ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. Using Basic Proportionality theorem prove that $\dfrac{AO}{BO} = \dfrac{CO}{DO}.$

Trapezium ABCD is shown in the figure below:

Consider △OAB and △OCD,

∠AOB = ∠COD [Vertically opposite angles are equal]
∠OBA = ∠ODC [Alternate angles are equal]
∠OAB = ∠OCD [Alternate angles are equal]

Therefore, by AA rule of similarity △OAB ~ △OCD,

$\Rightarrow \dfrac{AO}{CO} = \dfrac{BO}{DO} \\[1em] \Rightarrow \dfrac{AO}{BO} = \dfrac{CO}{DO} \text{ (On cross-multiplication)}$

Hence, proved that $\dfrac{AO}{BO} = \dfrac{CO}{DO}.$