In a quadrilateral ABCD the diagonals AC and BD meet at O, and the measurements are AO = 2 cm, OC = 5 cm, BO = 3 cm and OD = 7.5 cm. What does the converse of the proportionality result let you conclude about ABCD?

$\frac{AO}{OC}=\frac{2}{5}=0.4$ and $\frac{BO}{OD}=\frac{3}{7.5}=0.4$, so the diagonals cut each other in the same ratio. Reading the proportionality backwards gives $AB$ parallel to $DC$, i.e. ABCD is a trapezium. It need not be a parallelogram (that would need $AO=OC$) nor a rhombus. The third option is wrong because the ratios are in fact equal, so something definitely does follow.