Chapter 2: Reasoning and Proof

Inductive Reasoning

Learning Objectives

Recognize visual patterns and number patterns.
Extend and generalize patterns.
Write a counterexample to a pattern rule.

Introduction

You learned about some of the basic building blocks of geometry in Chapter 1. Some of these are points, lines, planes, rays, and angles. In this section we will begin to study ways we can reason about these building blocks.

One method of reasoning is called inductive reasoning. This means drawing conclusions based on examples.

Visual Patterns

Some people say that mathematics is the study of patterns. Let’s look at some visual patterns. These are patterns made up of shapes.

Example 1

A dot pattern is shown below.

A. How many dots would there be in the bottom row of a fourth pattern?

There will be $4$ dots. There is one more dot in the bottom row of each figure than in the previous figure. Also, the number of dots in the bottom row is the same as the figure number.

B. What would the total number of dots be in the bottom row if there were $6$ patterns?

There would be a total of $21$ dots. The rows would contain $1, 2, 3, 4, 5,$ and $6$ dots.

The total number of dots is $1 + 2 + 3 + 4 + 5 + 6 = 21$ .

Example 2

Next we have a pattern of squares and triangles.

A. How many triangles would be in a tenth illustration?

There will be $22$ triangles. There are $10$ squares, with a triangle above and below each square. There is also a triangle on each end of the figure. That makes $10 + 10 + 1 + 1 = 22$ triangles in all.

B. One of the figures would contain $34$ triangles. How many squares would be in that figure?

There will be $16$ squares. Take off one triangle from each end. This leaves $32$ triangles. Half of these $32$ triangles, or $16$ triangles, are above and $16$ triangles are below the squares. This means there are $16$ squares.

To check: With $16$ squares, there is a triangle above and below each square, making $2 \times 16 = 32$ squares. Add one triangle for each end and we have $32 + 1 + 1 = 34$ triangles in all.

C. How can we find the number of triangles if we know the figure number?

Let $n$ be the figure number. This is also the number of squares. $2n$ is the number of triangles above and below the squares. Add $2$ for the triangles on the ends.

If the figure number is $n$ , then there are $2n\ +\ 2$ triangles in all.

Example 3

Now look at a pattern of points and line segments.

For two points, there is one line segment with those points as endpoints.

For three noncollinear points (points that do not lie on a single line), there are three line segments with those points as endpoints.

A. For four points, no three points being collinear, how many line segments with those points as endpoints are there?

$6$ . The segments are shown below.

B. For five points, no three points being collinear, how many line segments with those points as endpoints are there?

$10$ . When we add a 5th point, there is a new segment from that point to each of the other four points. We can draw the four new dashed segments shown below. Together with the six segments for the four points in part A, this makes $6 + 4 = 10$ segments.

Number Patterns

You are already familiar with many number patterns. Here are a few examples.

Example 4 – Positive Even Numbers

The positive even numbers form the pattern $2, 4, 6, 8, 10, 12, \ldots.$

What is the $19^{th}$ positive even number?

The answer is $38$ . Each positive even number is $2$ more than the preceding one. You could start with $2$ , then add $2$ , $18\;\mathrm{times}$ , to get the $19^{th}$ number. But there is an easier way, using more advanced mathematical thinking. Notice that the $3^{rd}$ even number is $2 \times 3$ , the $4^{th}$ even number is $2 \times 4$ , and so on. So the $19^{th}$ even number is $2 \times 19 = 38.$

Example 5 – Odd Numbers

Odd numbers form the pattern $1, 3, 5, 7, 9, 11, \ldots.$

A. What is the $34^{th}$ odd number?

The answer is $67$ . We can start with $1$ and add $2, 33\;\mathrm{times}$ . $1 + 2 \times 33 = 1 + 66 = 67$ . Or, we notice that each odd number is $1$ less than the corresponding even number. The $34^{th}$ even number is $2 \times 34 = 68$ (example 4), so the $34^{th}$ odd number is $68 - 1 = 67$ .

B. What is the $n^{th}$ odd number?

$2n-1$ . The $n^{th}$ even number is $2n$ (example 4), so the $n^{th}$ odd number is $2n-1.$

Example 6 – Square Numbers

Square numbers form the pattern $1, 4, 9, 16, 25, \ldots.$

These are called square numbers because $1 = 1^2, 4 = 2^2, 9 = 3^2, 16 = 4^2, 25 = 5^2, \ldots.$

A. What is the $10^{th}$ square number?

The answer is $100$ . The $10^{th}$ square number is $10^2 = 100$ .

B. The $n^{th}$ square number is $441$ . What is the value of $n$ ?

The answer is $21$ . The $21^{st}$ square number is $21^2 = 441$ .

Conjectures and Counterexamples

A conjecture is an “educated guess” that is often based on examples in a pattern. Examples suggest a relationship, which can be stated as a possible rule, or conjecture, for the pattern.

Numerous examples may make you strongly believe the conjecture. However, no number of examples can prove the conjecture. It is always possible that the next example would show that the conjecture does not work.

Example 7

Here’s an algebraic equation.

$t = (n-1)(n-2)(n-3)$

Let’s evaluate this expression for some values of $n$ .

$n & =1;t=(n-1)(n-2)(n-3)=0 \times (-1) \times (-2)=0\ n & = 2; t = (n-1)(n-2)(n-3) = 1 \times 0 \times (-1) = 0\ n & = 3; t = (n-1)(n-2)(n-3) = 2 \times 1 \times 0 = 0$

These results can be put into a table.

$&n& &1& &2& &3\ &t& &0& &0& &0&$

After looking at the table, we might make this conjecture:

The value of $(n-1)(n-2)(n-3)$ is $0$ for any whole number value of $n$ .

However, if we try other values of $n$ , such as $n = 4$ , we have

$(n-1)(n-2)(n-3) = 3 \times 2 \times 1 = 6$

Obviously, our conjecture is wrong. For this conjecture, $n = 4$ is called a counterexample, meaning that this value makes the conjecture false. (Of course, it was a pretty poor conjecture to begin with!)

Example 8

Ramona studied positive even numbers. She broke some positive even numbers down as follows:

$8 = 3 + 5& &14 = 5 + 9& &36 = 17 + 19& &82 = 39 + 43&$

What conjecture might be suggested by Ramona’s results?

Ramona made this conjecture:

“Every positive even number is the sum of two different positive odd numbers.”

Is Ramona’s conjecture correct? Can you find a counterexample to the conjecture?

The conjecture is not correct. A counterexample is $2$ . The only way to make a sum of two odd numbers that is equal to $2$ is: $2 = 1 + 1$ , which is not the sum of different odd numbers.

Example 9

Artur is making figures for a graphic art project. He drew polygons and some of their diagonals.

Based on these examples, Artur made this conjecture:

If a convex polygon has $n$ sides, then there are $n-3$ diagonals from any given vertex of the polygon.

Is Artur’s conjecture correct? Can you find a counterexample to the conjecture?

The conjecture appears to be correct. If Artur draws other polygons, in every case he will be able to draw $n-3$ diagonals if the polygon has $n$ sides.

Notice that we have not proved Artur’s conjecture. Many examples have (almost) convinced us that it is true.

Lesson Summary

In this lesson you worked with visual and number patterns. You extended patterns to beyond the given items and used rules for patterns. You also learned to make conjectures and to test them by looking for counterexamples, which is how inductive reasoning works.

Points to Consider

Inductive reasoning about patterns is a natural way to study new material. But we saw that there is a serious limitation to inductive reasoning: No matter how many examples we have, examples alone do not prove anything. To prove relationships, we will learn to use deductive reasoning, also known as logic.

Review Questions

How many dots would there be in the fourth pattern of each figure below?

What is the next number in the following number pattern? $5, 8, 11, 14$
What is the tenth number in this number pattern? $3, 6, 11, 18$

The table below shows a number pattern.

$& n && 1 && 2 && 3 && 4 && 5 \ & t && 3 && 8 && 15 && 24 && 35$

What is the value of $t$ when $n = 6$ ?
What is the value of $n$ when $t = 99$ ?
Is $145$ a value of $t$ in this pattern? Explain your answer.

Give a counterexample for each of the following statements.

If $n$ is a whole number, then $n^2 > n.$
Every prime number is an odd number.
If $AB = 5$ and $BC = 2$ , then $AC = 7$ .

Review Answers

$9$
$20$
$13$
$17$
$102$
$48$
$9$
No. Values of $t$ are $3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, \ldots$ or, $t = n^2 + 2n;$ there is no value of $n$ that makes $t = 145.$
$1,$ because $1^2 = 1.$
$2,$ because $2$ is prime but not odd.
Any set of points where $A,B,$ and $C$ are not collinear.

Conditional Statements

Learning Objectives

Recognize if-then statements.
Identify the hypothesis and conclusion of an if-then statement.
Write the converse, inverse, and contrapositive of an if-then statement.
Understand a biconditional statement.

Introduction

In geometry we reason from known facts and relationships to create new ones. You saw earlier that inductive reasoning can help, but it does not prove anything. For that we need another kind of reasoning. Now you will begin to learn about deductive reasoning, the kind of reasoning used throughout mathematics and science.

If-Then Statements

In geometry, and in ordinary life, we often make conditional, or if-then, statements.

