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.

Explanation:

Step1: Identify given congruent segments

Given \( \overline{WZ}\cong\overline{ZS}\cong\overline{SD}\cong\overline{DW}\). So, for the first statement \( \overline{WZ}\cong\overline{SD}\) (since \( \overline{WZ}\cong\overline{ZS}\cong\overline{SD}\cong\overline{DW}\) and we can pick \( \overline{WZ}\) and \( \overline{SD}\) as two of the congruent segments).

Step2: Identify common side

\( \overline{ZD}\cong\overline{ZD}\) (by the reflexive property of congruence, which states that any segment is congruent to itself).

Step3: Apply SSS (Side - Side - Side) congruence criterion

In \( \triangle WZD\) and \( \triangle SDZ\), we have \( \overline{WZ}\cong\overline{SD}\) (from Step 1), \( \overline{ZD}\cong\overline{ZD}\) (from Step 2), and \( \overline{DW}\cong\overline{ZS}\) (given \( \overline{WZ}\cong\overline{ZS}\cong\overline{SD}\cong\overline{DW}\)). So, by SSS (if three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent), \( \triangle WZD\cong\triangle SDZ\).

Answer:

The first blank in the statement \( \overline{WZ}\cong\) should be filled with \( \overline{SD}\). The proof of \( \triangle WZD\cong\triangle SDZ\) is completed by using the SSS (Side - Side - Side) congruence criterion with \( \overline{WZ}\cong\overline{SD}\), \( \overline{ZD}\cong\overline{ZD}\) (reflexive property), and \( \overline{DW}\cong\overline{ZS}\) (given).