Mastering Geometric Proofs: Theorems, Conjectures, and Logical Reasoning

	$n$	Number of Cubes
Figure $1$	$0$	$1 + 3 \cdot 0 = 1$
Figure $2$	$1$	$1 + 3 \cdot 1 = 4$
Figure $3$	$2$	$1 + 3 \cdot 2 = 7$
Figure $4$	$3$	$1 + 3 \cdot 3 = 10$
Figure $121$	$120$	$1 + 3 \cdot 120 = 361$

n

Number of Cubes

Figure

1

0

1 + 3 \cdot 0 = 1

Figure

2

1

1 + 3 \cdot 1 = 4

Figure

3

2

1 + 3 \cdot 2 = 7

Figure

4

3

1 + 3 \cdot 3 = 10

Figure

121

120

1 + 3 \cdot 120 = 361

Card	Green Triangles
$1$	$1$
$2$	$3$
$3$	$6$
$4$	$10$

Card

Green Triangles

1

1

2

3

3

6

4

10

Card	Green Triangles
$1$	$1$
$2$	$3$
$3$	$6$
$4$	$10$

Card

Green Triangles

1

1

2

3

3

6

4

10

Card	Green Triangles
$1$	$1 = 1$
$2$	$1 + 2 = 3$
$3$	$1 + 2 + 3 = 6$
$4$	$1 + 2 + 3 + 4 = 10$

Card

Green Triangles

1

1 = 1

2

1 + 2 = 3

3

1 + 2 + 3 = 6

4

1 + 2 + 3 + 4 = 10

	Definition
Inductive Reasoning	The process of finding patterns in specific observations and writing a conclusion
Deductive Reasoning	The process of reaching logical conclusions from given statements

Definition

Inductive Reasoning

The process of finding patterns in specific observations and writing a conclusion

Deductive Reasoning

The process of reaching logical conclusions from given statements

Odd Numbers	Sum
$- 21$	$- 13$	$- 21 + (- 13) = - 34$
$- 7$	$3$	$- 7 + 3 = 4$
$5$	$- 5$	$5 + (- 5) = 0$
$1$	$5$	$1 + 5 = 6$
$11$	$51$	$11 + 51 = 62$
$33$	$45$	$33 + 45 = 78$

Odd Numbers

Sum

- 21

- 13

- 21 + (- 13) = - 34

- 7

3

- 7 + 3 = 4

5

- 5

5 + (- 5) = 0

1

5

1 + 5 = 6

11

51

11 + 51 = 62

33

45

33 + 45 = 78

Sum of Any Three Consecutive Integers
Observation I	Observation II	Observation III
$2 + 3 + 4 = 9 = 3 \cdot 3$	$7 + 8 + 9 = 24 = 3 \cdot 8$	$13 + 14 + 15 = 42 = 3 \cdot 14$
Conjecture: The sum of any three consecutive integers is three times the second number.

Sum of Any Three Consecutive Integers

Observation I

Observation II

Observation III

2 + 3 + 4 = 9 = 3 \cdot 3

7 + 8 + 9 = 24 = 3 \cdot 8

13 + 14 + 15 = 42 = 3 \cdot 14

Conjecture: The sum of any three consecutive integers is three times the second number.

Goldbach's Conjecture
Every even whole number greater than $2$ is the sum of two prime numbers.

Proof Methods
Visual Styles	Proof Types
Paragraph Proof Two-Column Proof	Direct Proof

Proof Methods

Visual Styles

Proof Types

Paragraph Proof

Two-Column Proof

Direct Proof

If $n$ is an odd number, then $n^{2}$ is also an odd number.

$n$	$n^{2}$	Is $n^{2}$ odd?
$1$	$1$	Yes
$3$	$9$	Yes
$5$	$25$	Yes
$7$	$49$	Yes

n

n^{2}

n^{2}

odd?

1

1

Yes

3

9

Yes

5

25

Yes

7

49

Yes

Statement	Reason
$n$ is odd	Given.
$n = 2 k + 1$	Every odd number is equal to twice an integer plus $1 .$
$n^{2} = (2 k + 1)^{2}$	Raise the equation to the power of $2 .$
$n^{2} = 4 k^{2} + 4 k + 1$	Expand the square.
$n^{2} = 2 (2 k^{2} + 2 k) + 1$	Factor out $2 .$
$n^{2}$ is odd	It is written as twice an integer plus $1 .$

Statement

Reason

n

is odd

Given.

n = 2 k + 1

Every odd number is equal to twice an integer plus

1 .

n^{2} = (2 k + 1)^{2}

Raise the equation to the power of

2 .

n^{2} = 4 k^{2} + 4 k + 1

Expand the square.

n^{2} = 2 (2 k^{2} + 2 k) + 1

Factor out

2 .

n^{2}

is odd

It is written as twice an integer plus

1 .