Statement 1: If the weather is nice, I’ll wash the car. (“Then” is implied even if not stated.)
Statement 2: If you work overtime, then you’ll be paid time-and-a-half.
Statement 3: If $2$ divides evenly into $x$ , then $x$ is an even number.
Statement 4: If a triangle has three congruent sides, it is an equilateral triangle. (“Then” is implied; this is a definition.)
Statement 5: All equiangular triangles are equilateral. (“If” and “then” are both implied.)

An if-then statement has two parts.

The “if” part is called the hypothesis.
The “then” part is called the conclusion.

For example, in statement 2 above, the hypothesis is “you work overtime.” The conclusion is “you’ll be paid time-and-a-half.”

Look at statement 1 above. Even though the word “then” is not actually present, the statement could be rewritten as: If the weather is nice, then I’ll wash the car. This is the meaning of statement 1. The hypothesis is “the weather is nice.” The conclusion is “I’ll wash the car.”

Statement 5 is a little more complicated. “If” and “then” are both implied without being stated. Statement 5 can be rewritten as: If a triangle is equiangular, then it is equilateral.

What is meant by an if-then statement? Suppose your friend makes the statement in statement 2 above, and adds another fact.

If you work overtime, then you’ll be paid time-and-a-half.
You worked overtime this week.

If we accept these statements, what other fact must be true? Combining these two statements, we can state with no doubt:

You’ll be paid time-and-a-half this week.

Let’s analyze statement 1, which was rewritten as: If the weather is nice, then I’ll wash the car. Suppose we accept statement 1 and another fact: I’ll wash the car.

Can we conclude anything further from these two statements? No. Even if the weather is not nice, I might wash the car. We do know that if the weather is nice I’ll wash the car. We don’t know whether or not I might wash the car even if the weather is not nice.

Converse, Inverse, and Contrapositive of an If-Then Statement

Look at statement 1 above again.

If the weather is nice, then I’ll wash the car.

This can be represented in a diagram as:

If $p$ then $q$ .

$p = \text{the weather is nice}& &q = \text{I'll wash the car}$

“If $p$ then $q$ ” is also written as

$p \rightarrow q$

Notice that conditional statements, hypotheses, and conclusions may be true or false. $p,$ $q,$ and the statement “If $p,$ then $q$ ” may be true or false.

In deductive reasoning we sometimes study statements related to a given if-then statement. These are formed by using $p,$ $q,$ and their opposites, or negations (“not”). Note that “not $p$ ” is written in symbols as $\lnot p$ .

$p,q,$ $\lnot p,$ and $\lnot q$ can be combined to produce new if-then statements.

The converse of $p \rightarrow q$ is $q \rightarrow p.$
The inverse of $p \rightarrow q$ is $\lnot p \rightarrow \lnot q.$
The contrapositive of $p \rightarrow q$ is $\lnot q \rightarrow \lnot p.$

Now let’s go back to statement 1: If the weather is nice, then I’ll wash the car.

$p \rightarrow q& &p & = \text{the weather is nice}\ & &q &= \text{I’ll wash the car}\ &&\lnot p &= \text{the weather is not nice}\ &&q &= \text{I’ll wash the car (or I wash the car)}\ &&\lnot q &= \text{I won’t wash the car (or I don’t wash the car)}$

Converse: $q \rightarrow p\;$ If I wash the car, then the weather was nice.

Inverse: $\lnot p \rightarrow \lnot q\;$ If the weather is not nice, then I won’t wash the car.

Contrapositive: $\lnot q \rightarrow \lnot p\;$ If I don’t wash the car, then the weather is not nice.

Notice that if we accept statement 1 as true, then the converse and inverse may, or may not, be true. But the contrapositive is true. Another way to say this is: The contrapositive is logically equivalent to the original if-then statement. In future work you may be asked to prove an if-then statement. If it’s easier to prove the contrapositive, then you can do this since the statement and its contrapositive are equivalent.

Example 1

Statement:

If $n > 2,$ then $n^2 > 4$ . True.

Converse:

If $n^2 > 4$ , then $n > 2$ . False.

A counterexample is $n = -3$ , where $n^2 = 9 > 4$ but $n = -3$ is not $> 2$

Inverse:

If $n$ is not $> 2$ then $n^2$ is not $> 4$ . False.

A counterexample is $n = -3,$ where $n$ is not $> 2$ but $n^2$ = $9 > 4$

Contrapositive:

If $n^2$ is not $> 4$ , then $n$ is not $> 2$ . True.

If $n^2$ is not $> 4$ , then $-2 < n < 2$ and $n$ is not $> 2$

Example 2

Statement: If $AB = BC,$ then $B$ is the midpoint of $AC$ . False (as shown below).

Needs $AB=BC$

Converse: If $B$ is the midpoint of $\overline{AC}$ , then $AB = BC$ . True.

Inverse: If $AB \neq BC$ , then $B$ is not the midpoint of $\overline{AC}$ . True.

Contrapositive: If $B$ is not the midpoint of $\overline{AC}$ , then $AB \neq BC$ False (see the diagram above).

Biconditional Statements

You recall that the converse of “If $p$ then $q$ ” is “If $q$ then $p$ .” When these two are combined, we have a biconditional statement.

Biconditional: $p\ \rightarrow\ q$ and $q\ \rightarrow\ p$

In symbols, this is written as: $p\ \leftrightarrow\ q$

We read $p\ \leftrightarrow\ q$ as: “ $p$ if and only if $q$ ”

Example 3

True statement: $m\angle{ABC} > 90^\circ$ if and only if $\angle{ABC}$ is an obtuse angle.

You can break this down to say:

If $m\angle{ABC} > 90^\circ$ then $\angle{ABC}$ is an obtuse angle and if $\angle{ABC}$ is an obtuse angle then $m\angle{ABC} > 90^\circ$ .

Notice that both parts of this biconditional are true; the biconditional itself is true.

You most likely recognize this as the definition of an obtuse angle.

Geometric definitions are biconditional statements that are true.

Example 4

Let $p$ be $x < 10$

Let $q$ be $2 x < 50$

a. Is $p$ $\rightarrow$ $q$ true?

Yes.

$p\ \rightarrow\ q$ is if $x < 10$ then $2 x < 50.$

From algebra we know that if $x < 10$ then $2 x < 2 (10)$ and $2x < 20.$ If $2x < 20$ , then we know that $2x < 50.$

So if $x < 10$ then $2x < 50$ , or $p\ \rightarrow\ q,$ is true.

b. Is $q \ \rightarrow\ p$ true?

No.

$q\ \rightarrow\ p$ is if $2x < 50$ , then $x < 10.$

From algebra we know that if $2x < 50$ , then $x < 25.$

However, $x < 25$ does not guarantee that $x < 10.$

$x$ can be less than $25$ but still not less than $10$ , for example if $x$ is $20$ .

So if $2x < 50$ , then $x < 10$ , or $q\ \rightarrow\ p$ , is false.

c. Is $p \leftrightarrow q$ true?

No.

$p\ \leftrightarrow\ q$ is $x < 10$ if and only if $2x < 50.$

We saw above that the if part of this statement, which is

If $2x < 50$ then $x < 10.$

This statement is false. One counterexample is $x = 20$ .

Note that if either $p\ \rightarrow\ q$ or $q\ \rightarrow\ p$ is false, then $p\ \leftrightarrow\ q$ is false.

Lesson Summary

In this lesson you have learned how to express mathematical and other statements in if-then form. You also learned that each if-then statement is linked to variations on the basic theme of “If $p$ then $q$ .” These variations are the converse, inverse, and contrapositive of the if-then statement. Biconditional statements combine the statement and its converse into a single “if and only if” statement. Definitions are an important type of biconditional, or if-and-only-if, statement.

Points to Consider

We called points, lines, and planes the building blocks of geometry. We will soon see that hypothesis, conclusion, as well as if-then and if-and-only-if statements are the building blocks that deductive reasoning, or logic, is built on. This type of reasoning will be used throughout your study of geometry. In fact, once you understand logical reasoning you will find that you apply it to other studies and to information you encounter all your life.

Review Questions

Write the hypothesis and the conclusion for each statement.

If $2$ divides evenly into $x$ , then $x$ is an even number.
If a triangle has three congruent sides, it is an equilateral triangle.
All equiangular triangles are equilateral.
What is the converse of the statement in exercise 1 above? Is the converse true?
What is the inverse of the statement in exercise 2 above? Is the inverse true?
What is the contrapositive of the statement in exercise 3? Is the contrapositive true?
The converse of a statement about collinear points , , and is: If and , then is the midpoint of .
- What is the statement?
- Is it true?
What is the inverse of the inverse of if $p$ then $q$ ?
What is the one-word name for the converse of the inverse of an if-then statement?
What is the one-word name for the inverse of the converse of an if-then statement?

For each of the following biconditional statements:

Write $p$ in words.
Write $q$ in words.
Is $p \rightarrow q$ true?
Is $q \rightarrow p$ true?
Is $p \leftrightarrow q$ true?

Note that in these questions, $p$ and $q$ could be reversed and the answers would be correct.

