To solve the problem given [tex]\( XY = WZ \)[/tex], we need to prove [tex]\( XW = YZ \)[/tex] step-by-step using logical and mathematical properties. Here is the detailed solution:
1. Statement 1: [tex]\(XY = WZ\)[/tex]
- Reason: This is given in the problem.
2. Statement 2: [tex]\(YW = YW\)[/tex]
- Reason: Reflexive Property. Any quantity is equal to itself.
3. Statement 3: [tex]\(XY + YW = YW + WZ\)[/tex]
- Reason: Given that [tex]\(XY = WZ\)[/tex] and adding the same quantity [tex]\(YW\)[/tex] to both sides of the equation using the Addition Property of Equality.
4. Statement 4: [tex]\(XY + YW = XW\)[/tex]
- Reason: This follows from the Segment Addition Postulate, which states that if [tex]\(Y\)[/tex] is between [tex]\(X\)[/tex] and [tex]\(W\)[/tex], then [tex]\(XY + YW = XW\)[/tex].
5. Statement 5: [tex]\(YW + WZ = YZ\)[/tex]
- Reason: This also follows from the Segment Addition Postulate, which tells us that if [tex]\(W\)[/tex] is between [tex]\(Y\)[/tex] and [tex]\(Z\)[/tex], then [tex]\(YW + WZ = YZ\)[/tex].
6. Statement 6: [tex]\(XW = YZ\)[/tex]
- Reason: By substituting the expressions from Statements 4 and 5 into Statement 3, we get [tex]\(XY + YW = YZ\)[/tex], which simplifies directly to [tex]\(XW = YZ\)[/tex].
So, the blank in Statement 2 should be filled with:
- Reflexive Property
Thus, the completed table will look like this:
[tex]\[
\begin{array}{|l|l|}
\hline \text{Statement} & \text{Reason} \\
\hline (1) XY = WZ & (1) \text{Given} \\
\hline (2) YW = YW & (2) \text{Reflexive Property} \\
\hline (3) XY + YW = YW + WZ & (3) \text{Addition Property of Equality} \\
\hline (4) XY + YW = XW & (4) \text{Segment Addition Postulate} \\
\hline (5) YW + WZ = YZ & (5) \text{Segment Addition Postulate} \\
\hline (6) XW = YZ & (6) \text{Transitive Property} \\
\hline
\end{array}
\][/tex]
