Lesson 4 ~40 min Geometrical Figures C · Path +85 XP

Proving Properties of Triangles and Quadrilaterals

A definition says the least that identifies a figure. Everything else you know about that figure is a property, and every property has a proof behind it. This lesson supplies the proofs.

Today's hook: Why do the base angles of an isosceles triangle come out equal? Not because someone decided they should. The definition of an isosceles triangle mentions only two equal SIDES, and the equal angles are a consequence, provable in four lines using nothing but a congruence test from Lesson 2.

0/5QUESTS

Think First

warm-up

Write down everything you know about a parallelogram. Now cross out the one fact that is its definition, and look at what is left. Every remaining line is something that must be PROVED from that definition. Pick one and try to see how the proof would begin.

Record your answer in your workbook.

The Big Idea

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A definition is the minimum that identifies a figure. A property is anything else that is always true of it, and it earns its place by being proved. The proof almost always follows the same route: draw a diagonal or an axis, get two triangles, prove them congruent, and read the property off the matching parts.

$$\text{definition} \;\longrightarrow\; \text{draw a line} \;\longrightarrow\; \text{congruent triangles} \;\longrightarrow\; \text{property}$$

That route is worth learning as a route, not as a set of separate results. Faced with an unfamiliar property to prove, the first question is always "which line should I draw to make two triangles?" For a quadrilateral it is a diagonal; for an isosceles triangle it is the line from the apex to the base.

$\triangle ABM \equiv \triangle ACM$

Draw the diagonal

A quadrilateral proof nearly always starts by splitting it into two triangles.

Definition first

Begin from the defining property only. Using another property may be circular.

Read off matching parts

Once congruent, every matching side and angle is available with one reason.

What You'll Master

objectives

Know

The formal definitions of the isosceles and equilateral triangle and of the special quadrilaterals
The properties of a parallelogram, rectangle, rhombus and square, and that each is a proved result

Understand

Why a property must be proved from the definition rather than assumed alongside it
Why drawing a diagonal is the standard first move in a quadrilateral proof

Can Do

Prove that the base angles of an isosceles triangle are equal, and its converse
Prove the standard properties of a parallelogram from its definition
Prove a property of a rhombus or rectangle, using the parallelogram results already established

Words You Need

vocabulary

PropertyA fact that is always true of a figure, proved from its definition rather than assumed.

ConverseA statement with its condition and conclusion swapped. The converse of a true statement needs its own proof.

ParallelogramA quadrilateral with both pairs of opposite sides parallel.

RhombusA quadrilateral with four equal sides.

RectangleA parallelogram with one right angle.

BisectTo cut exactly in half. Diagonals that bisect each other cross at their common midpoint.

Base Angles of an Isosceles Triangle

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The definition supplies only the two equal sides. The equal angles are a property, and here is the proof.

Given: $\triangle ABC$ with $AB = AC$.
To prove: $\angle B = \angle C$.

Construction: let $M$ be the midpoint of $BC$, and join $AM$.

Statement	Reason
$AB = AC$	given
$BM = CM$	$M$ is the midpoint of $BC$
$AM = AM$	common side
$\triangle ABM \equiv \triangle ACM$	SSS
$\angle B = \angle C$	matching angles of congruent triangles

Notice the shape of it. One construction line, one congruence test, and the property drops out as a matching part. Almost every proof in this lesson has that shape.

Converses Need Their Own Proof

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The result just proved says: equal sides give equal base angles. The converse says: equal base angles give equal sides. That is a different statement, and proving one does not prove the other.

Given: $\triangle ABC$ with $\angle B = \angle C$. To prove: $AB = AC$.

Let $AD$ bisect $\angle A$, with $D$ on $BC$. Then $\angle BAD = \angle CAD$ (construction), $\angle B = \angle C$ (given), and $AD = AD$ (common). Two angles and a matching side give $\triangle ABD \equiv \triangle ACD$ (AAS), so $AB = AC$ (matching sides).

Both directions being true is what lets you use "isosceles" as a two-way street: from equal sides you may claim equal angles, and from equal angles you may claim equal sides. Many geometric statements are not like this, so the two-way use has to be earned each time.

A converse that fails

"A square is a rectangle" is true. Its converse, "a rectangle is a square", is false. Truth in one direction guarantees nothing about the other.

