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Math Question

GUEST, GUEST A proof with some missing statements and reasons is shown. Given: \( \quad P Q R S \) is a parallelogram. \( \overline{P Q} \cong \overline{Q R} \) Prove: \( \quad P Q R S \) is a rhombus. \begin{tabular}{|l|l|} \hline \multicolumn{1}{|c|}{ Statement } & \multicolumn{1}{|c|}{ Reason } \\ \hline \( 1 . \) & \( 1 . \) Given \\ \hline \( 2 . \) & \( 2 . \) Given \\ \hline \( 3 . \) & \( 3 . \) \\ \hline \( 4 . \) & \( 4 . \) \\ \hline \( 5 . \overline{P Q} \cong \overline{Q R} \cong \overline{R S} \cong \overline{S P} \) & \( 5 . \) \\ \hline \( 6 . P Q R S \) is a rhombus. & \( 6 . \) \\ \hline \end{tabular} Drag the correct statement from the statements column and the correct reason from the reasons column to the table to complete line 3 of the proof. \begin{tabular}{|l|l|} \hline \multicolumn{1}{|c|}{ Statements } \\ \hline\( \overline{P Q} \cong \overline{S R} \) and \( \overline{P S} \cong \overline{Q R} \) & \multicolumn{1}{|c|}{ Reasons } \\ \hline\( \overline{P T} \cong \overline{T R} \) and \( \overline{S T} \cong \overline{T Q} \) & \begin{tabular}{l} Diagonals of a parallelogram bisect each other. \\ \hline Opposite angles of a parallelogram are congruent. \\ \hline\( \triangle P T Q \cong \triangle Q T R \) & Opposite sides of a parallelogram are congruent. \\ \hline\( \angle S P Q \cong \angle Q R S \) & Side-Side-Side \\ \hline \end{tabular} \end{tabular}

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