On the hanging sign, the three angles

TABLE 1.6 Further Algebraic Properties of Equality a, b, and c are real numbers Reflexive Property: a = a . Symmetric Property: If a = b , then b = a . Distributive Property: a b + c = a ⭈ b + a ⭈ c . Substitution Property: If a = b , then a replaces b in any equation. Transitive Property: If a = b and b = c , then a = c . TABLE 1.5 Properties of Equality a, b, and c are real numbers Addition Property of Equality: If a = b , then a + c = b + c . Subtraction Property of Equality: If a = b , then a - c = b - c . Multiplication Property of Equality: If a = b , then a c = b c . Division Property of Equality: If a = b and c Z 0, then = . b c a c Reminder Additional properties and techniques of algebra are found in Appendix A. To deal with this question, you must ask “What” it is that is known Given and “Why” the conclusion Prove should follow from this information. Completing the proof often requires deducing several related conclusions and thus several intermediate “whys”. In correctly piecing together a proof, you will usually scratch out several con- clusions and reorder them. Each conclusion must be justified by citing the Given hy- pothesis, a previously stated definition or postulate, or a theorem previously proved. Selected properties from algebra are often used as reasons to justify statements. For instance, we use the Addition Property of Equality to justify adding the same number to each side of an equation. Reasons found in a proof often include the properties found in Tables 1.5 and 1.6. As we discover in Example 1, some properties can be used interchangably. EXAMPLE 1 Which property of equality justifies each conclusion? a If 2x - 3 = 7, then 2x = 10. b If 2x = 10, then x = 5. Solution a Addition Property of Equality; added 3 to each side of the equation. b Multiplication Property of Equality; multiplied each side of the equation by . OR Division Property of Equality; divided each side of the equation by 2. 쮿 1 2 Before considering geometric proof, we study algebraic proof, in which each state- ment in a sequence of steps is supported by the reason why we can make that statement claim. The first claim in the proof is the Given problem; and the sequence of steps must conclude with a final statement representing the claim to be proved called the Prove statement . Study Example 2. Then cover the reasons and provide the reason for each state- ment. With statements covered, find the statement corresponding to each reason. EXAMPLE 2 GIVEN: 2x - 3 + 4 = 10 PROVE: x = 6 PROOF Statements Reasons