Properties of a Parallelogram

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The definition supplies only the parallels: a quadrilateral with both pairs of opposite sides parallel. Three properties follow, and one proof delivers the first two.

Given: $ABCD$ with $AB \parallel DC$ and $AD \parallel BC$. Draw the diagonal $AC$.

$\angle BAC = \angle DCA$ (alternate angles, $AB \parallel DC$); $\angle BCA = \angle DAC$ (alternate angles, $AD \parallel BC$); $AC = CA$ (common). So $\triangle ABC \equiv \triangle CDA$ (AAS).

From that single congruence: $AB = CD$ and $BC = DA$ (matching sides), so opposite sides are equal; and $\angle B = \angle D$ (matching angles), so opposite angles are equal.

The third property needs the other diagonal. With both drawn, meeting at $M$: $AB = DC$ (just proved), $\angle ABM = \angle CDM$ and $\angle BAM = \angle DCM$ (alternate angles), so $\triangle ABM \equiv \triangle CDM$ (AAS), giving $AM = CM$ and $BM = DM$. The diagonals bisect each other.

Building on What Is Already Proved

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Once the parallelogram results exist, they may be used as reasons. That is what makes the rectangle and rhombus proofs short.

A rectangle's diagonals are equal. A rectangle is a parallelogram with one right angle, and it can be shown all four angles are then right angles. In $\triangle ABC$ and $\triangle DCB$: $AB = DC$ (opposite sides of a parallelogram), $\angle ABC = \angle DCB = 90°$, and $BC = CB$ (common). So the triangles are congruent by SAS, and $AC = DB$.

A rhombus's diagonals are perpendicular. A rhombus has four equal sides, so it is a parallelogram and its diagonals bisect each other at $M$. In $\triangle AMB$ and $\triangle AMD$: $AB = AD$ (equal sides), $BM = DM$ (diagonals bisect), $AM = AM$ (common). By SSS the triangles are congruent, so $\angle AMB = \angle AMD$. These are adjacent angles on the straight line $BD$, so they sum to $180°$ and each is $90°$.

Order matters

You may only cite a result already proved. Using "the diagonals bisect each other" before proving it would make the argument circular.

Common Pitfalls

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Starting a parallelogram proof from "opposite sides are equal" when that is the thing being proved.

Fix: begin from the DEFINITION, the parallels, and nothing else. Anything else must already have been proved earlier in your working.

Assuming a statement's converse holds because the statement does.

Fix: a converse is a separate claim needing its own proof. "A square is a rectangle" is true; the reverse is not.

Forgetting the construction line, and trying to prove a quadrilateral property with no triangles present.

Fix: draw a diagonal and say so. "Construction: join $AC$" is a legitimate and expected line of a proof.

Watch Me Solve It · 3 examples

Watch Me Solve It · Equilateral from isosceles

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PROBLEM

Prove that an equilateral triangle, defined as a triangle with three equal sides, has three equal angles of $60°$ each.

1

Use the definition twice

$AB = AC \Rightarrow \angle B = \angle C$

$BA = BC \Rightarrow \angle A = \angle C$

Each pair of equal sides makes the triangle isosceles in a different way, and the base-angles property applies to each.
2

Combine

$\angle A = \angle B = \angle C$

Both results share $\angle C$, so all three angles are equal.
3

Bring in the angle sum

$\angle A + \angle B + \angle C = 180°$

Angle sum of a triangle.
4

Solve

$3 \times \angle A = 180° \Rightarrow \angle A = 60°$

Since all three are equal, each is a third of $180°$.

AnswerAll three angles are $60°$

Watch Me Solve It · Opposite angles of a parallelogram

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PROBLEM

$ABCD$ is a parallelogram. Prove that $\angle A = \angle C$.

1

State the definition and construct

$AB \parallel DC, \quad AD \parallel BC$

$\text{Construction: join } BD$

Only the parallels are given. The diagonal creates two triangles to work with.
2

Get two pairs of angles from the parallels

$\angle ABD = \angle CDB \quad \text{(alternate angles, } AB \parallel DC)$

$\angle ADB = \angle CBD \quad \text{(alternate angles, } AD \parallel BC)$
3

Add the common side and name the test

$BD = DB \quad \text{(common)}$

$\triangle ABD \equiv \triangle CDB \quad \text{(AAS)}$

Two angles and the side between them.
4

Read off the property

$\angle A = \angle C \quad \text{(matching angles of congruent triangles)}$

$A$ matches $C$ in the congruence statement.

