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Sagot :
To complete the two-column proof, we need to identify the appropriate statement and reason for line 4. Let's review the given proof and map out each line to understand the logical flow and identify the correct statement and reason for line 4.
From the paragraph proof:
1. It states that [tex]\(CE = CD + DE\)[/tex] and [tex]\(DF = EF + DE\)[/tex]. This is recorded in the statements with the reason being segment addition.
2. It is given that [tex]\(CD = EF\)[/tex].
3. By substituting [tex]\(EF\)[/tex] for [tex]\(CD\)[/tex] in the equation [tex]\(DF = EF + DE\)[/tex], we get [tex]\(DF = CD + DE\)[/tex].
4. Since [tex]\(CE = CD + DE\)[/tex] and [tex]\(DF = CD + DE\)[/tex], both [tex]\(CE\)[/tex] and [tex]\(DF\)[/tex] equate to the same expression. Hence, by the transitive property of equality, [tex]\(CE = DF\)[/tex].
5. It is given that [tex]\(AB = CE\)[/tex].
6. By the transitive property of equality, [tex]\(AB = DF\)[/tex].
So, for line 4, the statement [tex]\(CE = DF\)[/tex] is deduced from the previous statements, and the corresponding reason is the transitive property of equality, since both quantities equate to the same expression.
Therefore, the completed two-column proof is:
\begin{tabular}{|c|c|}
\hline Statements & Reasons \\
\hline \begin{tabular}{l}
1. \begin{tabular}{l}
[tex]$C E=C D+D E$[/tex] \\
[tex]$D F=E F+D E$[/tex]
\end{tabular}
\end{tabular} & 1. segment addition \\
\hline 2. [tex]$C D=E F$[/tex] & 2. given \\
\hline 3. [tex]$D F=C D+D E$[/tex] & 3. substitution property of equality \\
\hline 4. [tex]$CE=DF$[/tex] & 4. transitive property of equality \\
\hline 5. [tex]$A B=C E$[/tex] & 5. given \\
\hline 6. [tex]$A B=D F$[/tex] & 6. transitive property of equality \\
\hline
\end{tabular}
The statement and reason that belong in line 4 are:
- Statement: [tex]\(CE = DF\)[/tex]
- Reason: Transitive property of equality
From the paragraph proof:
1. It states that [tex]\(CE = CD + DE\)[/tex] and [tex]\(DF = EF + DE\)[/tex]. This is recorded in the statements with the reason being segment addition.
2. It is given that [tex]\(CD = EF\)[/tex].
3. By substituting [tex]\(EF\)[/tex] for [tex]\(CD\)[/tex] in the equation [tex]\(DF = EF + DE\)[/tex], we get [tex]\(DF = CD + DE\)[/tex].
4. Since [tex]\(CE = CD + DE\)[/tex] and [tex]\(DF = CD + DE\)[/tex], both [tex]\(CE\)[/tex] and [tex]\(DF\)[/tex] equate to the same expression. Hence, by the transitive property of equality, [tex]\(CE = DF\)[/tex].
5. It is given that [tex]\(AB = CE\)[/tex].
6. By the transitive property of equality, [tex]\(AB = DF\)[/tex].
So, for line 4, the statement [tex]\(CE = DF\)[/tex] is deduced from the previous statements, and the corresponding reason is the transitive property of equality, since both quantities equate to the same expression.
Therefore, the completed two-column proof is:
\begin{tabular}{|c|c|}
\hline Statements & Reasons \\
\hline \begin{tabular}{l}
1. \begin{tabular}{l}
[tex]$C E=C D+D E$[/tex] \\
[tex]$D F=E F+D E$[/tex]
\end{tabular}
\end{tabular} & 1. segment addition \\
\hline 2. [tex]$C D=E F$[/tex] & 2. given \\
\hline 3. [tex]$D F=C D+D E$[/tex] & 3. substitution property of equality \\
\hline 4. [tex]$CE=DF$[/tex] & 4. transitive property of equality \\
\hline 5. [tex]$A B=C E$[/tex] & 5. given \\
\hline 6. [tex]$A B=D F$[/tex] & 6. transitive property of equality \\
\hline
\end{tabular}
The statement and reason that belong in line 4 are:
- Statement: [tex]\(CE = DF\)[/tex]
- Reason: Transitive property of equality
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