Mixed Exercises 8. PROOF Five lights, A, B, C, D, and E, are aligned in a row. The middle light is the midpoint of the segment between the second and fourth lights and also the midpoint of the segment between the first and last lights. a. Draw a figure to illustrate the situation. b. Complete this proof. Given: Cis the midpoint of BD and AE . Prove: AB = DE Statement Reason 1. Cis the midpoint of BD and AE. 1. Given 2. ВС — CD and 2. ? 3. АС — АВ+ ВС, СЕ — CD + DE 3. 4. АС — ВС — АВ 4. ? 5. 5. Substitution Property 6. CE – CD = DE 6. ? 7. АВ 3D СЕ — CD 7. Symmetric Property of Equality 8. ? 8. ?

College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter9: Sequences, Probability And Counting Theory
Section9.5: Counting Principles
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Mixed Exercises
8. PROOF Five lights, A, B, C, D, and E, are aligned in a row. The middle light is the midpoint of the segment between the
second and fourth lights and also the midpoint of the segment between the first and last lights.
a. Draw a figure to illustrate the situation.
b. Complete this proof.
Given: Cis the midpoint of BD and AE .
Prove: AB =
DE
Statement
Reason
1. Cis the midpoint of BD and AE.
1. Given
2. BC = CD and
2.
3. АС — АВ + ВС, СЕ — CD + DE
3.
4. АС — ВС %3D АВ
4.
?
|
5.
5. Substitution Property
6. CE – CD = DE
6.
?
7. АВ 3D СЕ — CD
7. Symmetric Property of Equality
8.
8.
O v i 11:13
Transcribed Image Text:M Dashboard | McGraw-Hill A my.mheducation.com/secure/student/urn:com.mheducation.openlearning:enterprise.identity.organization:prod.global:organization:071e84b0-02fd-4f94-9165-e4340309dcd2/urn:com.mheducation.openlearning:enterprise.roster:prod.us-east-1:s. R * : + 154 of 305 -> Mixed Exercises 8. PROOF Five lights, A, B, C, D, and E, are aligned in a row. The middle light is the midpoint of the segment between the second and fourth lights and also the midpoint of the segment between the first and last lights. a. Draw a figure to illustrate the situation. b. Complete this proof. Given: Cis the midpoint of BD and AE . Prove: AB = DE Statement Reason 1. Cis the midpoint of BD and AE. 1. Given 2. BC = CD and 2. 3. АС — АВ + ВС, СЕ — CD + DE 3. 4. АС — ВС %3D АВ 4. ? | 5. 5. Substitution Property 6. CE – CD = DE 6. ? 7. АВ 3D СЕ — CD 7. Symmetric Property of Equality 8. 8. O v i 11:13
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