A U.S. citizen can vote if and only if he or she is $18$ or more years old.
A whole number is prime if and only if it is an odd number.
Points are collinear if and only if there is a line that contains the points.
$x + y = 17$ if only if $x = 8$ and $y = 9$

Review Answers

Hypothesis: $2$ divides evenly into $x$ ; conclusion: $x$ is an even number.
Hypothesis: A triangle has three congruent sides; conclusion: it is an equilateral triangle.
Hypothesis: A triangle is equiangular; conclusion: the triangle is equilateral.
If $x$ is an even number, then $2$ divides evenly into $x$ . True.
If a triangle does not have three congruent sides, then it is not an equilateral triangle. True.
If a triangle is not equilateral, then it is not equiangular. True.
If $B$ is the midpoint of $\overline{AC}$ , then $AB = 5$ and $BC = 5$ . False ( $AB$ and $BC$ could both be $6$ , $7$ , etc.).
If $p$ then $q.$
Contrapositive
Contrapositive
$p =$ he or she is $18$ or more years old; $q =$ a U. S. citizen can vote; $p \rightarrow q$ is true; $q \rightarrow p$ is true; $p \leftrightarrow q$ is true.
$p =$ a whole number is an odd number; $q =$ a whole number is prime; $p \rightarrow q$ is false; $q \rightarrow p$ is false; $p \leftrightarrow q$ is false.
$p =$ a line contains the points; $q =$ the points are collinear; is $p \rightarrow q$ is true; $q \rightarrow p$ is true; $p \leftrightarrow q$ is true.
$p = x = 8$ and $y = 9$ ; $q = x + y = 17$ ; $p \rightarrow q$ is true; $q \rightarrow p$ is false; $p \leftrightarrow q$ is false.

Deductive Reasoning

Learning Objectives

Recognize and apply some basic rules of logic.
Understand the different parts that inductive reasoning and deductive reasoning play in logical reasoning.
Use truth tables to analyze patterns of reasoning.

Introduction

You began to study deductive reasoning, or logic, in the last section, when you learned about if-then statements. Now we will see that logic, like other fields of knowledge, has its own rules. When we follow those rules, we will expand our base of facts and relationships about points, lines, and planes. We will learn two of the most useful rules of logic in this section.

Direct Reasoning

We all use logic—whether we call it that or not—in our daily lives. And as adults we use logic in our work as well as in making the many decisions a person makes every day.

Which product should you buy?
Who should you vote for?
Will this steel beam support the weight you place on it?
What will be your company’s profit next year?

Let’s see how common sense leads to the two most basic rules of logic.

Example 1

Suppose Bea makes the following statements, which are known to be true.

If Central High School wins today, they will go to the regional tournament.

Central High School does win today.

Common sense tells us that there is an obvious logical conclusion if these two statements are true:

Central High School will go to the regional tournament.

Example 2

Here are two true statements.

$5$ is an odd number.

Every odd number is the sum of an even and an odd number.

Based on only these two true statements, there is an obvious further conclusion:

$5$ is the sum of an even and an odd number.

(This is true, since $5 = 2 + 3$ ).

Example 3

Suppose the following two statements are true.

If you love me let me know, if you don’t then let me go. (A country music classic. Lyrics by John Rostill.)
You don’t love me.

What is the logical conclusion?

Let me go.

There are two statements in the first line. The second one is:

If you don’t (love me) then let me go.

You don’t love me is stated to be true in the second line.

Based on these true statements,“Let me go” is the logical conclusion.

Now let’s look at the structure of all of these examples, using the $p$ and $q$ symbols that we used earlier.

Each of the examples has the same form.

$& p \rightarrow q \\ & p$

conclusion: $q$

A more compact form of this argument, (logical pattern) is:

$& p \rightarrow q \ & p \ & \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \ & q$

To state this differently, we could say that the true statement $q$ follows automatically from the true statements $p \rightarrow q$ and $p$ .

This reasoning pattern is one of the basic rules of logic. It’s called the law of detachment.

Law of Detachment

Suppose $p$ and $q$ are statements. Then given

$p \rightarrow q$ and $p$

You can conclude

$q$

Practice saying the law of detachment like this: “If $p \rightarrow q$ is true, and $p$ is true, then $q$ is true.”

Example 4

Here are two true statements.

If $\angle{A}$ and $\angle{B}$ are a linear pair, then $m\angle{A} + m\angle{B} = 180^\circ$ .

$\angle{A}$ and $\angle{B}$ are a linear pair.

What conclusion do we draw from these two statements?

$m \angle{A} + m \angle{B} = 180^\circ.$

The next example is a warning not to turn the law of detachment around.

Example 5

Here are two true statements.

If $\angle{A}$ and $\angle{B}$ are a linear pair, then $m \angle{A} + m \angle{B} = 180^\circ.$

$m \angle{A} = 90^\circ$ and $m \angle{B} = 90^\circ.$

What conclusion can we draw from these two statements?

None! These statements are in the form

$p \rightarrow q$

$q$

Note that since $m \angle{A} = 90^\circ$ and $m \angle{B} = 90^\circ$ , we also know that $m \angle{A} + m \angle{B} = 180^\circ$ , but this does not mean that they are a linear pair.

The law of detachment does not apply. No further conclusion is justified.

You might be tempted to conclude that $\angle{A}$ and $\angle{B}$ are a linear pair, but if you think about it you will realize that would not be justified. For example, in the rectangle below $m \angle{A} = 90^\circ$ and $m \angle{B} = 90^\circ$ (and $m \angle{A} + m \angle{B}=180^\circ$ , but $\angle{A}$ , and $\angle{B}$ are definitely NOT a linear pair.

Now let’s look ahead. We will be doing some more complex deductive reasoning as we move ahead in geometry. In many cases we will build chains of connected if-then statements, leading to a desired conclusion. Start with a simplified example.

Example 6

Suppose the following statements are true.

1. If Pete is late, Mark will be late.

2. If Mark is late, Wen will be late.

3. If Wen is late, Karl will be late.

To these, add one more true statement.

4. Pete is late.

One clear consequence is: Mark will be late. But make sure you can see that Wen and Karl will also be late.

Here’s a symbolic form of the statements.

$p \rightarrow q$
$q \rightarrow r$
$r \rightarrow s$
$& p \ & \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\ & s$

Our statements form a “chain reaction.” Each “then” becomes the next “if” in a chain of statements. The chain can consist of any number of connected statements. Once we add the true $p$ statement as above, we know that the conclusion (the then part) of the last statement is justified.

Another way to look at this is to imagine a chain of dominoes. The dominoes are the linked if-then statements. Once the first domino falls, each domino knocks the next one over, and the last domino falls. $p$ is the tipping over of the first domino. The final conclusion of the last if-then statement is the last domino.

This is called the law of syllogism. A formal statement of this rule of logic is given below.

Law of Syllogism

Suppose $a_1, a_2, \ldots, a_{n-1}$ , and $a_{n}$ are statements. Then given that $a_1$ is true and that you have the following relationship:

$& a_1 \ \rightarrow \ a_2\ & a_2 \ \rightarrow \ a_3\ & \vdots\ & a_{n-1} \ \rightarrow \ a_n$

Then, you can conclude

$a_{1} \rightarrow a_{n}$

Inductive vs. Deductive Reasoning

You have now worked with both inductive and deductive reasoning. They are different but not opposites. In fact, they will work together as we study geometry and other mathematics.

How do these two kinds of reasoning complement (strengthen) each other? Think about the examples you saw earlier in this chapter.

Inductive reasoning means reasoning from examples. You may look at a few examples, or many. Enough examples might make you suspect that a relationship is true always, or might even make you sure of this. But until you go beyond the inductive stage, you can’t be absolutely sure that it is always true.

That’s where deductive reasoning enters and takes over. We have a suggestion arrived at inductively. We then apply rules of logic to prove, beyond any doubt, that the relationship is true always. We will use the law of detachment and the law of syllogism, and other logic rules, to build these proofs.

Symbolic Notation and Truth Tables

Logic has its own rules and symbols. We have already used letters like $p$ and $q$ to represent statements: for the negation (“not”), and the arrow $\rightarrow$ to indicate if-then. Here are two more symbols we can use.

$\land & = \text{and}\\ \lor &= \text{or}$

Truth tables are a way to analyze statements in logic. Let’s look at a few simple truth tables.

Example 1

How is $\lnot p$ related to $p$ logically? We make a truth table to find out. Begin with all the possible truth values of $p$ . This is very simple; $p$ can be either true $(T)$ , or false $(F)$ .

$p$
$T$
$F$

Next we write the corresponding truth values for $\lnot p$ . $\lnot p$ has the opposite truth value as $p$ . If $p$ is true, then $\lnot p$ is false, and vice versa. Complete the truth table by filling in the $\lnot p$ column.

$p$	$\lnot p$
$T$	$F$
$F$	$T$

Now we construct truth tables for slightly more complex logic.

Example 2

Draw a truth table for $p$ and $q$ written $p \land q$ .

Begin by filling in all the $T/F$ combinations possible for $p$ and $q$ .

$p$	$q$	$p \land q$
$T$	$T$
$T$	$F$
$F$	$T$
$F$	$F$

How can $p$ and $q$ be true? Common sense tells us that $p$ and $q$ is false whenever either $p$ or $q$ is false. We complete the last column accordingly.

$p$	$q$	$p \land q$
$T$	$T$	$T$
$T$	$F$	$F$
$F$	$T$	$F$
$F$	$F$	$F$

Another way to state the meaning of the truth table is that $p \land q$ is true only when $p$ is true and $q$ is true.

