Answered

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Write the following paragraph proof as a two-column proof.

Given: [tex]\(AB = CD\)[/tex] and [tex]\(BC = DE\)[/tex]
Prove: [tex]\(AC = CE\)[/tex]

| Statements | Reasons |
|---------------------------------------|-------------------------------------|
| 1. [tex]\(AB = CD\)[/tex] | 1. Given |
| 2. [tex]\(BC = DE\)[/tex] | 2. Given |
| 3. [tex]\(AB + BC = CD + BC\)[/tex] | 3. Addition Property of Equality |
| 4. [tex]\(CD + BC = CD + DE\)[/tex] | 4. Substitution Property of Equality |
| 5. [tex]\(AB + BC = CD + DE\)[/tex] | 5. Substitution Property of Equality |
| 6. [tex]\(AB + BC = AC\)[/tex] | 6. Segment Addition Postulate |
| 7. [tex]\(CD + DE = CE\)[/tex] | 7. Segment Addition Postulate |
| 8. [tex]\(AC = CE\)[/tex] | 8. Substitution Property of Equality |

Sagot :

GinnyAnswer
Certainly! Below is the two-column proof for the given argument:

[tex]\[ \begin{array}{|c|c|} \hline \textbf{Statements} & \textbf{Reasons} \\ \hline 1. \, AB = CD & 1. \, \text{Given} \\ \hline 2. \, BC = DE & 2. \, \text{Given} \\ \hline 3. \, AB + BC = CD + BC & 3. \, \text{Addition property of equality} \\ \hline 4. \, AB + BC = CD + DE & 4. \, \text{Substitution (using BC = DE)} \\ \hline 5. \, AB + BC = AC & 5. \, \text{Segment addition} \\ \hline 6. \, CD + DE = CE & 6. \, \text{Segment addition} \\ \hline 7. \, AC = CE & 7. \, \text{Substitution (replacing AB + BC with AC and CD + DE with CE)} \\ \hline \end{array} \][/tex]

This two-column proof corresponds to the given paragraph explanation and logically shows each step and its justification, leading to the conclusion that [tex]\( AC = CE \)[/tex].
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