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 OA/OC = OB/OD.

Given: ABCD is a trapezium with AB || DC. O is the point of intersection of two diagonals.
To Prove: OA/OC = OB/OD
Proof: In ΔAOB and ΔDOC
∠BAO = ∠OCD (Alternate Angles)
∠ABO = ∠ODC (Alternate Angles)
∠AOB = ∠DOC (Vertically opposite angles)
∴ By AAA criterion of similarity, ΔAOB ≅ ΔDOC
∴ OA/OC = OB/OD (Corresponding Sides of Similar Triangles)