Let’s do the same for $p$ or $q$ . Before we do that, we need to clarify which “or” we mean in mathematics. In ordinary speech, or is sometimes used to mean, “this or that, but not both.” This is called the exclusive or (it excludes or keeps out both). In mathematics, or means “this, that, or both this and that.” This is called the inclusive or. Knowing that or is inclusive makes the truth table an easy job.

Example 2

$5 = 2 + 3$ or $5 > 6$ is true, because $5 = 2 + 3$ is true.

$5 < 6$ or $6 < 5$ is true, because $5 < 6$ is true.

$5 = 2 + 3$ or $5 < 6$ is true because $5 = 2 + 3$ is true and $5 < 6$ is true.

$5 = 2 + 4$ or $5 > 6$ is false because $5 = 2 + 4$ is false and $5 > 6$ is false.

Example 3

Draw a truth table for $p$ or $q$ , which is written $p \lor q$ .

Begin by filling in all the $T/F$ combinations possible for $p$ and $q$ . Keeping in mind the definition of or above (inclusive), fill in the third column. $p$ or $q$ will only be false when both $p$ and $q$ are false; it is true otherwise.

$p$	$q$	$p \lor q$
$T$	$T$	$T$
$T$	$F$	$T$
$F$	$T$	$T$
$F$	$F$	$F$

Lesson Summary

Do we all have our own version of what is logical? Let’s hope not—we wouldn’t be able to agree on what is or isn’t logical! To avoid this, there are agreed-on rules for logic, just like there are rules for games. The two most basic rules of logic that we will be using throughout our studies are the law of detachment and the law of syllogism.

Points to Consider

Rules of logic are universal; they apply to all fields of knowledge. For us, the rules give a powerful method for proving new facts that are suggested by our explorations of points, lines, planes, and so on. We will structure a specific format, the two-column proof, for proving these new facts. In upcoming lessons you will write two-column proofs. The facts or relationships that we prove are called theorems.

Review Questions

Must the third sentence be true if the first two sentences are true? Explain your answer.

People who vote for Jane Wannabe are smart people.

I am a smart person.

I will vote for Jane Wannabe.

If Rae is the driver today then Maria is the driver tomorrow.

Ann is the driver today.

Maria is not the driver tomorrow.

All equiangular triangles are equilateral.

$\triangle ABC$ is equiangular.

$\triangle ABC$ is equilateral.

What additional statement must be true if the given sentences are true?

If West wins, then East loses.
If North wins, then West wins.
If $x > 5$ then $x > 3$ .
If $x > 3$ then $y > 7$ .

$x = 6$ .

Fill in the truth tables.

$p$	$\lnot p$	$p \land \lnot p$
$T$
$F$

$p$	$\lnot p$	$p \lor \lnot p$
$T$
$F$

$p$	$q$	$\lnot p$	$\lnot q$	$\lnot p \land \lnot q$
$T$	$T$
$T$	$F$
$F$	$T$
$F$	$F$

$p$	$q$	$\lnot q$	$q \lor \lnot q$	$p \land (q \lor \lnot q)$
$T$	$T$
$T$	$F$
$F$	$T$
$F$	$F$

When is $p \lor\ q \lor\ r$ true?
For what values of $x$ is the following statement true?

$x \ge 2$ or $x^2<4$

For what values of $x$ is the following statement true?

$x \ge 2$ or $x^2<4$

Review Answers

No (converse error).
No (inverse error).
Yes.
If North wins, then East loses.
$y > 7$ . (also $x > 3$ )

$p$	$\lnot p$	$p \land \lnot p$
$T$	$F$	$F$
$F$	$T$	$F$

Note that
$p \land \lnot p$
is
never
true.

$p$	$\lnot p$	$p \lor \lnot p$
$T$	$F$	$T$
$F$	$T$	$T$

Note that
$p \lor \lnot p$
is
always
true.

$p$	$q$	$\lnot p$	$\lnot q$	$\lnot p \land \lnot q$
$T$	$T$	$F$	$F$	$F$
$T$	$F$	$F$	$T$	$F$
$F$	$T$	$T$	$F$	$F$
$F$	$F$	$T$	$T$	$T$

Note that
$\lnot p \land \lnot q$
is true only when
$p$
and
$q$
are
both
false.

$p$	$q$	$\lnot q$	$q \lor \lnot q$	$p\land (q \lor \lnot q)$
$T$	$T$	$F$	$T$	$T$
$T$	$F$	$T$	$T$	$T$
$F$	$T$	$F$	$T$	$F$
$F$	$F$	$T$	$T$	$F$

$p \lor q \lor r$ is always true except when $p,\ q$ , and $r$ are all false.
$x > -2$
none, $\emptyset.$

Algebraic Properties

Learning Objectives

Identify and apply properties of equality.
Recognize properties of congruence “inherited” from the properties of equality.
Solve equations and cite properties that justify the steps in the solution.
Solve problems using properties of equality and congruence.

Introduction

We have begun to assemble a toolbox of building blocks of geometry (points, lines, planes) and rules of logic that govern deductive reasoning. Now we start to expand our geometric knowledge by applying logic to the geometric building blocks. We’ll make a smooth transition as some fundamental principles of algebra take on new life when expressed in the context of geometry.

Properties of Equality

All things being equal, in mathematics the word “equal” means “the same as.” To be precise, the equal sign $=$ means that the expression on the left of the equal sign and the expression on the right represent the same number. So equality is specifically about numbers—numbers that may be expressed differently but are in fact the same.

Some examples:

$12 - 5 = 7$
$\frac{372+372+372+372} {4} = 300+70+2$
$1.5(40 + 60) = 150$

Basic properties of equality are quite simple and you are probably familiar with them already. They are listed here in formal language and then translated to common sense terms.

Properties of Equality

For all real numbers $a, b$ , and $c$ :

Reflexive Property: $a = a$

That is, any number is equal to itself, or the same as itself.

Example: $25 = 25$

Symmetric Property: If $a = b$ then $b = a$ .

You can read an equality left to right, or right to left.

Example: If $8a = 32$ then $32 = 8a$

Example: If $m \angle{P} + m \angle{Q} = 180$ , then $180 = m \angle{P} + m \angle{Q}$ .

Sometimes it is more convenient to write $b = a$ than $a = b$ . The symmetric property allows this.

Transitive Property: If $a = b$ and $b = c$ then $a = c$ .

Translation: If there is a “chain” of linked equations, then the first number is equal to the last number. (You can prove that this applies to more than two equalities in the review questions.)

Example: If $a + 4 = 10$ and $10 = 6 + 4$ , then $a + 4 = 6 + 4$ .

As a reminder, here are some properties of equality that you used heavily when you learned to solve equations in algebra.

Substitution Property: If $a = b,$ then $b$ can be put in place of $a$ anywhere or everywhere.

Example: Given that $a = 9$ and that $a - c = 5$ . Then $9 - c = 5$ .

Addition Property of Equality: If $a = b$ , then $a + c = b + c$ .

Translation: You can add the same number to both sides of an equation and retain the equivalency.

Example: If $m\angle{A} + 30 = 90$ , then $m\angle{A} + 30 + -30 = 90 + -30$ .

Multiplication Property of Equality: If $a = b$ , then $ac = bc.$

Translation: You can multiply the same number on both sides of an equation and retain the equivalency.

Example: If $3x = 18$ , then $\frac{1}{3} (3x) = \frac{1}{3} (18)$ .

Keep in mind that these are properties about numbers. As you go further into geometry, you can apply the properties of equality to anything that is a number: lengths of segments and angle measures, for example.

Properties of Congruence

Let’s review the definitions of congruent segments and angles.

Congruent Segments: $\overline{MN}$ $\cong$ $\overline{PQ}$ if and only if $MN = PQ$ .

Remember that, although $\overline{MN}$ and $\overline{PQ}$ are segments, $MN$ and $PQ$ are lengths of those segments, meaning that $MN$ and $PQ$ are numbers. The properties of equality apply to $MN$ and $PQ$ .

Congruent Angles: $\angle{F}$ $\cong$ $\angle{G}$ if and only if $m\angle{F}$ = $m\angle{G}$

The comment above about segment lengths also applies to angle measures. The properties of equality apply to $m\angle{F}$ and $m\angle{G}$ .

Any statement about congruent segments or congruent angles can be translated directly into a statement about numbers. This means that each property of equality has a corresponding property of congruent segments and a corresponding property of congruent angles.

Here are some of the basic properties of equality and the corresponding congruence properties.

Given that $x, y,$ and $z$ are real numbers.

Reflexive Property of Equality: $x = x$

Reflexive Property of Congruence of Segments: $\overline{MN}\ \cong\ \overline{MN}$

Reflexive Property of Congruence of Angles: $\angle{P}\ \cong\ \angle{P}$

Symmetric Property of Equality: If $x = y$ , then $y = x$ .

Symmetric Property of Congruence of Segments: If $\overline{MN}\ \cong\ \overline{PQ}$ , then $\overline{PQ}\ \cong$ $\overline{MN}.$

Symmetric Property of Congruence of Angles: If $\angle{P}\ \cong\ \angle{Q}$ , then $\angle{Q}\ \cong\ \angle{P}$ .

Transitive Property of Equality: If $x = y$ and $y = z$ , then $x = z$

Transitive Property of Congruence of Segments

If $\overline{MN}\ \cong\ \overline{PQ}$ and $\overline{PQ}\ \cong\ \overline{ST}$ then $\overline{MN}\ \cong\ \overline{ST}$

Transitive Property of Congruence of Angles

If $\angle{P}\ \cong\ \angle{Q}$ , and $\angle{Q}\ \cong\ \angle{R}$ then $\angle{P}\ \cong\ \angle{R}$ .

