Select the correct answer from each drop-down menu.
Given: [tex]$CD = EF \quad AB = CB$[/tex]
Prove: [tex]$AB = DF$[/tex]
The length [tex]$CE = CD + DE$[/tex] and [tex]$DF = EF + DE$[/tex] by segment addition. It was given that [tex]$CD = EF$[/tex], and applying the substitution property of equality gives [tex]$DF = CD + DE$[/tex]. Since both [tex]$CE$[/tex] and [tex]$DF$[/tex] equal the same quantity, [tex]$CE = DF$[/tex] by the transitive property of equality. It was also given that [tex]$AB = CB$[/tex]. Applying the transitive property of equality again, [tex]$AB = DF$[/tex].
Use the paragraph proof to complete the two-column proof.
What statement and reason belong in line 4?
[tex]\[
\begin{tabular}{|l|l|}
\hline
\multicolumn{1}{|c|}{Statements} & \multicolumn{1}{c|}{Reasons} \\
\hline
1. $CE = CD + DE$ & 1. Segment addition \\
$DF = EF + DE$ & \\
\hline
2. $CD = EF$ & 2. Given \\
\hline
3. $DF = CD + DE$ & 3. Substitution property of equality \\
\hline
4. $CE = DF$ & 4. Transitive property of equality \\
\hline
5. $AB = CB$ & 5. Given \\
\hline
6. $AB = DF$ & 6. Transitive property of equality \\
\hline
\end{tabular}
\][/tex]