1. 2x

- 3 + 4 = 10 1. Given 2. 2x - 6 + 4 = 10 2. Distributive Property 3. 2x - 2 = 10 3. Substitution 4. 2x = 12 4. Addition Property of Equality 5. x = 6 5. Division Property of Equality NOTE 1: Alternatively, Step 5 could use the reason Multiplication Property of Equality multiply by . Division by 2 led to the same result. NOTE 2: The fifth step is the final step because the Prove statement has been made and justified. 쮿 1 2 Exs. 5–7 Exs. 1–4 Discover In the diagram, the wooden trim pieces are mitered cut at an angle to be equal and to form a right angle when placed together. Use the properties of algebra to explain why the measures of ⬔1 and ⬔2 are both 45°. What you have done is an informal “proof.” 1 2 1 2 ANSWER m⬔ 1 + m⬔ 2 = 90°. Because m⬔ 1 = m⬔ 2, we see that m⬔ 1 + m⬔ 1 = 90°. Thus, 2 ⭈ m⬔ 1 = 90°, and, dividing by 2, we see that m⬔ 1 = 45°. Then m⬔ 2 = 45°also. The Discover activity at the left suggests that a formal geometric proof also exists. The typical format for a problem requiring geometric proof is GIVEN: ________ [Drawing] PROVE: ________ Consider this problem: GIVEN: A -P-B on Figure 1.54 PROVE: AP = AB - PB First consider the Drawing Figure 1.54, and relate it to any additional information described by the Given. Then con- sider the Prove statement. Do you understand the claim, and does it seem reasonable? If it seems reasonable, the interme- diate claims must be ordered and supported to form the contents of the proof. Because a proof must begin with the Given and conclude with the Prove, the proof of the preced- ing problem has this form: PROOF Statements Reasons 1. A-P-B on 1. Given 2. ? 2. ? . . . . . . ?. AP = AB - PB ?. ? AB AB A B P Figure 1.54 A B P Figure 1.55 EXAMPLE 3 GIVEN: A-P-B on Figure 1.55 PROVE: AP = AB - PB PROOF Statements Reasons 1. A-P-B on 1. Given 2. AP + PB = AB 2. Segment-Addition Postulate 3. AP = AB - PB 3. Subtraction Property of Equality 쮿 AB AB To construct the proof, you must glean from the Drawing and the Given that AP + PB = AB In turn, you deduce through subtraction that AP = AB - PB. The complete proof prob- lem will have the appearance of Example 3, which follows the first of several “Strategy for Proof ” features used in this textbook. Some of the properties of inequality that are used in Example 4 are found in Table 1.7. While the properties are stated for the “greater than” relation ⬎, they are valid also for the “less than” relation . STRATEGY FOR PROOF 왘 The First Line of Proof General Rule: The first statement of the proof includes the “Given” information; also, the first reason is Given. Illustration: See the first line in the proof of Example 3. TABLE 1.7 Properties of Inequality a, b, and c are real numbers Addition Property of Inequality: If a b, then a + c b + c . Subtraction Property of Inequality: If a b, then a - c b - c. SAMPLE PROOFS Consider Figure 1.56 and this problem: GIVEN: MN PQ PROVE: MP NQ To understand the situation, first study the Drawing Figure 1.56 and the related Given. Then read the Prove with reference to the drawing. Constructing the proof requires that you begin with the Given and end with the Prove. What may be confusing here is that the Given involves MN and PQ, whereas the Prove involves MP and NQ. However, this is easily remedied through the addition of NP to each side of the inequality MN PQ; see step 2 in the proof of Example 4. M P N Q Figure 1.56 Exs. 8–10 M P N Q Figure 1.57 EXAMPLE 4 GIVEN: MN ⬎ PQ Figure 1.57 PROVE: MP ⬎ NQ PROOF Statements Reasons 1. MN ⬎ PQ 1. Given 2. MN + NP ⬎ NP + PQ 2. Addition Property of Inequality 3. But MN + NP = MP and 3. Segment-Addition Postulate NP + PQ = NQ 4. MP ⬎ NQ 4. Substitution NOTE: The final reason may come as a surprise. However, the Substitution Axiom of Equality allows you to replace a quantity with its equal in any statement— including an inequality See Appendix A.3 for more information. 쮿 General Rule: The final statement of the proof is the “Prove” statement. Illustration: See the last statement in the proof of Example 5. STRATEGY FOR PROOF 왘 The last statement of the proof EXAMPLE 5 Study this proof, noting the order of the statements and reasons. GIVEN: bisects ⬔RSU bisects ⬔USW Figure 1.58 PROVE: m ⬔RST + m⬔VSW = m⬔TSV PROOF Statements Reasons 1. bisects ⬔RSU 1. Given 2. m ⬔RST = m ⬔TSU 2. If an angle is bisected, then the measures of the resulting angles are equal. 3. bisects ⬔USW 3. Same as reason 1 4. m ⬔VSW = m ⬔USV 4. Same as reason 2 5. m ⬔RST + m ⬔VSW = 5. Addition Property of Equality m ⬔TSU + m ⬔USV use the equations from statements 2 and 4 6. m ⬔TSU + m ⬔USV = m ⬔TSV 6. Angle-Addition Postulate 7. m ⬔RST + m ⬔VSW = m ⬔TSV 7. Substitution 쮿 SV ST SV ST W U S R T V Exs. 11, 12 Figure 1.58 Exercises 1.5 In Exercises 1 to 6, which property justifies the conclusion of the statement?

1. If 2x

= 12, then x = 6.

2. If x

+ x = 12, then 2x = 12.

3. If x

+ 5 = 12, then x = 7.

4. If x - 5

= 12, then x = 17.

5. If

= 3, then x = 15.

6. If 3x - 2

= 13, then 3x = 15. In Exercises 7 to 10, state the property or definition that justifies the conclusion the “then” clause.

7. Given that ⬔s 1 and 2 are

supplementary, then m ⬔1 + m ⬔2 = 180°.

8. Given that m ⬔3

+ m ⬔4 = 180°, then ⬔s 3 and 4 are supplementary.

9. Given ⬔RSV and as

shown, then m ⬔RST + m ⬔TSV = m ⬔RSV.

10. Given that m ⬔RST

= m ⬔TSV, then bisects ⬔RSV. In Exercises 11 to 22, use the Given information to draw a conclusion based on the stated property or definition. ST ST x 5 In Exercises 23 and 24, fill in the missing reasons for the algebraic proof.

