Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that $\dfrac{OA}{OC} = \dfrac{OB}{OD}$.

Trapezium ABCD is shown in the figure below:

In Δ AOB and Δ COD,

∠AOB = ∠COD (Vertically opposite angles are equal)

∠BAO = ∠DCO (Alternate angles are equal)

⇒ Δ AOB ∼ Δ COD (By A.A. axiom)

We know that,

Ratio of corresponding sides of a similar triangle are proportional.

Hence, proved that $\dfrac{OA}{OC} = \dfrac{OB}{OD}$.