D. Trapezium
If the diagonals of a quadrilateral divide each other proportionally (i.e., $\dfrac{AO}{BO} = \dfrac{CO}{DO}$), then the quadrilateral is a trapezium.
Exercise 6.2, Q.10 directly states: "The diagonals of a quadrilateral ABCD intersect at O such that $\frac{AO}{BO} = \frac{CO}{DO}$, show that ABCD is a trapezium." This is the converse of the trapezium-diagonal property. In a parallelogram, diagonals bisect each other (equal parts, not just proportional parts), so option A is a special case only when the ratio is 1:1. The general condition points to a trapezium.