23. Given: 3x - 5

= 21 Prove: x = 12 PROOF Statements Reasons 1. 3x - 5 = 21 1. ? 2. 3x - 15 = 21 2. ? 3. 3x = 36 3. ? 4. x = 12 4. ? A B M Exercises 11, 12 R S T V Exercises 9, 10

11. Given: A

-M-B; Segment-Addition Postulate

12. Given: M

is the midpoint of ; definition of midpoint

13. Given: m

⬔1 = m ⬔2; definition of angle bisector

14. Given: bisects

⬔DEF; definition of angle bisector

15. Given: ⬔s 1 and 2 are

complementary; definition of complementary angles

16. Given: m

⬔1 + m ⬔2 = 90°; definition of complementary angles

17. Given: 2x - 3

= 7; Addition Property of Equality

18. Given: 3x

= 21; Division Property of Equality

19. Given: 7x

+ 5 - 3 = 30; Substitution Property of Equality

20. Given:

= 0.5 and 0.5 = 50; Transitive Property of Equality

21. Given: 32x - 1

= 27; Distributive Property

22. Given:

= -4; Multiplication Property of Equality x 5 1 2 EG AB 1 2 D E G F Exercises 13–16

24. Given: 2x

+ 9 = 3 Prove: x = -3 PROOF Statements Reasons

1. 2x

+ 9 = 3 1. ? 2. 2x = -6 2. ? 3. x = -3 3. ? In Exercises 25 and 26, fill in the missing statements for the algebraic proof.

25. Given: 2x

+ 3 - 7 = 11 Prove: x = 6 PROOF Statements Reasons 1. ? 1. Given 2. ? 2. Distributive Property 3. ? 3. Substitution Addition 4. ? 4. Addition Property of Equality 5. ? 5. Division Property of Equality

26. Given:

+ 3 = 9 Prove: PROOF Statements Reasons 1. ? 1. Given 2. ? 2. Subtraction Property of Equality 3. ? 3. Multiplication Property of Equality x = 30 x 5

30. Given:

⬔ABC and See figure for Exercise 29. Prove: m ⬔ABD = m ⬔ABC - m⬔DBC PROOF Statements Reasons 1. ⬔ABC and 1. ? 2. m ⬔ABD + m ⬔DBC 2. ? = m ⬔ABC 3. m ⬔ABD = m ⬔ABC 3. ? - m⬔DBC In Exercises 31 and 32, fill in the missing statements and reasons.

31. Given: M-N-P-Q

on Prove: MN + NP + PQ = MQ PROOF Statements Reasons 1. ? 1. ? 2. MN + NQ = MQ 2. ? 3. NP + PQ = NQ 3. ? 4. ? 4. Substitution Property of Equality

32. Given:

⬔TSW with and Prove: m ⬔TSW = m ⬔TSU + m ⬔USV + m ⬔VSW SV SU MQ BD BD In Exercises 27 to 30, fill in the missing reasons for each geometric proof.

27. Given: D

-E-F on Prove: DE = DF - EF PROOF Statements Reasons 1. D-E-F on 1. ? 2. DE + EF = DF 2. ? 3. DE = DF - EF 3. ?

28. Given:

E is the midpoint of Prove: DE = DF PROOF Statements Reasons 1. E is the midpoint of 1. ? 2. DE = EF 2. ? 3. DE + EF = DF 3. ? 4. DE + DE = DF 4. ? 5. 2DE = DF 5. ? 6. DE = DF 6. ?

29. Given: bisects

⬔ABC Prove: m ⬔ABD = m ⬔ABC 1 2 BD 1 2 DF 1 2 DF Í DF Í DF D F E Exercises 27, 28 PROOF Statements Reasons 1. ? 1. ? 2. m ⬔TSW = m ⬔TSU 2. ? + m ⬔USW 3. m ⬔USW = m ⬔USV 3. ? + m ⬔VSW 4. ? 4. Substitution Property of Equality B A D C Exercises 29, 30 PROOF Statements Reasons 1. bisects ⬔ABC 1. ? 2. m ⬔ABD = m ⬔DBC 2. ? 3. m ⬔ABD + m ⬔DBC 3. ? = m ⬔ABC 4. m ⬔ABD + m ⬔ABD 4. ? = m ⬔ABC 5. 2m ⬔ABD = m ⬔ABC 5. ? 6. m ⬔ABD = m ⬔ABC 6. ? 1 2 BD T U V W S M P N Q