Question

Diagonals AC and BD of a trapezium ABCD with AB ∥ DC intersect each other at the point O. Using similarity criterion for two triangles, show that \(\frac{OA}{ OC} = \frac{OB}{OD}\)

Answer 1

Given: OC is the point of intersection of AC and BD in the trapezium ABCD, with AB ∥ DC. 

Required to prove: \(\frac{OA}{ OC} = \frac{OB}{OD}\)

We know that, 

In ΔAOB and ΔCOD 

∠AOB = ∠COD [Vertically Opposite Angles] 

∠OAB = ∠OCD    [Alternate angles] 

Then, ΔAOB ∼ ΔCOD 

Therefore, \(\frac{OA}{ OC} = \frac{OB}{OD}\)   [Corresponding sides are proportional]