If the diagonals of a quadrilateral divide each other proportionally , prove that itAskis a trapezium

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

In ∆AOD and ∆BOC

∠AOD =  ∠COB  (Vertically  opposite angles)

Thus ∆AOD ~ ∆BOC  (by SAS similarity creation)

⇒ ∠OAD = ∠OCB                ...  (1)

Now transversal AC intesect AD and BC such that ∠CAD = ∠ACB        (from (1))  (Alternate opposite angle)

So AD || BC

Hence ABCD is a trapezium.