Using Congruence Properties in Equations

When you solve equations in algebra you use properties of equality. You might not write out the logical justification for each step in your solution, but you know that there is an equality property that justifies that step.

Let’s see how we can use the properties of congruence to justify statements in deductive reasoning. Abbreviated names of the properties can be used.

Example 1

Given points $A, B$ , and $C$ , with $AB = 8 , BC = 17$ , and $AC = 20$ .

Are $A$ , $B$ , and $C$ collinear?

$AB + BC & = AB + BC \text{(reflexive)}. && \text{Why do we want this?}\ &&&\text{So that we can bring in the numbers that are}\ AB\ \text{and}\ BC. \ AB + BC & = 8 + 17 && \text{Justification is substitution of}\ 8\ \text{for}\ AB\ \text{and}\ 17\ \text{for}\ BC \ AB + BC & = 25 && 8 + 17 = 25;\ \text{This is arithmetic.}\ &&&\text{No justification is needed as long as the arithmetic is correct}. \ 25 & \neq 20 && \text{More arithmetic}. \ AB + BC & \neq AC && \text{Substituting}\ AB + BC\ \text{for}\ 25\ \text{and}\ AC\ \text{for}\ 20. \ & A, B,\ \text{and}\ C\ \text{are\ not\ collinear}. && \text{Segment addition postulate}.\ &&& A, B,\ \text{and}\ C\ \text{are collinear if and only if}\ AB + BC = AC.$

Example 2

Given that $m\angle{A} + m\angle{B} = 100^\circ$ and $\angle{B} = 40^\circ$ .

Prove that $\angle{A}$ is an acute angle.

$m\angle{A} + m\angle{B} & = 100,\ m\angle{B} = 40. && \text{These are the given facts}. \ m\angle{A} + 40 & = 100. && \text{Substitute}\ 40\ \text{for}\ m\angle{B}\ \text{using the transitive property}. \ m\angle{A} + 40 + (-40) & = 100 + (-40). && \text{Addition property of equality; add} -40\ \text{to both sides}. \ m\angle{A} & = 60. && \text{Arithmetic}. \ 60 & < 90. && \text{More arithmetic}. \ m\angle{A} & < 90^\circ. && \text{Substitute}\ m\angle{A}\ \text{for}\ 60. \ & \angle{A}\ \text{is an acute angle}. && \text{Definition. An angle is acute if and only if its measure is between}\ 0^\circ\ \text{and}\ 90^\circ.$

The deductive reasoning scheme in example 2 is called a proof. The final statement must be true if the given information is true.

Lesson Summary

We built on our previous knowledge of properties of equality to derive corresponding properties of congruence. This enabled us to test statements about congruence, and to create new properties and relationships about congruence. We had our first introduction, in informal terms, to logical proof.

Points to Consider

In the examples and review questions, terms like given, prove, and reason were used. In upcoming lessons we’ll see how to identify the given facts, how to draw a diagram to represent a statement that we need to prove, and how to organize proofs more formally. As we move ahead we’ll prove many important geometric relationships called theorems. We have now established the framework of logic that we’ll use repeatedly in future work.

Review Questions

Given: $x, y$ , and $z$ are real numbers.

Use the given property or properties of equality to fill in the blank in each of the following questions.

Symmetric: If $x = 3$ , then ___________________.
Reflexive: If $x + 2 = 9$ , then ___________________.
Transitive: If $y = 12$ and $x = y$ then __________________.
Symmetric: If $x + y = y + z$ , then ____________________.
Reflexive: If $x + y = y + z$ , then _____________________.
Substitution: If $x = y - 7$ and $x = z + 4$ , then _________________________.
Use the transitive property of equality to write a convincing logical argument (a proof) that the statement below is true.

If $a = b$ and $b = c$ and $c = d$ and $d = e$ , then $a=e$ .

Note that this chain could be extended with additional links.

Let $M$ be the relation “is the mother of.” Let $B$ be the relation “is the brother of.”

Is $M$ symmetric? Explain your answer.
Is $B$ symmetric? Explain your answer.
Is $M$ transitive? Explain your answer.
Is $B$ transitive? Explain your answer.
Let $w, x, y,$ and $z$ be real numbers. Prove: If $w = y$ and $x = z$ , then ${w + x} = {y+z}.$
The following statement is not true. “Let $A, B, C, D, E,$ and $F$ be points. If $AB=DE$ and $BC=EF$ , then $AC=DF.$ ” Draw a diagram with these points shown to provide a counterexample.

Review Answers

$3 = x.$
$x + 2 = 9.$
$x = 12.$
$y + z = x + y.$
$x + y = y + z.$
$z + 4 = y + 7$ ( or $y - 7 = z + 4$ ).
If $a = b$ and $b = c$ then $a = c$ (transitive property). If $a = c$ and $c = d$ then $a = d$ (transitive property). If $a = d$ and $d = e$ then $a = e$ (transitive property).
No. If Maria is the mother of Juan, it does NOT follow that Juan is the mother of Maria!
Yes. For example, if Bill is Frank’s brother, then Frank is Bill’s brother.
No. If $M$ were transitive, then “Maria is Fern’s mother and Fern is Gina’s mother” would lead to “Maria is Gina’s mother.” However, Maria would actually have to be Gina’s grandmother!
Yes. If Bill is Frank’s brother and Frank is Greg’s brother, then Bill is Greg’s brother. You might say the brother of my brother is (also) my brother.
$w = y$ and $x = z$ (given); $w + x = w + x$ (reflexive); $w + x = y + z$ (substitute $y$ for $w$ and $z$ for $x$ ).
Below is an example:

A correct response is a diagram showing:
- $AB = DE$ .
- $BC=EF$ .
- $AC \neq DF$ .
If $A$ , $B$ , and $C$ are collinear and $D$ , $E$ , and $F$ are not collinear then the conditions are satisfied.

Diagrams

Learning Objectives

Provide the diagram that goes with a problem or proof.
Interpret a given diagram.
Recognize what can be assumed from a diagram and what can not be.
Use standard marks for segments and angles in diagrams.

Introduction

Geometry is about objects such as points, lines, segments, rays, planes, and angles. If we are to solve problems about these objects, our work is made much easier if we can represent these objects in diagrams. In fact, for most of us, diagrams are absolutely essential for problem solving in geometry.

Basic Postulates—Another Look

Just as undefined terms are building blocks that other definitions are built on, postulates are the building blocks of logic. We’re now ready to restate some of the basic postulates in slightly more formal terms, and to use diagrams.

Postulate 1 Through any two distinct points, there is exactly one line.

Comment: Any two points are collinear.

Postulate 2 There is exactly one plane that contains any three noncollinear points.

Comment: Sometimes this is expressed as: “Three noncollinear points determine a plane.”

Postulate 3 If two points are in a plane, then the whole line through those two points is in the plane.

Postulate 4 If two distinct lines intersect, then the intersection is exactly one point.

Comments: Some lines intersect, some do not. If lines do intersect, it is in only one point, otherwise one or both “lines” would have to curve, which lines do not do.

Postulate 5 If two distinct planes intersect, then the intersection is exactly one line.

Comments: Some planes intersect, some do not. Think of a floor and a ceiling as models for planes that do not intersect. If planes do intersect, it is in a line. Think of the edge of a box (a line) formed where two sides of the box (planes) meet.

Postulate 6 The Ruler Postulate: The points on a line can be assigned real numbers, so that for any two points, one corresponds to $0$ and the other corresponds to a nonzero real number.

Comments: This is how a number line and a ruler work. This also means we can measure any segment.

Postulate 7 The Segment Addition Postulate: Points $P, Q,$ and $R$ are collinear if and only if $PQ + QR = PR$ .

Comment: If $P, Q,$ and $R$ are not collinear, then $PQ + QR > PR$ . We saw examples of this fact in earlier sections of this chapter.

Postulate 8 The Protractor Postulate: If rays in a plane have a common endpoint, $0$ can be assigned to one ray and a number between $0$ and $180$ can be assigned to each of the other rays.

Comment: This means that any angle has a (degree) measure.

Postulate 9 The Angle Addition Postulate: Let $P, Q, R,$ and $S$ be points in a plane. $S$ is in the interior of $\angle{PQR}$ if and only if $m\angle{PQR}$ + $m\angle{SQR}$ = $m\angle{PQR}$ .

Comment: If an angle is made up of other angles, the measures of the component angles can be added to get the measure of the “big” angle.

Postulate 10 The Midpoint Postulate: Every line segment has exactly one midpoint.

Comments: If $M$ is a point on $\overline{AB}$ and $AM = MB$ there is not another point on $\overline{AB}$ , let’s say point $N$ ,with $AN = NB$ . The midpoint of a segment is unique.

Postulate 11 The Angle Bisector Postulate: Every angle has exactly one bisector.

Comments: The bisector of an angle is a ray. If $\overrightarrow{B P}$ bisects , $\angle{ABC}$ there is not another ray that bisects the angle. The bisector (ray) of an angle is unique.