Answer$\angle A = \angle C$

Watch Me Solve It · A diagonal of a rhombus bisects its angles

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PROBLEM

$ABCD$ is a rhombus. Prove that the diagonal $AC$ bisects $\angle BAD$.

1

Use the definition

$AB = BC = CD = DA \quad \text{(definition of a rhombus)}$

Four equal sides is all that is given.
2

Construct and list the facts

$\text{Join } AC$

$AB = AD, \quad CB = CD, \quad AC = AC \text{ (common)}$

The diagonal splits the rhombus into two triangles sharing that diagonal.
3

Prove the triangles congruent

$\triangle ABC \equiv \triangle ADC \quad \text{(SSS)}$

Three pairs of equal sides.
4

Read off the conclusion

$\angle BAC = \angle DAC \quad \text{(matching angles)}$

Those two angles together make $\angle BAD$, so $AC$ cuts it in half.

Answer$\angle BAC = \angle DAC$, so $AC$ bisects $\angle BAD$

Brain Trainer · 4 problems

Brain Trainer · Definition or property?

4 problems

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

1 "A quadrilateral with both pairs of opposite sides parallel." Definition or property of a parallelogram?

It is the minimum that identifies the figure.Definition
2 "The diagonals bisect each other." Definition or property of a parallelogram?

It must be proved from the parallels.Property
3 What construction starts almost every quadrilateral proof?

Two triangles are needed before any congruence test can be used.Draw a diagonal
4 The base angles of an isosceles triangle are equal. State the converse.

Swap the condition and the conclusion.Equal base angles give equal sides

Complete in your workbook.

Multiple Choice · 5 questions

MC1

Definition versus property

+10 XP

For a rhombus, "the diagonals are perpendicular" is:

MC2

Where the base-angles proof starts

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To prove the base angles of an isosceles triangle are equal, the standard construction is to:

MC3

Converses

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"If a quadrilateral is a square, then it is a rectangle." The converse of this statement is:

MC4

Which reason is available

+10 XP

While proving that the opposite sides of a parallelogram are equal, you may NOT use:

MC5

Reusing proved results

+10 XP

A proof that a rectangle's diagonals are equal may legitimately cite:

Short Answer · 3 questions

Prove a triangle property

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SHORT ANSWER

$\triangle ABC$ has $AB = AC$, and $D$ is the point on $BC$ such that $AD \perp BC$.
(a) Prove that $\triangle ABD \equiv \triangle ACD$.
(b) Hence prove that $D$ is the midpoint of $BC$.
(c) State which congruence test you used and explain why SSS would not have been available.

Write your working in your book.

Prove a quadrilateral property

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SHORT ANSWER

$ABCD$ is a parallelogram, defined as a quadrilateral with both pairs of opposite sides parallel. The diagonals meet at $M$.
(a) Prove that $AB = CD$.
(b) Using your result from (a), prove that $AM = CM$.
(c) Explain why part (b) had to come after part (a) rather than before it.

Write your working in your book.

Definition, property, or neither

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SHORT ANSWER

For each statement, say whether it is the definition of the figure, a property that must be proved, or false. Justify each answer.
(a) A rectangle is a parallelogram with one right angle.
(b) The diagonals of a rectangle bisect its angles.
(c) A rhombus has four equal sides.
(d) The diagonals of a parallelogram are equal.

Write your working in your book.

Stretch Challenge · The midpoint theorem

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CHALLENGE

In $\triangle ABC$, $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$.
(a) Prove that $\triangle ADE \sim \triangle ABC$, using a similarity test from Lesson 3.
(b) Hence prove that $DE \parallel BC$ and that $DE = \tfrac{1}{2} BC$.
(c) State the converse of this theorem, and say whether proving the theorem establishes it.

Quick Review

recap

Definition

The minimum that identifies the figure

Property

Everything else, and each one is proved

The route

Draw a diagonal, prove congruence, read it off

Converse

A separate claim, needing its own proof

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