ABCD is a trapezium where AB \(\parallel\) CD. The diagonals AC and BD intersect at point O. Prove that: AO × OD = BO × OC.

Q.ABCD is a trapezium where AB \(\parallel\) CD. The diagonals AC and BD intersect at point O. Prove that: AO × OD = BO × OC.

Assume ABCD is a trapezium where AB is parallel to CD, and the diagonals AC and BD intersect at point O. To prove: AO × OD = BO × OC Proof: In triangles ∆AOB and ∆COD, ∠OAB = ∠OCD [AB ∥ CD and AC is a transversal] ∠AOB = vertically opposite ∠COD Remaining ∠OBA = ∠ODC ∴ Triangles ∆AOB and ∆COD are similar. ∴ \(\frac{AO}{OB} = \frac{OC}{OD}\) That is, AO × OD = BO × OC (Proved)