Using Diagrams

Now we apply our definitions and postulates to a geometric figure. When measures are given on a figure, we can assume that the measurements on the figure are correct. We can also assume that:

Points that appear to be collinear are collinear.
Lines, rays, or segments that appear to intersect do intersect.
A ray that appears to be in the interior of an angle is in the interior of the angle.

We cannot assume the following from a diagram:

That lines, segments, rays, or planes are parallel or perpendicular.
That segments or angles are congruent.

These must be stated or indicated in the diagram.

The diagram below shows some segment and angle measures.

Example 1

A. Is $M$ the midpoint of $\overline{AB}?$ Explain your answer.

No. $M$ is on $\overline{AB}$ , but $AM \neq MB$ .

B. Is $Q$ the midpoint of $\overline{AD}?$ Explain your answer.

Yes. $Q$ is on $\overline{AD}$ , and $AQ$ = $QD$ .

C. Name an angle bisector and the angle that it bisects.

$\overrightarrow{P N}$ bisects $\angle{MPC}$ .

D. Fill in the blank: $m\angle{AMP} = m\angle{AMQ} + m\angle$ _______.

$QMP$

E. Is $\overrightarrow{M Q}$ the bisector of $\angle{AMP}?$ Explain your answer.

No. If $\overrightarrow{M Q}$ bisected $\angle{AMP}$ , then $m\angle{AMQ}$ would be $45^\circ.$ That would make $AQ = AM,$ but $AQ \neq AM.$

Sometimes we use special marks in diagrams. Tick marks show congruent segments. Arc marks show congruent angles. Right angle marks show right angles and perpendicular lines and segments.

When these signs are used, the relationships they represent become part of the given information for a problem.

Example 2

Blue blobs are dots for points and single and double arc marks to show equal angles.

Based on the marks on the diagram, we know that:

$BE = CH$ (single tick marks).
$BC = FG$ (double tick marks).
$m\angle{BEF} = m\angle{CHG}$ (single arc marks).
$m\angle{ABE} = m\angle{DCH}$ (double arc marks).
$\overline{BF} \perp \overline{EF}.$

Lesson Summary

As we move forward toward more formal reasoning, we have reviewed the basic postulates and expressed them more formally. We saw that most geometric situations involve diagrams. In diagrams we can assume some facts, and we cannot assume others.

Points to Consider

In upcoming lessons you will organize your reasoning pattern into the two-column proof. This is a traditional pattern that still works very well today. It gives us a clear, direct format, and uses the basic rules of logic that we saw in earlier lessons. We will prove many important geometric relationships called theorems throughout the rest of this geometry course.

Review Questions

Use the diagram to answer questions 1-8.

Name a right angle.
Name two perpendicular lines (not segments).
Given that $EF = GH,$ is $EG = FH$ true? Explain your answer completely.
Given that $\overleftrightarrow{BC} \| \overleftrightarrow{FG},$ is $BCGF$ a rectangle? Explain your answer informally. (Note: This is a new question. Do not assume that the given from a previous question is included in this question.)
Fill in the blanks:
$m\angle{ABF} = m\angle{ABE} + m\angle{\underline{\;\;\;\;\;}}$ . Why?

$m\angle{DCG} = m\angle{DCH} + m\angle{\underline{\;\;\;\;\;}}$ . Why?
Fill in the blanks:
$AB + \underline{\;\;\;\;\;} & = AC \ \underline{\;\;\;\;\;} + CD & = BD$
Given that $\angle{EBF}\cong\angle{HCG}$ , prove $\angle{ABF}\cong\angle{DCG}.$
Given that $AB = CD$ , prove: $AC = BD.$

What geometric objects does the real-world model suggest?

Model: two railroad tracks
Model: a floor and a ceiling
Model: two lines on a piece of graph paper
Model: referee’s arms when signaling a touchdown
Model: capital letter $L$
Model: the spine of a book where the front and back covers join

Review Answers

$\angle{BFG}$
$\overleftrightarrow{BF}$ and $\overleftrightarrow{EH}$
Yes
$EF & = GH\ \text{Given} \ EF + FG & = EF + FG\ \text{Reflexive} \ EF + FG & = GH + FG\ \text{Substitution} \ EF + FG & = EG , GH + FG = FH\ \text{Segment addition postulate} \ EG & = FH\ \text{Substitution}$
Yes. It’s given that $\overline{BC}\ \cong\ \overline{FG}$ (so $BC = FG$ ). Since $\overleftrightarrow{BC}\ \|\ \overleftrightarrow{FG}$ and $\overline{FG}\ \perp\ \overline{BF},$ then $\overline{BC}\ \perp\ \overline{BF}\ CG$ must be equal to $BF$ , and this would make $BCGF$ a rectangle.
$& EBF.\ & HCG.$
$& BC.\ & BC.$
$& \angle{EBF}\ \cong\ \angle{HCG}\ \text{Given} \\ &\angle{ABE}\ \cong\ \angle{DCH}\ \text{Given} \ &m\angle{ABE}\ = m\angle{ABE}\ +\ m\angle{EBF}\ \text{Angle Addition Postulate} \ &m\angle{DCG}\ = m\angle{DCH}\ +\ m\angle{HCG}\ \text{Angle Addition Postulate} \ &m\angle{ABF}\ = m\angle{DCH}\ +\ m\angle{HCG}\ \text{Substitution} \ &m\angle{ABF}\ = m\angle{DCG}\ \text{Substitution}\ \ &\angle{ABF}\ \cong\ \angle{DCG}\ \text{Definition of congruent angles}$
$AB &= CD\ \text{Given} \ AB + BC & = AB + BC\ \text{Reflexive} \ AB + BC & = CD + BC\ \text{Substitution} \\ AB + BC & = AC, CD + BC = BD\ \text{Segment addition postulate} \ AC & = BD\ \text{Substitution}$
Parallel lines
Parallel planes
Parallel or perpendicular lines
Parallel lines or segments
Perpendicular segments
Intersecting planes

Two-Column Proof

Learning Objectives

Draw a diagram to help set up a two-column proof.
Identify the given information and statement to be proved in a two-column proof.
Write a two-column proof.

Introduction

You have done some informal proofs in earlier sections. Now we raise the level of formality higher. In this section you will learn to write formal two-column proofs. You’ll need to draw a diagram, identify the given and prove, and write a logical chain of statements. Each statement will have a reason, such as a definition, postulate, or previously proven theorem, that justifies it.

Given, Prove, and Diagram

Example 1

Write a two-column proof for the following:

If $A$ , $B$ , $C$ , and $D$ are points on a line, in the given order, and $AB = CD$ , then $AC = BD$ .

Comments: The if part of the statement contains the given. The then part is the section that you must prove. A diagram should show the given facts.

We start with the given, prove, and a diagram.

Given: $A, B, C$ , and $D$ are points on a line in the order given. $AB = CD$ .
Prove: $AC = BD$ .

$4$ points on the line; $AB = CD$

Now it’s time to start with the given. Then we use logical reasoning to reach the statement we want to prove. Often (not always) the proof starts with the given information.

In the two column format, Statements go on the left side, and Reasons for each statement on the right. Reasons are generally definitions, postulates, and previously proved statements (called theorems).

Statement	Reason
1. $AB = CD$	Given
2. $A, B, C$ , and $D$ are collinear in that order	Given
3. $BC = BC$	Reflexive
4. $AC = AB + BC$ and $BD = CD + BC$	Segment Addition Postulate
5. $AB + BC = CD + BC$	Addition Property of Equality
6. $AC = BD$	Substitution

$AC = BD$ is what we were given to prove, and we’ve done it.

Example 2

Write a two-column proof of the following:

Given: $\overrightarrow{B F}$ bisects $\angle{ABC}; \angle{ABD} \cong \angle{CBE}$
Prove: $\angle{DBF} \cong \angle{EBF}$

Statement	Reason
1. $\overrightarrow{B F}$ bisects $\angle{ABC}$	Given
2. $m\angle{ABE}= m\angle{CBF}$	Definition of Angle Bisector
3. $m\angle{ABF}= m\angle{ABD} + m\angle{DBF}$	Angle Addition Postulate
4. $m\angle{CBF}= m\angle{CBE} + m\angle{EBF}$	Angle Addition Postulate
5. $m\angle{ABD} + m\angle{DBF} = m\angle{CBE} + m\angle{EBF}$	Substitution
6. $\angle{ABD} \cong \angle{CBE}$	Given
7. $m\angle{CBE}+ m\angle{DBF}= m\angle{CBE}+ m\angle{EBF}$	Substitution
8. $m\angle{DBF}= m\angle{EBF}$	Subtract $m\angle{CBE}$ from both sides (Reminder: Angle measures are all real numbers, so properties of equality apply.)
9. $\angle{DBF} \cong \angle{EBF}$	Definition of congruent angles

This is the end of the proof. The last statement is the requirement made in the proof above. This is the signal that the proof is completed.

Lesson Summary

In this section you have seen two examples illustrating the format of two-column proofs. The format of two-column proofs is the same regardless of the specific details. Geometry originated many centuries ago using this same kind of deductive reasoning proof.

Points to Consider

You will see and write many two-column proofs in future lessons. The framework will stay the same, but the details will be different. Some of the statements that we prove are important enough that they are identified by their names. You will learn about many theorems and use them in proofs and problem solving.

Review Questions

Use the diagram below to answer questions 1-10.

Which of the following can be assumed to be true from the diagram? Answer yes or no.

