Given that ABCD is a quadrilateral and diagonals AC and BD intersect at O such that

IN Δ AOD and ΔBOC

∠ AOD = ∠COB

Thus Δ AOC ∼ ΔBOC (SAS similarity criterion)

⇒ ∠OAD = ∠ OCB ……………..1

Now transversal AC intersect AD and BC such the ∠CAD = ∠ACB

(from …..1 ) (alternate opposite angles)

So AD ∥ BC

Hence ABCD is a trapezium