Pythagorean Theorem
If $a triangle has a right angle,$ then $the$ $hypotenuse$ $squared$ $equals$ $the$ $sum$ $of$ $the$ $squares$ $of$ $the$ $legs .$

Let $C$ be a point on $\overline{A E},$ $B$ be a point on $\overline{A C},$ and $D$ be a point on $\overline{C E},$ such that $\overline{A B} ≅ \overline{D E}$ and $\overline{B C} ≅ \overline{C D} .$ Prove that $\overline{A C} ≅ \overline{C E} .$

Given	Prove
$\overline{A B} ∥ \overline{C E}$	Angles $α$ and $β$ are supplementary $⇕$ $α + β = 18 0^{\circ}$

Given

Prove

\overline{A B} ∥ \overline{C E}

Angles

α

and

β

are supplementary

⇕

α + β = 18 0^{\circ}

If $M$ is the midpoint of $\overline{A B},$ then $A B = 2 A M .$

$2$ is an irrational number.

	Description
Indirect Proof or Proof by Contradiction	A claim is proven by showing how the opposite conclusion of the claim creates a contradiction.

Description

Indirect Proof or Proof by Contradiction

A claim is proven by showing how the opposite conclusion of the claim creates a contradiction.

	Description
Proof by Contrapositive	A claim is proven by stating and proving the contrapositive of the claim.
Proof by Mathematical Induction	One special case is shown to be true. Next, it is shown that if the statement is true for any special case, then it is also true for some other special case.
Coordinate Proof	It is used to prove geometric statements by placing geometric figures in a coordinate plane and assigning variables to the coordinates of points.

Description

Proof by Contrapositive

A claim is proven by stating and proving the contrapositive of the claim.

Proof by Mathematical Induction

One special case is shown to be true. Next, it is shown that if the statement is true for any special case, then it is also true for some other special case.

Coordinate Proof

It is used to prove geometric statements by placing geometric figures in a coordinate plane and assigning variables to the coordinates of points.

Complete the following paragraph proof by pairing each missing part with the appropriate expression or phrase.

Given : Prove : A B = 2 B C M is the midpoint of \overline{A B} \overline{A M} ≅ \overline{B C}

Proof: Since

M

is the midpoint of

\overline{A B},

it divides

\overline{A B}

into two equal parts. That is,

A M = M B .

Applying the Segment Addition Postulate,

\underline{(I)} = 2 A M .

Next, by the

\underline{(II)},

\underline{(III)} = 2 B C .

Simplifying the equation gives

A M = B C .

Consequently,

\overline{A M} ≅ \overline{B C}

by the

\underline{(IV)} .

Let's write the proof ourselves so we can determine the missing parts of the given proof. Let's start by making a diagram that illustrates the given information. We have to consider a segment AB, its midpoint M, and a point C such that 2BC is equal to AB.

Since M is the midpoint, it divides AB into two equal parts — that is, AM=MB. Then, the length of AB is equal to the sum of AM and MB by the Segment Addition Postulate.

From this last equation, we can say that the first missing part is AB. (I) → AB Remember that it was given that AB=2BC. Therefore, we can substitute 2AM for AB into the given equation. 2AM = 2BC This equation is true thanks to the Substitution Property of Equality. We have just determined the second and third missing parts! (II) & → Substitution Property of Equality (III) & → 2AM Let's continue the proof even though we know what the fourth missing part is. Simplifying the equation 2AM=2BC gives that AM=BC. That is, AM has the same length as BC. Therefore, by the definition of congruent segments, AM≅BC. This gives us the fourth missing part. (IV) → definition of congruent segments

Proof: Since M is the midpoint of AB, it divides AB into two equal parts. That is, AM=MB. Applying the Segment Addition Postulate, AB=2AM. Next, by the Substitution Property of Equality, 2AM=2BC. Simplifying the equation gives AM=BC. Consequently, AM≅BC by the definition of congruent segments.

In the following diagram, $\overline{A E} ∥ \overline{F G},$ $\overline{B C} ∥ \overline{D H},$ and $m ∠ G F H = 6 0^{\circ} .$

The following two-column table summarizes the proof that the measure of $∠ A B C$ is $12 0^{\circ} .$

Statements	Reasons
$\overline{A E} ∥ \overline{F G},$ $\overline{B C} ∥ \overline{D H},$ $m ∠ G F H = 6 0^{\circ}$	a.
$∠ E D F ≅ ∠ G F H$	Two parallel lines and a transversal generate congruent corresponding angles
b.	Definition of congruent angles
$m ∠ E D F = 6 0^{\circ}$	c.
$∠ A B C$ and $∠ E D F$ are supplementary angles	Two parallel lines and a transversal generate supplementary same-side exterior angles
d.	Definition of supplementary angles
e.	Substitution Property of Equality
$m ∠ A B C = 12 0^{\circ}$	f.

