1. 5.1 Two-Column and Paragraph Proofs
Date:
5.1 Two-Column and Paragraph Proofs
Example 2: Write a Two-Column Proof
Given: AB CD, AB = 5x + 2, CD = 3x + 22
Prove: x = 10
Statements
Reasons
1. AB CD, AB = 5x + 2, CD = 3x + 22 1.
2. AB = CD 2.
3. 5x + 2 = 3x + 22 3.
4. 2x + 2 = 22 4.
5. 2x = 20 5.
6. x = 10 6.

2. 5.1 Two-Column and Paragraph Proofs
Date:
5.1 Two-Column and Paragraph Proofs
Example 2: Write a Two-Column Proof
Given: AB CD, AB = 5x + 2, CD = 3x + 22
Prove: x = 10
Statements
Reasons
1. AB CD, AB = 5x + 2, CD = 3x + 22
1. Given
2. AB = CD 2.
3. 5x + 2 = 3x + 22 3.
4. 2x + 2 = 22 4.
5. 2x = 20 5.
6. x = 10 6.

3. 5.1 Two-Column and Paragraph Proofs
Date:
5.1 Two-Column and Paragraph Proofs
Example 2: Write a Two-Column Proof
Given: AB CD, AB = 5x + 2, CD = 3x + 22
Prove: x = 10
Statements
Reasons
1. AB CD, AB = 5x + 2, CD = 3x + 22
1. Given
2. AB = CD
2. Definition Congruent Segments
3. 5x + 2 = 3x + 22 3.
4. 2x + 2 = 22 4.
5. 2x = 20 5.
6. x = 10 6.

4. 5.1 Two-Column and Paragraph Proofs
Date:
5.1 Two-Column and Paragraph Proofs
Example 2: Write a Two-Column Proof
Given: AB CD, AB = 5x + 2, CD = 3x + 22
Prove: x = 10
Statements
Reasons
1. AB CD, AB = 5x + 2, CD = 3x + 22
1. Given
2. AB = CD
2. Definition Congruent Segments
3. 5x + 2 = 3x + 22
3. Substitution Prop. Of Equality
4. 2x + 2 = 22 4.
5. 2x = 20 5.
6. x = 10 6.

5. 5.1 Two-Column and Paragraph Proofs
Date:
5.1 Two-Column and Paragraph Proofs
Example 2: Write a Two-Column Proof
Given: AB CD, AB = 5x + 2, CD = 3x + 22
Prove: x = 10
Statements
Reasons
1. AB CD, AB = 5x + 2, CD = 3x + 22
1. Given
2. AB = CD
2. Definition Congruent Segments
3. 5x + 2 = 3x + 22
3. Substitution Prop. Of Equality
4. 2x + 2 = 22
4. Subtraction Prop. Of Equality
5. 2x = 20 5.
6. x = 10 6.

6. 5.1 Two-Column and Paragraph Proofs
Date:
5.1 Two-Column and Paragraph Proofs
Example 2: Write a Two-Column Proof
Given: AB CD, AB = 5x + 2, CD = 3x + 22
Prove: x = 10
Statements
Reasons
1. AB CD, AB = 5x + 2, CD = 3x + 22
1. Given
2. AB = CD
2. Definition Congruent Segments
3. 5x + 2 = 3x + 22
3. Substitution Prop. Of Equality
4. 2x + 2 = 22
4. Subtraction Prop. Of Equality
5. 2x = 20
5. Subtraction Prop. Of Equality
6. x = 10 6.

7. 5.1 Two-Column and Paragraph Proofs
Date:
5.1 Two-Column and Paragraph Proofs
Example 2: Write a Two-Column Proof
Given: AB CD, AB = 5x + 2, CD = 3x + 22
Prove: x = 10
Statements
Reasons
1. AB CD, AB = 5x + 2, CD = 3x + 22
1. Given
2. AB = CD
2. Definition Congruent Segments
3. 5x + 2 = 3x + 22
3. Substitution Prop. Of Equality
4. 2x + 2 = 22
4. Subtraction Prop. Of Equality
5. 2x = 20
5. Subtraction Prop. Of Equality
6. x = 10
6. Division Prop. Of Equality

8. Example 3a: Write a Paragraph Proof
Given: PQ PS; PQ = 3y; PS = 48
Prove: y > 15

9. Example 3a: Write a Paragraph Proof
Given: PQ PS; PQ = 3y; PS = 48
Prove: y > 15
Since it is given that PQ PS, then PQ = PS by Definition of Congruent Segments.

10. Example 3a: Write a Paragraph Proof
Given: PQ PS; PQ = 3y; PS = 48
Prove: y > 15
Since it is given that PQ PS, then PQ = PS by Definition of Congruent Segments. Since PQ = 3y and PS = 48, then 3y = 48 by Substitution Property of Equality.

11. Example 3a: Write a Paragraph Proof
Given: PQ PS; PQ = 3y; PS = 48
Prove: y > 15
Since it is given that PQ PS, then PQ = PS by Definition of Congruent Segments. Since PQ = 3y and PS = 48, then 3y = 48 by Substitution Property of Equality. Since y = 16 by Division Property of Equality, then y must be greater than 15. 

12. c. No, y > 15 means that y could equal 17 and 3(17) ≠ 48.

