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

Given:
Trapezium \(ABCD\) with \(AB \parallel DC\)
Diagonals \(AC\) and \(BD\) intersect at point \(O\)

To prove:
\[ \frac{OA}{OC} = \frac{OB}{OD} \]

Proof:
Since \(AB \parallel DC\), by properties of trapezium,
the triangles \(\triangle AOB\) and \(\triangle COD\) are similar.

**Reason:**
- \(\angle AOB = \angle COD\) (Vertically opposite angles)
- \(\angle OAB = \angle OCD\) (Alternate interior angles since \(AB \parallel DC\))

Therefore, by AA similarity criterion,
\[ \triangle AOB \sim \triangle COD \]

From similarity of triangles,
\[ \frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{DC} \]

Hence proved:
\[ \frac{OA}{OC} = \frac{OB}{OD} \]