Given : Quadrilateral $ABCD$ in which diagonals $AC$ and $BD$ intersect at $O$ and $\dfrac{AO}{OC} = \dfrac{BO}{OD}$.
To Prove : $ABCD$ is a trapezium, i.e. $AB \parallel DC$.

Construction : Through $O$, draw $OE \parallel DC$ meeting $AD$ at $E$.

Step 1: In $\triangle ADC$, since $OE \parallel DC$, by Basic Proportionality Theorem :
$$\frac{AE}{ED} = \frac{AO}{OC} \qquad \ldots (i)$$
Step 2: Given
$$\frac{AO}{OC} = \frac{BO}{OD} \qquad \ldots (ii)$$
Step 3: From $(i)$ and $(ii)$ :
$$\frac{AE}{ED} = \frac{BO}{OD}$$
Step 4: In $\triangle DAB$, the points $E$ on $DA$ and $O$ on $DB$ divide the sides in the same ratio. So by the CONVERSE of Basic Proportionality Theorem :
$$EO \parallel AB$$
Step 5: But by construction $EO \parallel DC$. Hence
$$AB \parallel DC$$
Step 6: $ABCD$ is a quadrilateral with one pair of opposite sides parallel.
Therefore $ABCD$ is a trapezium. Hence proved.