Question

given: \\(\overline{xy} \cong \overline{uv}\\), \\(\overline{yz} \cong \overline{tu}\\)
prove: \\(\overline{xz} \cong \overline{tv}\\)

statements | reasons

1. \\(\overline{xy} \cong \overline{uv}\\), \\(\overline{yz} \cong \overline{tu}\\) | 1.
2. \\(xy = uv\\), \\(yz = tu\\) | 2.
3. \\(xy + yz = xz\\), \\(tu + uv = tv\\) | 3.
4. \\(uv + yz = xz\\), \\(yz + uv = tv\\) | 4.
5. \\(xz = tv\\) | 5.
6. \\(\overline{xz} \cong \overline{tv}\\) | 6.

Explanation:

Step1: Identify Given Information

The given information is \(\overline{XY}\cong\overline{UV}\) and \(\overline{YZ}\cong\overline{TU}\). This is the starting point (Given).

Step2: Use Definition of Congruent Segments

By the definition of congruent segments, if two segments are congruent, their lengths are equal. So \(XY = UV\) and \(YZ=TU\) (Definition of Congruent Segments).

Step3: Use Segment Addition Postulate

For the first line, \(X\), \(Y\), \(Z\) are collinear, so \(XY + YZ=XZ\) (Segment Addition Postulate). For the second line, \(T\), \(U\), \(V\) are collinear, so \(TU + UV = TV\) (Segment Addition Postulate).

Step4: Substitute Equal Lengths

Since \(XY = UV\) and \(YZ = TU\), substitute \(XY\) with \(UV\) in \(XY + YZ\) and \(TU\) with \(YZ\) in \(TU + UV\). So \(UV+YZ = XZ\) and \(YZ + UV=TV\) (Substitution Property of Equality).

Step5: Transitive Property of Equality

From \(UV + YZ=XZ\) and \(YZ + UV = TV\), we get \(XZ = TV\) (Transitive Property of Equality, or since both equal \(UV + YZ\)).

Step6: Use Definition of Congruent Segments Again

If \(XZ = TV\), then by the definition of congruent segments, \(\overline{XZ}\cong\overline{TV}\) (Definition of Congruent Segments).

Answer:

The reasons for each step are:

1. Given
2. Definition of Congruent Segments
3. Segment Addition Postulate
4. Substitution Property of Equality
5. Transitive Property of Equality (or Simplification from Step 4)
6. Definition of Congruent Segments

And we have proven \(\overline{XZ}\cong\overline{TV}\)