1. **State the problem:** Given that $PS \perp QR$ and $\triangle PQT \cong \triangle SRT$, prove that quadrilateral $PQSR$ is a rhombus. 2. **Recall definitions and properties:** - A rhombus is a quadrilateral with all sides equal in length. - Congruent triangles have corresponding sides and angles equal. - If diagonals are perpendicular and bisect each other, the quadrilateral is a rhombus. 3. **Analyze given information:** - $PS \perp QR$ means diagonals $PR$ and $QS$ intersect at right angles at point $T$. - $\triangle PQT \cong \triangle SRT$ implies: - $PQ = SR$ (corresponding sides) - $QT = RT$ (corresponding sides) - $PT = ST$ (corresponding sides) 4. **Use congruence to find side equalities:** - Since $\triangle PQT \cong \triangle SRT$, $PQ = SR$. - Given $PS \perp QR$, and $T$ is the intersection of diagonals, $T$ is the midpoint of both $PR$ and $QS$ (because congruent triangles share equal segments $PT = ST$ and $QT = RT$). 5. **Show all sides are equal:** - From congruence and midpoint properties: - $PQ = SR$ - $PS = QR$ (given or from congruence of triangles) 6. **Conclude:** - Since $PQ = SR$ and $PS = QR$, and all sides are equal, $PQSR$ is a rhombus. **Final answer:** Quadrilateral $PQSR$ is a rhombus because its sides are all equal and its diagonals are perpendicular and bisect each other.