The diagonals of a quadrilateral ABCD intersect each other at the point O such that . Show that ABCD is a trapezium.

Given: ABCD is a trapezium with AB || CD and the diagonals AC and BD intersect at 'O'.
To prove: =
Proof:      
In the figure consider the triangle OAB and OCD
 ∠DOC = ∠ΑOB   (Vertically opposite angles are equal)
since AB || DC,
∠DCO = ∠OAB   (Alternate angles are equal) 
∴ By AA corollary of similar triangles.
∴ Δ OAB ∼ Δ OCB When the two triangle are similar, the side are proportionally.
⇒ =
Hence proved.