The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\frac{AO}{BO} = \frac{CO}{DO}$. Show that $ABCD$ is a trapezium.

(N/A) Let us consider the following figure for the given question.
Draw a line $OE \parallel AB$ such that $E$ lies on $AD$.
In $\triangle ABD$,since $OE \parallel AB$,by using the Basic Proportionality Theorem $(BPT)$,we obtain:
$\frac{AE}{ED} = \frac{BO}{OD}$ $...(1)$
However,it is given that:
$\frac{AO}{BO} = \frac{CO}{DO}$
Rearranging the terms,we get:
$\frac{AO}{CO} = \frac{BO}{DO}$ $...(2)$
From equations $(1)$ and $(2)$,we obtain:
$\frac{AE}{ED} = \frac{AO}{CO}$
In $\triangle ADC$,since $\frac{AE}{ED} = \frac{AO}{OC}$,by the converse of the Basic Proportionality Theorem,we have:
$EO \parallel DC$
Since we constructed $OE \parallel AB$ and we proved $OE \parallel DC$,it follows that:
$AB \parallel DC$
Therefore,$ABCD$ is a trapezium because one pair of opposite sides is parallel.