$ABCD$ is a trapezium in which $AB \parallel DC$ and its diagonals intersect each other at the point $O$. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.

(N/A) Draw a line $EF$ through point $O$,such that $EF \parallel CD$.
In $\triangle ADC$,since $EO \parallel CD$,by using the Basic Proportionality Theorem $(BPT)$,we obtain:
$\frac{AE}{ED} = \frac{AO}{OC} \quad ...(1)$
Since $AB \parallel CD$ and $EF \parallel CD$,it follows that $EF \parallel AB$.
In $\triangle ABD$,since $EO \parallel AB$,by using the Basic Proportionality Theorem,we obtain:
$\frac{AE}{ED} = \frac{BO}{OD} \quad ...(2)$
From equations $(1)$ and $(2)$,we have:
$\frac{AO}{OC} = \frac{BO}{OD}$
Rearranging the terms,we get:
$\frac{AO}{BO} = \frac{OC}{OD}$