13. PRACTICE
Conjecture: An opinion or conclusion formed on the basis of incomplete information.
1. Given: JK LM
Prove:

14. PRACTICE
Conjecture: An opinion or conclusion formed on the basis of incomplete information.
1. Given: JK LM
Prove:
a. 4w + 1 = 6w – 6
w =

15. PRACTICE
Conjecture: An opinion or conclusion formed on the basis of incomplete information.
1. Given: JK LM
Prove:
a. 4w + 1 = 6w – 6
-2w + 1 = -6
-2w = -7
w = 7/2

16. Practice 1. Given: JK LM Prove: w = 7/2 b. Statements Reasons 1. Given2.
3. 4w + 1 = 6w – 6 3.
4. -2w + 1 = -6 4. 5.
5. Subtr. Prop. of =
6. -2w/-2 = -7/-2 6.
7. w = 7/2 7.

17. 2. Midpoint: The point that divides a line segment into 2 congruent parts.

18. 2a. What is the length of the course from point A to point W?

19. 2a. 5x – 110 = 2x + 100
x =

20. 2a. 5x – 110 = 2x + 100
x =
AW = (5x – 110)m
= 5( ) – 110 =

21. 2b.
Given w is the ____________ of ____, prove that ________. Since w is the ____________ of ____, AW WB. ___ = ___ by Definition of Congruent Segments so ________ = ________. 3x – 110 = 100 by _______________ Property of Equality. 3x = 210 by _______________ Property of Equality. x = 70 by _____________ Property of Equality. AW = __________ = 5(70) – 110 = 240 m by ______________ Property of Equality.

22. 2b.
Given w is the midpoint of AB, prove that AW = Since w is the ____________ of ____, AW WB. ___ = ___ by Definition of Congruent Segments so ________ = ________. 3x – 110 = 100 by _______________ Property of Equality. 3x = 210 by _______________ Property of Equality. x = 70 by _____________ Property of Equality. AW = __________ = 5(70) – 110 = 240 m by ______________ Property of Equality.

23. 2b.
Given w is the midpoint of AB, prove that AW = Since w is the midpoint of AB,
AW WB. ___ = ___ by Definition of Congruent Segments so ________ = ________. 3x – 110 = 100 by _______________ Property of Equality. 3x = 210 by _______________ Property of Equality. x = 70 by _____________ Property of Equality. AW = __________ = 5(70) – 110 = 240 m by ______________ Property of Equality.

24. 2b.
Given w is the midpoint of AB, prove that AW = Since w is the midpoint of AB,
AW WB. AW = WB by Definition of Congruent Segments so ________ = ________. 3x – 110 = 100 by _______________ Property of Equality. 3x = 210 by _______________ Property of Equality. x = 70 by _____________ Property of Equality. AW = __________ = 5(70) – 110 = 240 m by ______________ Property of Equality.

25. 2b.
Given w is the midpoint of AB, prove that AW = Since w is the midpoint of AB,
AW WB. AW = WB by Definition of Congruent Segments so 5X = 2X x – 110 = 100 by _______________ Property of Equality. 3x = 210 by _______________ Property of Equality. x = 70 by _____________ Property of Equality. AW = __________ = 5(70) – 110 = 240 m by ______________ Property of Equality.

26. 2b.
Given w is the midpoint of AB, prove that AW = Since w is the midpoint of AB,
AW WB. AW = WB by Definition of Congruent Segments so 5X = 2X x – 110 = 100 by Subtraction Property of Equality. 3x = 210 by _______________ Property of Equality. x = 70 by _____________ Property of Equality. AW = __________ = 5(70) – 110 = 240 m by ______________ Property of Equality.

27. 2b.
Given w is the midpoint of AB, prove that AW = Since w is the midpoint of AB,
AW WB. AW = WB by Definition of Congruent Segments so 5X = 2X x – 110 = 100 by Subtraction Property of Equality. 3x = 210 by Addition Property of Equality. x = 70 by _____________ Property of Equality. AW = __________ = 5(70) – 110 = 240 m by ______________ Property of Equality.

28. 2b.
Given w is the midpoint of AB, prove that AW = Since w is the midpoint of AB,
AW WB. AW = WB by Definition of Congruent Segments so 5X = 2X x – 110 = 100 by Subtraction Property of Equality. 3x = 210 by Addition Property of Equality. x = 70 by Division Property of Equality. AW = __________ = 5(70) – 110 = 240 m by ______________ Property of Equality.

29. 2b.
Given w is the midpoint of AB, prove that AW = Since w is the midpoint of AB,
AW WB. AW = WB by Definition of Congruent Segments so 5X = 2X x – 110 = 100 by Subtraction Property of Equality. 3x = 210 by Addition Property of Equality. x = 70 by Substitution Property of Equality. AW =
5x = 5(70) – 110 = 240 m by ______________ Property of Equality.

30. 2b.
Given w is the midpoint of AB, prove that AW = Since w is the midpoint of AB,
AW WB. AW = WB by Definition of Congruent Segments so 5X = 2X x – 110 = 100 by Subtraction Property of Equality. 3x = 210 by Addition Property of Equality. x = 70 by Substitution Property of Equality. AW =
5x = 5(70) – 110 = 240 m by Substitution Property of Equality.

31. 3. Given: MN PQ, MN = 5x – 10, PQ = 4x + 10 Prove: MN = 90. 1. Given
Statements
Reasons 1.
Given
2. MN = PQ 2. 3.
3. Substitution Prop. =
4. x – 10 = 10 4. 5.
5. Addition Prop. of =
MN = 5x – 10 6.
7. MN = 5(20) – 10 = 90 7.

32. 4. Given: PQ = 4(x – 3) + 1, QR = x + 10, x = 7 Prove: PQ QR 1. Given
Statements
Reasons 1.
Given
2. PQ = 4(7 – 3) + 1 = 17 2.
3. QR = = 17 3.
4. PQ = QR 4.
5. PQ QR 5.