$\overline{AD}\ \cong\ \overline{BC}$
$\overline{AB}\ \cong\ \overline{CD}$
$\overline{CD}\ \cong\ \overline{BC}$
$\overline{AB}\ \| \ \overline{CD}$
$\overline{AB}\ \perp\ \overline{AD}$
$\overrightarrow{A C}$ bisects $\angle{DAB}$
$m\angle{CAB} = 45^\circ$
$m\angle{DCA} = 45^\circ$
$ABCD$ is a square
$ABCD$ is a rectangle

Use the diagram below to answer questions 11-14.

Given: $X$ bisects $\overline{WZ},\ Y$ is the midpoint of $\overline{XZ}$ , and $WZ=12$ .

How many segments have two of the given points as endpoints?
What is the value of each of the following?
$WY$
$XZ$
$ZW$
Write a two-column proof for the following:

Given: $\overrightarrow{A C}$ bisects $\angle{DAB}$

Prove: $m\angle{BAC} = 45$

Review Answers

No
No
Yes
No
Yes
No
No
No
No
No
$6$
$9$
$6$
$12$

Statement	Reason
1. $\overrightarrow{A C}$ bisects $\angle{DAB}$	Given
2. $m\angle{DAC} = m\angle{BAC}$	Definition of angle bisector
3. $m\angle{DAC} + m\angle{BAC} = m\angle{DAB}$	Angle Addition Postulate
4. $\overline{AD} \perp \overline{AB}$	Given
5. $m\angle{DAB} = 90$	Definition of perpendicular segments
6. $m\angle{BAC} + m\angle{BAC} = 90$	Substitution
7. $2 m\angle{BAC}= 90$	Algebra (Distributive Property)
8. $m\angle{BAC} = 45$	Multiplication Property of Equality

Segment and Angle Congruence Theorems

Learning Objectives

Understand basic congruence properties.
Prove theorems about congruence.

Introduction

In an earlier lesson you reviewed many of the basic properties of equality. Properties of equality are about numbers. Angles and segments are not numbers, but their measures are numbers. Congruence of angles and segments is defined in terms of these numbers. To prove congruence properties, we immediately turn congruence statements into number statements, and use the properties of equality.

Equality Properties

Reminder: Here are some of the basic properties of equality. These are postulates—no proof needed. For each of these there is a corresponding property of congruence for segments, and one for angles. These are theorems—we’ll prove them.

Properties of Equality for real numbers $x, y$ , and $z$ .

Reflexive $x = x$
Symmetric If $x = y$ then $y = x$
Transitive If $x = y$ and $y = z$ , then $x = z$

These properties are convertibles; we can convert them quickly and easily into congruence theorems.

Note that diagrams are needed to prove the congruence theorems. They are about angles and segments...ALL angles and segments, wherever and whenever they are found. No special setting (diagram) is needed.

Segment Congruence Properties

In this section we’ll prove a series of segment theorems.

Reflexive: $\overline{AB}\ \cong\ \overline{AB}$

Statement	Reason
1. $AB = AB$	Reflexive Property of Equality
2. $\overline{AB}\ \cong\ \overline{AB}$	Definition of congruent segments

Symmetric: If $\overline{AB}$ $\cong$ $\overline{CD}$ , then $\overline{CD}$ $\cong$ $\overline{AB}$

Given: $\overline{AB}\ \cong\ \overline{CD}$

Prove: $\overline{CD}\ \cong\ \overline{AB}$

Statement	Reason
1. $\overline{AB}\ \cong\ \overline{CD}$	Given
2. $AB = CD$	Definition of congruent segments
3. $CD = AB$	Symmetric Property of Equality
4. $\overline{CD}\ \cong\ \overline{AB}$	Definition of congruent segments

Transitive: If $\overline{AB}\ \cong\ \overline{CD}$ and $\overline{CD}\ \cong\ \overline{EF}$ , then $\overline{AB}\ \cong\ \overline{EF}$

Given: $\overline{AB}\ \cong\ \overline{CD}; \overline{CD}\ \cong\ \overline{EF}$

Prove: $\overline{AB}\ \cong\ \overline{EF}$

Statement	Reason
1. $\overline{AB}\ \cong\ \overline{CD}; \overline{CD} \cong\ \overline{EF}$	Given
2. $AB = CD; CD = EF$	Definition of congruent segments
3. $AB = EF$	Transitive property of equality
4. $\overline{AB}\ \cong\ \overline{EF}$	Definition of congruent segments

Angle Congruence Properties

Watch for proofs of the Angle Congruence Properties in the Lesson Exercises.

Reflexive: $\angle {A}\ \cong\ \angle {A}$

Symmetric: If $\angle {A}\ \cong\ \angle {A}$ , then $\angle {B}\ \cong\ \angle {A}$

Transitive: If $\angle{A}\cong\angle {B}$ and $\angle {B}\ \cong\ \angle {C}$ , then $\angle {A}$ $\cong\ \angle {C}$

Lesson Summary

In this lesson we looked at old information in a new light. We saw that the properties of equality—reflexive, symmetric, transitive—convert easily into theorems about congruent segments and angles. In the next section we’ll move ahead into new ground. There we’ll get to use all the tools in our geometry toolbox to solve problems and to create new theorems.

Points to Consider

We are about to transition from introductory concepts that are necessary but not too “geometric” to the real heart of geometry. We needed a certain amount of foundation material before we could begin to get into more unfamiliar, challenging concepts and relationships. We have the definitions and postulates, and analogs of the equality properties, as the foundation. From here on out, we will be able to experience geometry on a richer and deeper level.

Review Questions

Prove the Segment Congruence Properties, in questions 1-3.

Reflexive: $\angle {A}\cong\angle {A}.$
Symmetric: If $\angle {A}\cong\angle {B}$ , then $\angle {B}\cong\angle {A}.$
Transitive: If $\angle {A}\cong\angle {B}$ and $\angle {B}\cong$ $\angle {C}$ , then $\angle {A}\cong\angle {C}.$
Is the following statement true? If it’s not, give a counterexample. If it is, prove it.

If $\angle{A}\cong\angle{B}$ and $\angle{C}\cong\angle{D}$ , then $m\angle{A} + m\angle{C}= m\angle{B} + m\angle{D}.$

Give a reason for each statement in the proof below.

If $A, B, C,$ and $D$ are collinear, and $\overline{AB}\cong\overline{CD}$ , then $\overline{AC}\cong\overline{BD}$ .

Given: $A, B, C$ , and $D$ are collinear, and $\overline{AB}\cong\overline{CD}.$

Prove: $\overline{AC}\cong\overline{BD}.$

Is the following statement true? Explain your answer. (A formal two-column proof is not required.)

Let $P, Q, R, S,$ and $T$ be points in a single plane. If $\overrightarrow{Q S}$ is in the interior of $\angle{PQR}$ , and $\overrightarrow{Q T}$ is in the interior of $\angle{PQS}$ , then $\overrightarrow{Q T}$ is in the interior of $\angle{PQR}$ .

Note that this is a bit like a Transitive Property for a ray in the interior of an angle.

Review Answers

Statement	Reason
A. $m\angle{A}= m\angle{A}$	Reflexive Property of Equality
B. $\angle{A}\cong\angle{A}$	Definition of Congruent Angles

Given:
$\angle{A}\cong\angle{B}$

Prove:

$\angle{B}\cong\angle{A}$

Statement	Reason
A. $\angle{A}\cong\angle{B}$	Given
B. $m\angle{A}= m\angle{B}$	Definition of Congruent Angles
C. $m\angle{B}= m\angle{A}$	Symmetric Property of Equality
D. $\angle{B}= \angle{A}$	Definition of Congruent Angles

Given:
$\angle{A} \cong \angle{B}$ and $\angle{B} \cong \angle{C}$

Prove:

$\angle{A} \cong \angle{C}$

Statement	Reason
A. $\angle{A} \cong \angle{B}$ and $\angle{B} \cong \angle{C}$	Given
B. $m\angle{A}= m\angle{B}$ and $m\angle{B}= m\angle{C}$	Definition of Congruent Angles
C. $m\angle{A}= m\angle{C}$	Transitive Property of Equality
D. $\angle{A} \cong \angle{C}$	Definition of Congruent Angles

Yes
Given:

$\angle{A} \cong \angle{B}$ and $\angle{C} \cong \angle{D}$

Prove:

$m\angle{A} + m\angle{C} = m\angle{B}+ m\angle{D}$

Statement	Reason
A. $\angle{A} \cong \angle{B}$ and $\angle{C} \cong \angle{D}$	Given
B. $m\angle{A} = m\angle{B}, m\angle{C}= m\angle{D}$	Definition of Congruent Angles
C. $m\angle{A} + m\angle{C} = m\angle{B}+ m\angle{C}$	Addition Property of Equality
D. $m\angle{A} + m\angle{C} = m\angle{B}+ m\angle{D}$	Substitution

Statement	Reason
$A, B, C$ , and $D$ are collinear	A._____ Given
$\overline{AB}\ \cong\ \overline{CD}$	B._____ Given
$AB = CD$	C._____ Definition of Congruent Segments
$AB + BC = CD + BC$	D._____ Addition Property of Equality
$AB + BC = BC + CD$	E._____ Commutative Property of Equality
$AB + BC = AC$	F._____ Definition of Collinear Points
$BC + CD = BD$	G._____ Definition of Collinear Points
$AC = BD$	H._____ Substitution Property of Equality
$\overline{AC}\ \cong\ \overline{BD}$	I._____ Definition of Congruent Segments

True. Since $\overrightarrow{Q S}$ is in the interior of $\angle{PQR},\ m\angle{PQS} + m\angle{SQR}\ = m\angle{PQR}$ Since $\overrightarrow{Q T}$ is in the interior of $\angle{PQS}$ , then $m\angle{PQT}\ + m\angle{TQS}\ = m\angle{PQS}$ . So
$(m\angle{PQT}\ + m\angle{TQS}) + m\angle{SQR}\ & = m\angle{PQR} \ m\angle{PQT} + (m\angle{TQS}\ + m\angle{SQR}) & = m\angle{PQR} \ m\angle{PQT}\ + m\angle{TQR}\ & = m\angle{PQR}$

$\overrightarrow{Q T}$ is in the interior of $\angle{PQR}$ by the angle addition property.

