To fill in the missing statement and reason in line 4 of the two-column proof, we need to follow the logical steps outlined in the paragraph proof and complete the reasoning for each step along the way.
Let's carefully crate the missing steps based on the given information:
1. [tex]\(CE = CD + DE\)[/tex] - by segment addition.
2. [tex]\(DF = EF + DE\)[/tex] - given.
3. [tex]\(CD = EF\)[/tex] - given substitution property of equality.
4. [tex]\(DF = CD + DE\)[/tex] - by applying substitution property of equality (substituting [tex]\(EF\)[/tex] with [tex]\(CD\)[/tex] in [tex]\(DF = EF + DE\)[/tex]).
5. [tex]\(CE = DF\)[/tex] - by transitive property of equality (since both [tex]\(CE\)[/tex] and [tex]\(DF\)[/tex] equal the same quantity [tex]\(CD + DE\)[/tex]).
6. [tex]\(AB = CE\)[/tex] - given.
7. [tex]\(AB = DF\)[/tex] - by transitive property of equality (substituting [tex]\(CE\)[/tex] with [tex]\(DF\)[/tex]).
Next, these steps need to match the format of the two-column proof. Specifically, we need to determine the missing step and its reason:
Line 4 should state the step we logically deduced: [tex]\( CE = DF \)[/tex], with the reason being the transitive property of equality.
Here's the complete two-column proof:
\begin{tabular}{|l|l|}
\hline \multicolumn{1}{|c|}{ Statements } & \multicolumn{1}{c|}{ Reasons } \\
\hline 1. [tex]$CE = CD + DE$[/tex] & 1. Segment addition \\
[tex]$DF = EF + DE$[/tex] & 2. Given \\
\hline 2. [tex]$CD = EF$[/tex] & 3. Substitution property of equality \\
\hline 3. [tex]$DF = CD + DE$[/tex] & 4. Substitution property of equality \\
\hline 4. [tex]$CE = DF$[/tex] & 5. Transitive property of equality \\
\hline 5. [tex]$AB = CE$[/tex] & 6. Given \\
\hline 6. [tex]$AB = DF$[/tex] & 7. Transitive property of equality \\
\hline
\end{tabular}
So, the missing statement and reason in line 4 are:
Statement: [tex]\(CE = DF\)[/tex]
Reason: Transitive property of equality
