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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.

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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.

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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.

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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.

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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.

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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.

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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

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Example 3a: Write a Paragraph Proof

Given: PQ PS; PQ = 3y; PS = 48 Prove: y > 15

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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.

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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.

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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.

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

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

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

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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

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Practice 1. Given: JK LM Prove: w = 7/2 b. Statements Reasons 1. Given

2. 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.

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

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

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2a. 5x – 110 = 2x + 100 x =

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

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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