Proofs About Angle Pairs

Learning Objectives

State theorems about special pairs of angles.
Understand proofs of the theorems about special pairs of angles.
Apply the theorems in problem solving.

Introduction

So far most of the things we have proven have been fairly straightforward. Now we have the tools to prove some more in-depth theorems that may not be so obvious. We’ll start with theorems about special pairs of angles. They are:

right angles
supplementary angles
complementary angles
vertical angles

Right Angle Theorem

If two angles are right angles, then the angles are congruent.

Given: $\angle{A}$ and $\angle{B}$ are right angles.

Prove: $\angle{A}\ \cong\ \angle{B}$

Statement	Reason
1. $\angle{A}$ and $\angle{B}$ are right angles.	Given
2. $m\angle{A} = 90,\ m\angle{B} = 90$	Definition of right angle
3. $m\angle{A} = m\angle{B}$	Substitution
4. $\angle{A}\ \cong\ \angle{B}$	Definition of congruent angles

Supplements of the Same Angle Theorem

If two angles are both supplementary to the same angle (or congruent angles) then the angles are congruent.

Comments: As an example, we know that if $\angle{A}$ is supplementary to a $30^\circ$ angle, then $m\angle{A}$ $= 150^\circ$ . If $\angle{B}$ is also supplementary to a $30^\circ$ angle, then $m\angle{B}=150^\circ$ too, and $m\angle{A}$ $= m\angle{B}.$

Given: $\angle{A}$ and $\angle{B}$ are supplementary angles. $\angle{A}$ and $\angle{C}$ are supplementary angles.

Prove: $\angle{B}$ $\cong$ $\angle{C}$

Statement	Reason
1. $\angle{A}$ and $\angle{B}$ are supplementary angles.	Given
2. $\angle{A}$ and $\angle{C}$ are supplementary angles.	Given
3. $m\angle{A} + m\angle{B} = 180,\ m\angle{A}+ m\angle{C} = 180$	Definition of Supplementary Angles
4. $m\angle{A}+ m\angle{B}= m\angle{A}+ m\angle{C}$	Substitution
5. $m\angle{B}= m\angle{C}$	Addition Property of Equality
6. $\angle{B}\cong\angle{C}$	Definition of Congruent Angles

Example 1

Given that $\angle{1} \cong \angle{4}$ , what other angles must be congruent?

Answer:

$\angle{C}\ \cong\ \angle{F}$ by the Right Angle Theorem, because they’re both right angles.

$\angle{2} \cong \angle{3}$ by the Supplements of the Same Angle Theorem and the Linear Pair Postulate: $\angle{1}$ and $\angle{2}$ are a linear pair, which makes them supplementary. $\angle{3}$ and $\angle{4}$ are also a linear pair, which makes them supplementary too. Then by Supplements of the Same Angle Theorem, $\angle{2} \cong \angle{3}$ because they’re supplementary to congruent angles $\angle{1}$ and $\angle{4}$ .

Complements of the Same Angle Theorem

If two angles are both complementary to the same angle (or congruent angles) then the angles are congruent.

Comments: Only one word is different in this theorem compared to the Supplements of the Same Angle Theorem. Here we have angles that are complementary, rather than supplementary, to the same angle.

The proof of the Complements of the Same Angle Theorem is in the Lesson Exercises, and it is very similar to the proof above.

Vertical Angles Theorem

Vertical Angles Theorem: Vertical angles are congruent.

Vertical angles are common in geometry problems, and in real life wherever lines intersect: cables, fence lines, highways, roof beams, etc. A theorem about them will be useful. The vertical angle theorem is one of the world’s briefest theorems. Its proof draws on the new theorems just proved earlier in this section.

Given: Lines $k$ and $m$ intersect.

Prove: $\angle{1}\cong\angle{3}$ , and $\angle{2} \cong \angle{4}$

Statement	Reason
1. Lines $k$ and $m$ intersect.	Given
2. $\angle{1}$ and $\angle{2}$ , $\angle{2}$ and $\angle{3}$ are linear pairs.	Definition of Linear Pairs
3. $\angle{1}$ and $\angle{2}$ are supplementary, and $\angle{2}$ and $\angle{3}$ are supplementary.	Linear Pair Postulate
4. $\angle{1} \cong \angle{3}$	Supplements of the Same Angle Theorem

This shows that $\angle{1} \cong \angle{3}$ . The same proof can be used to show that $\angle{2} \cong \angle{4}$ .

Example 2

Given: $\angle{2} \cong \angle{3},\ k \bot p$

Each of the following pairs of angles are congruent. Give a reason.

$\angle{1}$ and $\angle{5}$ answer: Vertical Angles Theorem

$\angle{1}$ and $\angle{4}$ answer: Complements of Congruent Angles Theorem

$\angle{2}$ and $\angle{6}$ answer: Vertical Angles Theorem

$\angle{3}$ and $\angle{7}$ answer: Vertical Angles Theorem

$\angle{6}$ and $\angle{7}$ answer: Vertical Angles Theorem and Transitive Property

$\angle{3}$ and $\angle{6}$ answer: Vertical Angles Theorem and Transitive Property

$\angle{4}$ and $\angle{5}$ answer: Complements of Congruent Angles Theorem

Example 3

Given: $\angle{1} \cong \angle{2},\ \angle{3} \cong \angle{4}$
Prove: $\angle{1} \cong \angle{4}$

Statement	Reason
1. $\angle{1}\ \cong\ \angle{2},\ \angle{3}\ \cong\ \angle{4}$	Given
2. $\angle{2} \cong \angle{3}$	Vertical Angles Theorem
3. $\angle{1} \cong \angle{4}$	Transitive Property of Congruence

Lesson Summary

In this lesson we proved theorems about angle pairs.

Right angles are congruent.
Supplements of the same, or congruent, angles are congruent.
Complements of the same, or congruent, angles are congruent.
Vertical angles are congruent.

We saw how these theorems can be applied in simple or complex figures.

Points to Consider

No matter how complicated or abstract the model of a real-world situation may seem, in the final analysis it can often be expressed in terms of simple lines, segments, and angles. We’ll be able to use the theorems of this section when we encounter complicated relationships in future figures.

Review Questions

Use the diagram to answer questions 1-3.

Given: $m\angle{1} = 60^\circ$

$m\angle{1}=m\angle{3}=60^\circ$

Fill in the blanks.

$m\angle{2}=$ _________
$m\angle{3}=$ _________
$m\angle{4}=$ _________
Fill in the reasons in the following proof.
Given: $\overline{AE} \bot \overline{EC}$ and $\overline{BE} \bot \overline{ED}$

Prove: $\angle{1} \cong \angle{3}$

Statement	Reason
$\overline{AE} \bot \overline{EC}$ and $\overline{BE} \bot \overline{ED}$	a.____
$\angle{A E C}$ and $\angle{B E D}$ are right angles	b.____
$m\angle{A E C} = m\angle{1} + m\angle{2}$ and $m\angle{B E D} = m\angle{2} + m\angle{3}$	c.____
$m\angle{A E C} = m\angle{B E D} = 90$	d.____
$m\angle{1} + m\angle{2} = m\angle{2} + m\angle{3} = 90^\circ$	e.____
$\angle{1}$ and $\angle{2}$ are complementary , $\angle{2}$ and $\angle{3}$ are complementary	f.____
$\angle{1}\ \cong\ \angle{3}$	g.___

Which of the following statements must be true? Answer Yes or No.
1. $\angle{1} \cong \angle{2}$
2. $\angle{2} \cong \angle{4}$
3. $\angle{5} \cong \angle{6}$
The following diagram shows a ray of light that is reflected from a mirror. The dashed segment is perpendicular to the mirror. $\angle{2}\ \cong\ \angle{3}$ .

$\angle{1}$ is called the angle of incidence; $\angle{4}$ is called the angle of reflection. Explain how you know that the angle of incidence is congruent to the angle of reflection.

Review Answers

$120^\circ$
$60^\circ$
$120^\circ$
1. Given
2. Definition of Perpendicular Segments
3. Angle Addition Postulate
4. Definition of Right Angle
5. Substitution (Transitive Property of Equality)
6. Definition of Complementary Angles
7. Complements of the Same Angle are Congruent
1. No
2. Yes
3. No
$\angle{1}$ and $\angle{2}$ are complementary; $\angle{3}$ and $\angle{4}$ are complementary . $\angle{1}$ $\cong$ $\angle{4}$ because they are complements of congruent angles $2$ and $3$ .