Chapter 2: Reasoning and Proof

High School Geometry · MathHub · 2026

Learning Objectives

2.1 Inductive Reasoning

Inductive reasoning uses patterns in specific cases to form a general conclusion called a conjecture. It is the basis of mathematical discovery — but inductive reasoning can lead to false conclusions.

Inductive Reasoning and Conjectures

Inductive reasoning: Observing a pattern and drawing a general conclusion.

Conjecture: A conclusion based on observed patterns. Not yet proven — it may be false.

Counterexample: A single specific case that proves a conjecture false. One counterexample is enough to disprove any conjecture.

Example 2.1 — Forming and Disproving a Conjecture

Pattern: 1, 4, 9, 16, 25, … (perfect squares)

Conjecture: "The difference between consecutive perfect squares is always odd."

4−1=3 ✓, 9−4=5 ✓, 16−9=7 ✓, 25−16=9 ✓ → Pattern holds so far.

Is the conjecture true? In general, $(n+1)^2 - n^2 = 2n+1$, which is always odd. This conjecture happens to be true — but inductive reasoning alone doesn't prove it; we need deductive reasoning.

A false conjecture example: "The expression $n^2 - n + 41$ is always prime." It works for $n=1$ through $n=40$, but at $n=41$: $41^2 - 41 + 41 = 41^2 = 1681$, which is not prime. Counterexample found.

TRY IT

A student conjectures: "The product of any two odd numbers is odd." Find three examples that support this, then determine if it's actually true or find a counterexample.

Show Answer
Examples: 3×5=15 (odd), 7×9=63 (odd), 11×3=33 (odd)
Truth: An odd number has the form (2a+1). Product: (2a+1)(2b+1) = 4ab+2a+2b+1 = 2(2ab+a+b)+1, which is always odd. Conjecture is TRUE — but we needed deductive reasoning to confirm it.

2.2 Conditional Statements

A conditional statement (also called an if-then statement) has the form: If p, then q, written $p \to q$. The part after "if" is the hypothesis (p); the part after "then" is the conclusion (q).

Related Conditionals

Key fact: A conditional and its contrapositive always have the same truth value. The converse and inverse always have the same truth value (but may differ from the original).

Example 2.2 — Writing All Four Related Conditionals

Original: "If it is raining, then the ground is wet." (True)

StatementFormTruth
If raining, then wet$p \to q$True
If wet, then raining (Converse)$q \to p$False*
If not raining, then not wet (Inverse)$\neg p \to \neg q$False*
If not wet, then not raining (Contrapositive)$\neg q \to \neg p$True

*A sprinkler could wet the ground without rain — counterexample disproves the converse.

Biconditional Statements

A biconditional statement combines a conditional and its converse: "p if and only if q" (written $p \leftrightarrow q$). It is true only when both $p$ and $q$ have the same truth value.

A biconditional is valid (true) when both the conditional and its converse are true. Definitions in geometry are always biconditionals.

Example 2.3 — Biconditional Statements

"An angle is a right angle if and only if its measure is 90°."

This is valid because: (1) If an angle is right, its measure is 90° ✓, and (2) If its measure is 90°, it is a right angle ✓.

Invalid biconditional attempt: "A figure is a square if and only if it has four sides." The converse fails — a rectangle has four sides but is not necessarily a square. So this biconditional is false.

Conditional p→q illustrated: when p is true, q must also be true

Figure 2.1 — Venn Diagram: Conditional Statement p → q

2.3 Deductive Reasoning

Deductive reasoning uses accepted facts, definitions, postulates, and theorems to reach conclusions that must be true. It is the foundation of mathematical proof.

Laws of Deductive Reasoning

Law of Detachment: If $p \to q$ is true and $p$ is true, then $q$ must be true.

Example: "If it rains, the game is cancelled." It is raining. Therefore, the game is cancelled.


Law of Syllogism: If $p \to q$ and $q \to r$ are both true, then $p \to r$ is true.

Example: "If I study, I pass. If I pass, I graduate." Therefore: "If I study, I graduate."

Example 2.4 — Law of Syllogism

Statement 1: If a number is divisible by 6, then it is divisible by 3.

Statement 2: If a number is divisible by 3, then the sum of its digits is divisible by 3.

Conclusion (Law of Syllogism): If a number is divisible by 6, then the sum of its digits is divisible by 3.

TRY IT

Given: (1) If $\angle A$ and $\angle B$ are supplementary, then $m\angle A + m\angle B = 180°$. (2) $\angle A$ and $\angle B$ are supplementary. What can you conclude? What law did you use?

Show Answer
Conclusion: $m\angle A + m\angle B = 180°$
Law used: Law of Detachment (the hypothesis of statement 1 is given as true in statement 2, so the conclusion follows).

2.4 Properties of Equality and Congruence

Proofs in geometry use properties of equality (for measures) and congruence (for geometric figures). These are the tools that justify each step.

Algebraic Properties of Equality

Properties of Congruence

2.5 Writing Two-Column Proofs

A two-column proof organizes statements and reasons in two columns. Each statement must be justified by a definition, property, postulate, or theorem.

Proof Strategy: Start by writing "Given" as your first statement. Identify what you need to prove. Work through the chain of logic — each new statement should follow from what you've established, ending with the "Prove" statement as your last line.

