Question

Prove that the diagonals of a trapezium divide each other proportionally.

Answer 1

Given: In trapezium ABCD, AB || CD. Diagonals AC and BD intersect at O.

In △AOB and △COD:

• ∠ABO = ∠CDO (alternate interior angles, since AB || CD)
• ∠AOB = ∠COD (vertically opposite angles)

⇒ △AOB ∼ △COD (by AA similarity)

Therefore, corresponding sides are proportional:

`(AO)/(OC) = (BO)/(OD)`

Hence proved that the diagonals of a trapezium divide each other proportionally.