P, Q, R and S are the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD, and PQRS always turns out to be a parallelogram. What extra condition on ABCD makes PQRS a rhombus?

PQ is a mid-segment of triangle ABC, so PQ = ½AC, and QR is a mid-segment of triangle BCD, so QR = ½BD. A parallelogram is a rhombus exactly when two adjacent sides are equal, that is when ½AC = ½BD, so the condition is AC = BD. Perpendicular diagonals is the other half of the same family of results — it makes PQRS a RECTANGLE, since PQ ∥ AC and QR ∥ BD would then meet at 90°, and that is the trap here. ABCD being a parallelogram or cyclic says nothing about the diagonals being equal in length.