1. 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.

2. 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.

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

4. 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

5. 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.

6. Section 2-2
Introduction to
Proofs

7. 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

8. 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.

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

10. 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.

11. 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

12. 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

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

14. 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

15. Angle Addition Postulate –
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.

16. 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

17. 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

18. 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