Answer :
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].
[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].