In quadrilateral PQRS the sides satisfy PQ = PS and RQ = RS, and both diagonals PR and QS are drawn. Which one of these statements is NOT true in general?

PQ = PS, RQ = RS and PR is common, so triangles PQR and PSR are congruent by SSS; CPCT then gives ∠QPR = ∠SPR, so PR bisects ∠QPS. Both P and R are equidistant from Q and S, so the whole line PR is the perpendicular bisector of QS. The reverse fails: QS cuts PR at right angles but bisects it only when PQ = QR as well, that is when the kite is a rhombus. Assuming BOTH diagonals bisect each other is the classic slip — that belongs to parallelograms.