC$
(b) $PQRS$ whose diagonals bisect each other at right angles
(c) $WXYZ$ with $WX \parallel ZY$ and $WZ = XY$
(d) $EFGH$, a parallelogram with $\angle E = 90°$

Write your working in your book.

Prove a test

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SHORT ANSWER

Prove that a quadrilateral with both pairs of opposite sides equal is a parallelogram.
(a) State what is given and what is to be proved.
(b) Give the proof in full, with a reason on every line.
(c) Explain why this needed proving at all, given that Lesson 4 proved the opposite sides of a parallelogram are equal.

Write your working in your book.

A combined problem

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SHORT ANSWER

In $\triangle ABC$, $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$. The point $F$ lies on the line $DE$ extended, such that $DE = EF$.
(a) Prove that $\triangle ADE \equiv \triangle CFE$.
(b) Hence prove that $DBCF$ is a parallelogram.
(c) Given $BC = 14$ cm, find $DF$, justifying your answer.

Write your working in your book.

Stretch Challenge · Which conditions are enough?

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CHALLENGE

A student proposes three new tests. For each, either prove it or disprove it with a counterexample.
(a) "A quadrilateral with two pairs of equal opposite angles is a parallelogram."
(b) "A quadrilateral with one pair of parallel sides and one pair of equal opposite angles is a parallelogram."
(c) "A quadrilateral whose diagonals bisect each other and are equal is a rectangle."

Quick Review

recap

Property

If it is this figure, then this is true

Test

If this is true, then it is this figure

Test 3

Equal AND parallel, on the same pair

Disproof

One counterexample is enough

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