Complete the proof by pairing each missing part with the appropriate expression or phrase.

Let's write the proof ourselves so we can determine the missing parts. Consider the given diagram again.

When we want to write a two-column proof, the first step is writing the given information. We are told that AE ∥ FG, BC∥ DH, and m∠ GFH = 60^(∘), so let's fill in the first row of the table.

Statements	Reasons
AE ∥ FG, BC∥ DH, m∠ GFH = 60^(∘)	a. Given

Since DH is a transversal that cuts a pair of parallel lines, the corresponding angles are congruent. This means that ∠ EDF and ∠ GFH are congruent.

Let's write this in the second row.

Statements	Reasons
AE ∥ FG, BC∥ DH, m∠ GFH = 60^(∘)	a. Given
∠ EDF ≅ ∠ GFH	Two parallel lines and a transversal generate congruent corresponding angles

By definition, congruent angles have the same measure. This means that m∠ EDF=m∠ GFH. Since we know that m∠ GFH=60^(∘), we can substitute 60^(∘) into the equation. From this, we can write two more rows of the proof.

Statements	Reasons
AE ∥ FG, BC∥ DH, m∠ GFH = 60^(∘)	a. Given
∠ EDF ≅ ∠ GFH	Two parallel lines and a transversal generate congruent corresponding angles
b. m∠ EDF= m∠ GFH	Definition of congruent angles
m∠ EDF = 60^(∘)	c. Substitution Property of Equality

We also have that AE is a transversal that cuts another pair of parallel lines.

Notice that ∠ ABC and ∠ EDF are same-side exterior angles. Since the same-side exterior angles are supplementary, ∠ ABC and ∠ EDF are also supplementary angles. As such, the sum of their measures is 180^(∘). We can add another couple of rows to the proof!

Statements	Reasons
AE ∥ FG, BC∥ DH, m∠ GFH = 60^(∘)	a. Given
∠ EDF ≅ ∠ GFH	Two parallel lines and a transversal generate congruent corresponding angles
b. m∠ EDF= m∠ GFH	Definition of congruent angles
m∠ EDF = 60^(∘)	c. Substitution Property of Equality
∠ ABC and ∠ EDF are supplementary angles	Two parallel lines and a transversal generate supplementary same-side exterior angles
d. m∠ ABC + m∠ EDF = 180^(∘)	Definition of supplementary angles

Next, we can substitute the equation written in the fourth row into this last equation. After that, we can subtract 60^(∘) from both sides to finally get the measure of ∠ ABC.

Statements	Reasons
AE ∥ FG, BC∥ DH, m∠ GFH = 60^(∘)	a. Given
∠ EDF ≅ ∠ GFH	Two parallel lines and a transversal generate congruent corresponding angles
b. m∠ EDF= m∠ GFH	Definition of congruent angles
m∠ EDF = 60^(∘)	c. Substitution Property of Equality
∠ ABC and ∠ EDF are supplementary angles	Two parallel lines and a transversal generate supplementary same-side exterior angles
d. m∠ ABC + m∠ EDF = 180^(∘)	Definition of supplementary angles
e. m∠ ABC + 60^(∘) = 180^(∘)	Substitution Property of Equality
m∠ ABC = 120^(∘)	f. Subtraction Property of Equality

Now that we have completed the proof, we can identify the missing parts. a.& → Given b.& → m∠ EDF = m∠ GFH c.& → Substitution Property of Equality d.& → m∠ ABC + m∠ EDF = 180^(∘) e.& → m∠ ABC + 60^(∘) = 180^(∘) f.& → Subtraction Property of Equality

In the following diagram, $\overline{A B}$ and $\overline{D E}$ are parallel, $m ∠ A = 5 0^{\circ},$ and $m ∠ B = 4 0^{\circ} .$

Triangle ABC and segment DE, passing through C, parallel to segment AB.

The following flowchart shows the proof that $∠ A C B$ is a right angle.

Consider the following four statements.

I . II . III . IV . m ∠ D C A = m ∠ A m ∠ E C B = m ∠ B Parallel lines and a transversal generate alternate interior angles that are congruent 5 0^{\circ} + m ∠ A C B + 4 0^{\circ} = 18 0^{\circ} Given

Complete the proof by pairing each missing part with the appropriate statement.

