To complete the proof, let's fill in the missing statements and reasons in the table.
Given: [tex]\(M-N-P-Q\)[/tex] on [tex]\(\overline{MQ}\)[/tex]
Prove: [tex]\(MN + NP + PQ = MQ\)[/tex]
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Statements} & \text{Reasons} \\
\hline
1. M - N - P - Q \text{ on } \overline{MQ} & 1. \text{Given} \\
2. MN + NQ = MQ & 2. \text{By the definition of a line segment} \\
3. NP + PQ = NQ & 3. \text{By the definition of a line segment} \\
4. MN + NP + PQ = MQ & 4. \text{Substitution Property of Equality} \\
\hline
\end{array}
\][/tex]
Detailed Explanation:
1. Statement 1: Given
[tex]\(M-N-P-Q\)[/tex] on [tex]\(\overline{MQ}\)[/tex] means that points M, N, P, and Q are collinear on the line segment from [tex]\(M\)[/tex] to [tex]\(Q\)[/tex].
2. Statement 2: [tex]\(MN + NQ = MQ\)[/tex]
Reason: By the definition of a line segment, the total distance [tex]\(MQ\)[/tex] can be expressed as the sum of the distances from [tex]\(M\)[/tex] to [tex]\(N\)[/tex] and from [tex]\(N\)[/tex] to [tex]\(Q\)[/tex]. This is because [tex]\(N\)[/tex] is between [tex]\(M\)[/tex] and [tex]\(Q\)[/tex].
3. Statement 3: [tex]\(NP + PQ = NQ\)[/tex]
Reason: Similarly, the distance [tex]\(NQ\)[/tex] can be expressed as the sum of the distances from [tex]\(N\)[/tex] to [tex]\(P\)[/tex] and from [tex]\(P\)[/tex] to [tex]\(Q\)[/tex]. This is because [tex]\(P\)[/tex] is between [tex]\(N\)[/tex] and [tex]\(Q\)[/tex].
4. Statement 4: [tex]\(MN + NP + PQ = MQ\)[/tex]
Reason: Substitution Property of Equality. Using statement 3, substitute [tex]\(NQ\)[/tex] in statement 2 with [tex]\(NP + PQ\)[/tex]:
[tex]\[
MN + (NP + PQ) = MQ
\][/tex]
Which simplifies to:
[tex]\[
MN + NP + PQ = MQ
\][/tex]
Thus, we have shown using the definitions of line segments and the substitution property of equality that [tex]\(MN + NP + PQ = MQ\)[/tex].
