Diagonals AC and BD of a trapezium ABCD with intersect each other at the point O. Using similarity criterion for two triangles, show that .

We have,

ABCD is a trapezium with AB || DC

In ΔAOB and ΔCOD <AOB = <COD (Vertically opposite angle)

<OAB = <OCD (Alternate interior angle)

Then, ΔAOB ~ΔCOD (By AA similarity)

So, OA/OC = OB/OD(Corresponding parts of similar triangle are proportional)