# Warm Up Identify the hypothesis and conclusion of each conditional; write the converse of the conditional; decide whether the conditional and converse.

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Warm Up Identify the hypothesis and conclusion of each conditional; write the converse of the conditional; decide whether the conditional and converse are true or false. If false, provide a counterexample. If you attend CB West, then you are in high school. If 4x = 20, then x = 5.

False If you attend CB West, then you are in high school.
Hypothesis: You attend CB West. Conclusion: You are in high school. Converse: If you are in high school, then you attend CB West. False Counterexample: You could attend CB South.

2. If 4x = 20, then x = 5. Hypothesis: 4x = 20. Conclusion: x = 5
Converse: If x = 5, then 4x = 20. True

HOMEWORK ANSWERS – PG.34 Hypothesis: 2x – 1 = 5 Conclusion: x = 3
Hypothesis: She’s smart Conclusion: I’m a genius Hypothesis: 8y = 40 Conclusion: y = 5 Hypothesis: S is the midpoint of RT Conclusion: RS = ½RT Hypothesis: m1 = m2 Conclusion: 1  2 Hypothesis: 1  2 Conclusion: m1 = m2

HOMEWORK ANSWERS – PG.35 Hypothesis: 3x – 7 = 32 Conclusion: x = 13
Hypothesis: I’m not tired Conclusion: I can’t sleep Hypothesis: You will Conclusion: I’ll try Hypothesis: m1 = 90 Conclusion: 1 is a right angle Hypothesis: a + b = a Conclusion: b = 0 Hypothesis: x = -5 Conclusion: x² = 25 B is between A and C if and only if AB + BC = AC mAOC = 180 if and only if AOC is a straight angle.

Section 2-2 Introduction to Proofs

Properties from Equality
Addition Property: If a = b, then a + c = b + c. Subtraction Property: If a = b, then a - c = b - c. Multiplication Property: If a = b, then ca = cb. Division Property: If a = b and c ¹ 0, then

Properties of Equality
Substitution Property: if a = b, then either a or b may be substituted for the other in any equation or inequality. Reflexive Property: a = a. Symmetric Property: if a = b, then b = a. Transitive Property: if a = b and b = c, then a = c.

Properties of Congruence
Reflexive Property: DE ; ÐD Symmetric Property: --If FG, then DE Transitive Property: --If ÐB and ÐC, then ÐC.

Introduction to Proofs
Given: Prove: What you know What you’re trying to prove Statements Reasons All reasons must be: 1. “Given” 2. Postulates 3. Theorems 4. Definitions 5. Properties All of your math steps go here.

Statements Reasons 1. Given 2. Addition Property 3. Division Property
Example 1: Given: 3x – 10 = 20 Prove: x = 10. Statements Reasons Given 1. 3x – 10 = 20 2. 3x – = 3x = 30 2. Addition Property 3. Division Property

Statements Reasons 1. Given 2. Substitution 4. Division Property
Example 2: Given: 3x + y = 22; y = 4 Prove: x = 6 Statements Reasons 1. 3x + y = 22; y = 4 Given 2. Substitution 2. 3x + 4 = 22 3. 3x = 18 3. Subtraction Property 4. x = 6 4. Division Property

NOTE: You will NEVER use the reason “PROVE” as a justification in your proofs.

Because you use postulates, properties, definitions, and theorems for your reasons in proofs, it is a good idea to review the vocabulary and postulates that we learned in Unit 1. Segment Addition Postulate – If point B is between points A and C, then AB + BC = AC. A B C

If point B lies in the interior of ÐAQC, then mÐAQB + mÐBQC = mÐAQC. If ÐAQC is a straight angle and B is a point not on AC, then mÐAQB + mÐBQC = 180. A Q C B A Q C B If you used this statement above, it would also be acceptable to use “definition of a linear pair” as your reason.

F T L A Statements Reasons FL = AT LA = LA FL + LA = AT + LA
Example 3: F T Given: FL = AT Prove: FA = LT L A Statements Reasons FL = AT LA = LA FL + LA = AT + LA FL + LA = FA LA + AT = LT 5. FA = LT Given Sometimes, this step is not necessary. You could just go right to the next step. 2. Reflexive Property 3. Addition Property 4. Segment Addition Postulate 5. Substitution

Statements Reasons mÐAOC = mÐBOD mÐAOC = mÐ1 + mÐ2 mÐBOD = mÐ2 + mÐ3
Example 4: Given: mÐAOC = mÐBOD Prove: mÐ1 = mÐ3 B A 1 2 3 D O Statements Reasons mÐAOC = mÐBOD mÐAOC = mÐ1 + mÐ2 mÐBOD = mÐ2 + mÐ3 mÐ1 + mÐ2 = mÐ2 + mÐ3 mÐ2 = mÐ2 5. mÐ = mÐ3 Given 2. Angle Addition Postulate 3. Substitution Sometimes, this step is not necessary. You could just go right to the next step. Reflexive 5. Subtraction Property

P R S Q T Statements Reasons 1. RS = PS 1. Given 2. RS + ST = PS + ST
Example 5: P Given: RS = PS and ST = SQ. Prove: RT = PQ R S Q T Statements Reasons RS = PS Given 2. RS + ST = PS + ST 2. Addition Property ST = SQ Given 4. RS + ST = PS + SQ 4. Substitution 5. RS + ST = RT PS + SQ = PQ 5. Segment Addition Postulate RT = PQ 6. Substitution

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