Let's follow the flow of the proof and find the missing parts. We will start with the top-left box, which contains the given information. Now, let's take a look at the box that the first box is pointing to. That box contains two pairs of congruent angles. ∠ DCA ≅ ∠ A ∠ ECB ≅ ∠ B a. Let's mark these four angles in the diagram. This may help us to determine the reason that justifies the congruence statements.

If we take AC as a transversal, we can see that ∠ DCA and ∠ A are alternate interior angles. Similarly, if we take BC as a transversal, ∠ ECB and ∠ B are alternate interior angles. In both cases, the alternate interior angles are congruent because AB∥DE. This is the reason that goes in the second box, which corresponds to statement II.

Blank	Statement
a.	II. Parallel lines and a transversal generate alternate interior angles that are congruent

Let's continue from where we left off. This time a statement is missing. b. Def. of Congruent Angles At this point, we know that the previous statement and the definition of congruent angles are used to derive this missing statement. The previous box contained statements about congruent angles, so let's apply the definition of congruence to those congruent angles. ∠ DCA ≅ ∠ A ∠ ECB ≅ ∠ B ⇓ m∠ DCA = m∠ A m∠ ECB = m∠ B The equations inside this last box fill in the second blank. These equations correspond to statement I.

Blank	Statement
a.	II. Parallel lines and a transversal generate alternate interior angles that are congruent
b.	I. m∠ DCA = m∠ A m∠ ECB = m∠ B

Let's move on to the third blank. m∠ A = 50^(∘) m∠ B = 40^(∘) c. This box has two main characteristics. It has no incoming arrow and the angle measures in it can be found in the diagram. This indicates that this box contains given information. Consequently, the third blank corresponds to statement IV.

Blank	Statement
a.	II. Parallel lines and a transversal generate alternate interior angles that are congruent
b.	I. m∠ DCA = m∠ A m∠ ECB = m∠ B
c.	IV. Given

Finally, the fourth blank corresponds to the equation obtained by substituting m∠ DCA = 50^(∘) and m∠ ECB = 40^(∘) into the equation m∠ DCA + m∠ ACB + m∠ ECB = 180^(∘). We know this from the arrows pointing to the box.

The resulting equation fills in the fourth blank. This equation corresponds to statement III.

Blank	Statement
a.	II. Parallel lines and a transversal generate alternate interior angles that are congruent
b.	I. m∠ DCA = m∠ A m∠ ECB = m∠ B
c.	IV. Given
d.	III. 50^(∘) + m∠ ACB + 40^(∘) = 180^(∘)

We have completed the flowchart proof!

Consider the following four figures.

Which of the following options corresponds to the fifth figure of the pattern?

Let's study the given pattern of figures and try to state a conjecture about the fifth figure. In other words, let's use inductive reasoning. We can see that each figure has a different number of slices, so let's start by counting the number of slices in each one.

The first figure has 2 slices, the second figure has 3, the third has 4, and so on. The number of slices increases by one from one figure to the next. From this, we can say that the fifth figure will have 6 slices. Since option A has 7 slices and option D has 8 slices, we can discard them.

Options B and D both have 6 slices. We need to identify another characteristic in the given pattern so we know which one is the fifth figure. Taking another look at the given pattern, we can see that the slices in each figure are the same size. In light of this, we can conclude that option C is the fifth figure in the pattern.

Statements	Reasons
$∠ 1$ and $∠ 2$ are vertical angles	Given
$∠ 1$ and $∠ 3$ form a linear pair $∠ 2$ and $∠ 3$ form a linear pair	From the diagram
$m ∠ 1 + m ∠ 3$ $= 18 0^{\circ}$ $m ∠ 2 + m ∠ 3$ $= 18 0^{\circ}$	Definition of linear pair
$m ∠ 1 + m ∠ 3 =$ $m ∠ 2 + m ∠ 3$	Substitution Property of Equality
$m ∠ 1 = m ∠ 2$	Subtraction Property of Equality

Geometric Proof

Catch-Up and Review

Which Is More Reliable?

Looking for Patterns

Inductive Reasoning

The Thief's Pattern

Hint

Solution

Logical Reasoning

Deductive Reasoning

An Odd Riddle

Answer

Hint

Solution

Using Inductive Reasoning

Using Deductive Reasoning

Conclusion

Unproven Statements

Conjecture

Extra

Justifying a Conjecture Logically

Proof

Direct Proof

Proven Statements

Theorem

Writing a Proof in a Paragraph

Paragraph Proof

The Detective's Hit

Answer

Hint

Solution

Writing a Proof in a Table

Two-Column Proof

Police Chase

Answer

Hint

Solution

Other Types of Proofs

Geometric Proof

Recommended exercises

	14 Theory slides
	11 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Statements	Reasons
$M$ is the midpoint of $\overline{A B}$	Given