1. The problem is to prove that if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. 2. Recall the definitions: - A parallelogram has opposite sides parallel: \(\overline{AB} \parallel \overline{DC}\) and \(\overline{AD} \parallel \overline{BC}\). - A rhombus is a parallelogram with all sides equal: \(AB = BC = CD = DA\). - The diagonals of a parallelogram intersect at their midpoints. 3. The statement to prove is: If \(\overline{AC} \perp \overline{BD}\) (diagonals are perpendicular) and the figure is a parallelogram, then all sides are equal, i.e., it is a rhombus. 4. Among the options, the correct rephrased statement is: \[\text{In quadrilateral } ABCD, \text{ if } \overline{AB} \parallel \overline{DC}, \overline{AD} \parallel \overline{BC}, \text{ and } \overline{AC} \perp \overline{BD}, \text{ then } AB = BC = CD = DA.\] This matches option C. 5. Explanation: The conditions \(\overline{AB} \parallel \overline{DC}\) and \(\overline{AD} \parallel \overline{BC}\) ensure the figure is a parallelogram. The additional condition \(\overline{AC} \perp \overline{BD}\) implies the diagonals are perpendicular, which is a property unique to rhombuses among parallelograms. Therefore, the parallelogram must be a rhombus with all sides equal. Final answer: Option C.