Question

In a trapezium \(ABCD\), \(AB \parallel DC\) and its diagonals intersect at \(O\). Prove that \[ \frac{OA}{OC} = \frac{OB}{OD} \]

Hint:

When diagonals of a trapezium intersect, the triangles formed are similar by AA similarity.

Answer 1

In trapezium \(ABCD\), by using properties of similar triangles formed by intersecting diagonals: \[ \triangle AOB \sim \triangle COD \] Therefore, corresponding sides are proportional: \[ \frac{OA}{OC} = \frac{OB}{OD} \]