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 $\frac{OA}{OC} =\frac{OB}{OD}$.

Let's draw a trapezium ABCD with AB || DC.

In ΔAOB and ΔCOD

∠AOB = ∠COD (vertically opposite angles)

∠BAO = ∠DCO (alternate interior angles)

⇒ ΔAOB ∼ ΔCOD (AA criterion)

Hence, $\frac{OA}{OC} =\frac{OB}{OD}$