1. **State the problem:** Given a diamond-shaped quadrilateral PQRS with diagonals PR and QS intersecting at T, where PR is the perpendicular bisector of QS. Given lengths: PS = 10, QT = 6, PT = 8, QR = 14. Find: a) $m \angle PTQ$ b) TS c) PQ d) RS 2. **Key properties and formulas:** - Since PR is the perpendicular bisector of QS, it means $\angle PTQ = 90^\circ$ because PR is perpendicular to QS at T. - T is the midpoint of QS, so $QT = TS$. - Use the Pythagorean theorem in triangles formed by these points. 3. **Find $m \angle PTQ$:** Since PR is perpendicular to QS at T, $m \angle PTQ = 90^\circ$. 4. **Find TS:** Since T is midpoint of QS, $TS = QT = 6$. 5. **Find PQ:** Triangle PQS is a diamond shape, so PS = PQ = 10 (opposite sides equal in a rhombus). 6. **Find RS:** Since QR = 14 and QS = QT + TS = 6 + 6 = 12, but QR is a side, not diagonal. We use the Pythagorean theorem in triangle QTR: - PT = 8, QT = 6, so PR = $\sqrt{PT^2 + TR^2}$. But we need RS, which equals PQ = 10 (since PQRS is a rhombus, all sides equal). **Final answers:** a) $m \angle PTQ = 90^\circ$ b) $TS = 6$ c) $PQ = 10$ d) $RS = 10$