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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]

Sagot :

GinnyAnsweridn
Here's the complete two-column proof with the correct statement and reason for line 4:

[tex]\[ \begin{array}{|c|c|} \hline \text{Statements} & \text{Reasons} \\ \hline 1. C E = C D + D E \quad D F = E F + D E & \text{segment addition} \\ \hline 2. C D = E F & \text{given} \\ \hline 3. D F = C D + D E & \text{substitution property of equality} \\ \hline 4. C E = D F & \text{transitive property of equality} \\ \hline 5. A B = C B & \text{given} \\ \hline 6. A B = D F & \text{transitive property of equality} \\ \hline \end{array} \][/tex]

In this proof:

- Line 1 states the segment addition properties for [tex]\( C E \)[/tex] and [tex]\( D F \)[/tex].
- Line 2 uses the given information [tex]\( C D = E F \)[/tex].
- Line 3 applies the substitution property of equality to replace [tex]\( E F \)[/tex] in the expression for [tex]\( D F \)[/tex].
- Line 4 concludes that since both [tex]\( C E \)[/tex] and [tex]\( D F \)[/tex] are equal to [tex]\( C D + D E \)[/tex], they are equal to each other by the transitive property of equality.
- Line 5 accepts the given relationship [tex]\( A B = C B \)[/tex].
- Line 6 uses the transitive property of equality to establish that [tex]\( A B = D F \)[/tex].
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