Example 2.5 — Two-Column Algebraic Proof

Given: $2(x + 3) = 14$

Prove: $x = 4$

StatementsReasons
1. $2(x + 3) = 14$Given
2. $2x + 6 = 14$Distributive Property
3. $2x = 8$Subtraction Property of Equality
4. $x = 4$Division Property of Equality

Example 2.6 — Two-Column Geometric Proof

Given: $\overline{AC}$ with $B$ between $A$ and $C$; $AB = CD$; $B$, $C$, $D$ are collinear with $C$ between $B$ and $D$

Prove: $AC = BD$

StatementsReasons
1. $B$ is between $A$ and $C$; $C$ is between $B$ and $D$Given
2. $AC = AB + BC$Segment Addition Postulate
3. $BD = BC + CD$Segment Addition Postulate
4. $AB = CD$Given
5. $AC = BC + CD$Substitution (steps 2 and 4)
6. $AC = BD$Transitive Property of Equality (steps 3 and 5)

Segment Addition Postulate: B between A and C means AB + BC = AC

Figure 2.2 — Segment Addition Postulate Visualization

TRY IT

Given: $m\angle 1 + m\angle 2 = 90°$ and $m\angle 2 = 35°$. Prove: $m\angle 1 = 55°$ using a two-column proof.

Show Answer
StatementsReasons
1. m∠1 + m∠2 = 90°Given
2. m∠2 = 35°Given
3. m∠1 + 35° = 90°Substitution (steps 1, 2)
4. m∠1 = 55°Subtraction Property of Equality

Complementary angles: when two angles sum to 90°

Figure 2.3 — Complementary Angles and Angle Addition

Practice Problems

1

Find a counterexample to disprove: "If $n$ is a whole number, then $n^2 > n$."

Show Solution
Try $n = 1$: $1^2 = 1$, not greater than 1. Also $n=0$: $0^2=0$, not greater than 0.
Counterexample: $n = 1$ (or $n = 0$). The conjecture is false.
2

Write the converse, inverse, and contrapositive of: "If a polygon is a square, then it has four sides." Determine the truth value of each.

Show Solution
Conditional: Square → 4 sides. True
Converse: If 4 sides, then square. False (rectangle is a counterexample)
Inverse: If not square, then not 4 sides. False
Contrapositive: If not 4 sides, then not square. True
3

Use the Law of Syllogism to write a new conditional: (1) "If it is Friday, then we have pizza." (2) "If we have pizza, then everyone is happy."

Show Solution
Conclusion: "If it is Friday, then everyone is happy." (Law of Syllogism — chain the middle link "we have pizza" away.)
4

Is the following a valid biconditional? "A triangle is equilateral if and only if all three angles are 60°." Explain.

Show Solution
Conditional: Equilateral triangle → all angles 60°. True
Converse: All angles 60° → equilateral. True (equal angles force equal sides by the Isosceles Triangle Theorem)
Both directions are true, so this is a valid biconditional.
5

Complete the proof. Given: $3x - 7 = 20$. Prove: $x = 9$. Fill in the missing reasons.

Show Solution
Step 1: $3x - 7 = 20$ — Given
Step 2: $3x = 27$ — Addition Property of Equality
Step 3: $x = 9$ — Division Property of Equality
6

Given that $\angle 1 \cong \angle 3$ and $\angle 3 \cong \angle 5$, what can you conclude? What property justifies your conclusion?

Show Solution
Conclusion: $\angle 1 \cong \angle 5$
Reason: Transitive Property of Congruence.
7

Write a two-column proof. Given: $\angle ABC$ is a right angle. $\angle ABD + \angle DBC = \angle ABC$. $\angle ABD = 40°$. Prove: $\angle DBC = 50°$.

Show Solution
1. ∠ABC is right → m∠ABC = 90° (Def. of right angle)
2. m∠ABD + m∠DBC = m∠ABC (Angle Addition Postulate)
3. m∠ABD + m∠DBC = 90° (Substitution, steps 1, 2)
4. m∠ABD = 40° (Given)
5. 40° + m∠DBC = 90° (Substitution, steps 3, 4)
6. m∠DBC = 50° (Subtraction Property of Equality)
8

Look at the pattern: 2, 5, 10, 17, 26, … Make a conjecture about the next two terms and describe the pattern.

Show Solution
Differences: 3, 5, 7, 9, … (odd numbers increasing by 2)
Next terms: 26 + 11 = 37; 37 + 13 = 50
Conjecture: The $n$th term is $n^2 + 1$ (so 1²+1=2, 2²+1=5, 3²+1=10, ✓).

📋 Chapter Summary

Types of Reasoning

Inductive Reasoning

Drawing a general conclusion from specific examples. Results in a conjecture. Not guaranteed to be true — one counterexample disproves it.

Deductive Reasoning

Using accepted facts (postulates, theorems) to prove new conclusions. The basis of geometric proof. If premises are true, the conclusion must be true.

Conditional Statements

Conditional

"If $p$, then $q$" — written $p \Rightarrow q$. The hypothesis is $p$; conclusion is $q$.

Converse

"If $q$, then $p$" — $q \Rightarrow p$. Not necessarily true even when the original is true.

Inverse

"If not $p$, then not $q$" — $\neg p \Rightarrow \neg q$. Same truth value as the converse.

Contrapositive

"If not $q$, then not $p$" — $\neg q \Rightarrow \neg p$. Always has the same truth value as the original conditional.

Proof Tips

  1. Mark the diagram — label given information and what you know
  2. Start from given — list all given statements first
  3. Work toward the goal — each step must follow from previous ones
  4. Cite a reason for every step — postulate, theorem, definition, or given

📘 Key Terms

ConjectureAn unproven statement based on observations. Requires a proof to become a theorem.
CounterexampleA single example that disproves a conjecture or conditional statement.
Biconditional"$p$ if and only if $q$" — both the conditional and its converse are true.
PostulateA statement accepted as true without proof. The starting assumptions of geometry.
TheoremA statement that has been proven using postulates, definitions, and previously proven theorems.
Two-Column ProofA structured proof format with statements in the left column and reasons in the right column.
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