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Complete the following proof by dragging and dropping the correct reason in the spaces below.

Given: [tex]$Q$[/tex] is between [tex]$P$[/tex] and [tex]$R$[/tex], [tex]$R$[/tex] is between [tex]$Q$[/tex] and [tex]$S$[/tex], [tex]$PR = QS$[/tex]
Prove: [tex]$PQ = RS$[/tex]

\begin{tabular}{|l|l|}
\hline \multicolumn{1}{|c|}{Statement} & Reason \\
\hline \multicolumn{1}{|c|}{Q is between P and R} & Given \\
\hline [tex]$R$[/tex] is between [tex]$Q$[/tex] and [tex]$S$[/tex] & Given \\
\hline [tex]$QR + RS = QS$[/tex] & Segment Addition Postulate \\
\hline [tex]$PR = QS$[/tex] & Given \\
\hline [tex]$PQ + QR = PR$[/tex] & Segment Addition Postulate \\
\hline [tex]$PQ + QR = QR + RS$[/tex] & Substitution Property of Equality \\
\hline [tex]$PQ + QR - QR = QR + RS - QR$[/tex] & Subtraction Property of Equality \\
\hline [tex]$PQ = RS$[/tex] & Simplify \\
\hline
\end{tabular}

\begin{tabular}{|c|c|}
\hline
Given & Segment Addition Postulate \\
Transitive Property & Subtraction Property of Equality \\
Reflexive Property & Substitution Property of Equality \\
Simplify & Definition of Congruent Segments \\
Addition Property of Equality & Definition of Midpoint \\
Multiplication Property of Equality & Division Property of Equality \\
\hline
\end{tabular}

Sagot :

GinnyAnswer
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].
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