2. Lesson 2.5 Proving Statements about Segments

3. VOCABULARY A true statement that follows as a result of other true statements is called a theorem. A two-column proof has numbered statements and reasons that show the logical order of an argument. A proof can be written in paragraph form, called a paragraph proof.

4. Using Congruence Use the diagram and the given information to complete the missing steps in the proof. Given: AD = 8, BC = 8, BC ≅ CD Prove: AD ≅ CD 1.AD=81.Given 2.BC=82.Given 3.AD=BC3. Transitive Prop of equality 4.AD ≅ BC4.Def of ≅ segments 5.BC ≅ CD5.Given 6.AD ≅ CD6. Transitive prop of congruence A B C D

5. Using Algebra a.Given: AC = 91 1.AC=911.Given 2.AB+BC=AC2.Segment addition postulate 3.x-10+4x+1=913.Substitution prop of equality 4.5x – 9 = 914.Simplify 5.5x = 1005.Addition prop of equality 6.x = 206.Division prop of equality A B C x-10 4x+1

6. Example 2 Given: DE ≅ EF, EF ≅ GF 1.DE ≅ EF, EF ≅ GF 1. Given 2. DE ≅ GF 2. Transitive property of congruence 3.DE = GF 3. Definition of congruence 4.3x – 5 = 7x – 21 4. Substitution prop of equality 5.3x + 16 = 7x 5. Addition prop of equality 6.16 = 4x 6. Subtraction prop of equality 7.x = 4 7. Division prop of equality DEF G (3x – 5) (7x – 21)

7. Next one… Given: EG ≅ HF 1.EG ≅ HF 1. Given 2.EG = EF + FG2. Segment addition postulate HF = FG + GH 3.EF+FG=GF+GH 3. Segment addition postulate 4.EF = GH 4. Subtraction prop of equality 5.4x = x + 95. Substitution prop of equality 6.3x = 96. Subtraction prop of equality 7.x = 37. Division prop of equality EG F H 4xx + 9

8. More… Given: IP ≅ JO, LM ≅ IP 1.IP ≅ JO, LM ≅ IP1. Given 2.JO ≅ LM 2. Transitive prop of congruence 3.JO = LM3. Def of congruent segments 4.8x–4=3x+11 4. Substitution prop of equality 5.5x – 4 = 115. Subtraction prop of equality 6.5x = 15 6. Addition prop of equality 7.x = 37. Division prop of equality IPJOLM (8x – 4)(3x + 11)

9. Given: LK=5, JK=5 JK ≅ JL Prove: LK ≅ JL 1. LK=5, JK=51. Given 2. JK ≅ JL2. Given 3. JK = JL3. Definition of congruence 4. 5 = JL4. Substitution prop of equality 5. LK = JL5. Substitution prop of equality 6. LK ≅ JL6. Definition of congruence J K L

10. Given: Q is the midpoint of PR Prove: PQ & QR are each ½ PR 1. Q is midpoint if PR1. Given 2. PQ = QR2. Definition of midpoint 3. PQ+QR=PR3. Segment Addition postulate 4. QR+QR=PR4. Substitution prop of equality 5. 2QR = PR5. Simplify 6. QR = ½ PR6. Division prop of equality 7. PQ = ½ PR7. Substitution prop of equality P Q R

11. Given: AB ≅ BC, BC ≅ CD Find BC 1. AB ≅ BC1. Given 2. BC ≅ CD2. Given 3. AB ≅ CD3. Transitive prop of congruence 4. AB = CD4. Definition of congruent segments 5. 3x–1=2x+35. Substitution prop of equality 6. x – 1 = 36. Subtraction prop of equality 7. x = 47. Addition prop of equality 8. AB = 118. Substitution prop of equality 9. AB = BC9. Definition of congruent segments 10. BC = 1110. Substitution prop of equality A BC D 3x - 1 2x + 3