Answer :
Certainly! Let's complete the proof step-by-step by filling in the correct reasons from the given options.
Here's the proof with the appropriate reasons:
[tex]\[ \begin{tabular}{|l|l|} \hline \multicolumn{1}{|c|}{ \textbf{Statement} } & \multicolumn{1}{|c|}{ \textbf{Reason} } \\ \hline Q is between P and R & Given \\ \hline $R$ is between $Q$ and $S$ & Given \\ \hline $Q R + R S = Q S$ & Segment Addition Postulate \\ \hline $P R = Q S$ & Given \\ \hline $P Q + Q R = Q R + R S$ & Segment Addition Postulate \\ \hline $P Q + Q R - Q R = Q R + R S - Q R$ & Subtraction Property of Equality \\ \hline $P Q = R S$ & Simplify \\ \hline \end{tabular} \][/tex]
Reasoning:
1. Given statements are directly used to establish Q's position between P and R and R's position between Q and S.
2. The Segment Addition Postulate is used twice to express [tex]$Q S$[/tex] and [tex]$P R$[/tex] in terms of their segments.
3. The Subtraction Property of Equality is used to isolate [tex]$PQ$[/tex] and [tex]$RS$[/tex] after subtracting [tex]$QR$[/tex] from both sides.
This proof completes by logically following from the given statements to reach the conclusion that [tex]$PQ = RS$[/tex].
Here's the proof with the appropriate reasons:
[tex]\[ \begin{tabular}{|l|l|} \hline \multicolumn{1}{|c|}{ \textbf{Statement} } & \multicolumn{1}{|c|}{ \textbf{Reason} } \\ \hline Q is between P and R & Given \\ \hline $R$ is between $Q$ and $S$ & Given \\ \hline $Q R + R S = Q S$ & Segment Addition Postulate \\ \hline $P R = Q S$ & Given \\ \hline $P Q + Q R = Q R + R S$ & Segment Addition Postulate \\ \hline $P Q + Q R - Q R = Q R + R S - Q R$ & Subtraction Property of Equality \\ \hline $P Q = R S$ & Simplify \\ \hline \end{tabular} \][/tex]
Reasoning:
1. Given statements are directly used to establish Q's position between P and R and R's position between Q and S.
2. The Segment Addition Postulate is used twice to express [tex]$Q S$[/tex] and [tex]$P R$[/tex] in terms of their segments.
3. The Subtraction Property of Equality is used to isolate [tex]$PQ$[/tex] and [tex]$RS$[/tex] after subtracting [tex]$QR$[/tex] from both sides.
This proof completes by logically following from the given statements to reach the conclusion that [tex]$PQ = RS$[/tex].