Question

complete the proof.
given: \\( \overline { w z } \cong \overline { z s } \cong \overline { s d } \cong \overline { d w } \\)
prove: \\( \triangle w z d \cong \triangle s d z \\)
complete the proof below.

statement	reasons
2. \\( \overline { w d } \cong \\)	2. given

Explanation:

Step1: Analyze the given information

We are given \( \overline{WZ}\cong\overline{ZS}\cong\overline{SD}\cong\overline{DW}\). For the first statement \( \overline{WZ}\cong\overline{SD}\), it is directly from the given equal - length segments.

Step2: Determine the second congruent segment

Since \( \overline{WZ}\cong\overline{ZS}\cong\overline{SD}\cong\overline{DW}\), and we need to prove \( \triangle WZD\cong\triangle SDZ\). In the SSS (Side - Side - Side) congruence criterion, we already have \( \overline{WZ}\cong\overline{SD}\) (from step 1). The common side for \( \triangle WZD\) and \( \triangle SDZ\) is \( \overline{ZD}\). And from the given \( \overline{WZ}\cong\overline{ZS}\cong\overline{SD}\cong\overline{DW}\), we know that \( \overline{WD}\cong\overline{SZ}\) is not relevant for the SSS of \( \triangle WZD\) and \( \triangle SDZ\), \( \overline{ZD}\cong\overline{ZD}\) (reflexive property, but we are looking for the segment from the given equal - length set). The second segment for SSS (after \( \overline{WZ}\cong\overline{SD}\)) is \( \overline{ZD}\). But if we consider the SSS for \( \triangle WZD\) and \( \triangle SDZ\) with the given \( \overline{WZ}\cong\overline{SD}\), and we know that \( \overline{ZD}\) is common. However, if we look at the structure of the proof (matching the given equal - length segments in the order of the triangles' sides), since \( \overline{WZ}\cong\overline{SD}\), and for the other two sides of the triangles \( \triangle WZD\) and \( \triangle SDZ\), we have \( \overline{ZD}\) common. But if we follow the given \( \overline{WZ}\cong\overline{ZS}\cong\overline{SD}\cong\overline{DW}\), for the second statement in the proof (to use SSS), we note that \( \overline{WD}\cong\overline{SZ}\) is not correct. The correct second segment (to use SSS for \( \triangle WZD\) and \( \triangle SDZ\)):
We know that \( \overline{WZ}\cong\overline{SD}\) (given), \( \overline{ZD}\cong\overline{ZD}\) (reflexive). But if we consider the order of the triangles \( \triangle WZD\) and \( \triangle SDZ\), and the given \( \overline{WZ}\cong\overline{ZS}\cong\overline{SD}\cong\overline{DW}\), the second segment (to complete the SSS) is \( \overline{ZD}\). But if we assume that the proof is using the given equal - length segments in pairs for the sides of the two triangles. Since \( \overline{WZ}\cong\overline{SD}\) (first pair), and the other pair of sides (for SSS) is \( \overline{ZD}\cong\overline{ZD}\) (reflexive). But if we look at the problem's given \( \overline{WZ}\cong\overline{ZS}\cong\overline{SD}\cong\overline{DW}\) and the need to prove \( \triangle WZD\cong\triangle SDZ\) using SSS:
The sides of \( \triangle WZD\) are \( \overline{WZ}\), \( \overline{ZD}\), \( \overline{DW}\) and the sides of \( \triangle SDZ\) are \( \overline{SD}\), \( \overline{ZD}\), \( \overline{SZ}\). Since \( \overline{WZ}\cong\overline{SD}\), \( \overline{DW}\cong\overline{SZ}\) (from \( \overline{WZ}\cong\overline{ZS}\cong\overline{SD}\cong\overline{DW}\)), and \( \overline{ZD}\cong\overline{ZD}\). But if we follow the proof structure (where the second statement is given), and considering the order of the triangles \( \triangle WZD\) and \( \triangle SDZ\) and the SSS criterion:
The second segment (to match the given in the proof) is \( \overline{ZD}\).

Answer:

\(\overline{ZD}\)