# Diagonals of trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

## Solution :

In triangles AOB and COD,

$$\angle$$ AOB = $$\angle$$ COD       (vertically opposite angles)

$$\angle$$ OAB = $$\angle$$ OCD        (corresponding angles)

$$\therefore$$ By AA similarity,

$$\triangle$$ AOB ~ $$\triangle$$ COD

$$\implies$$  $$area(\triangle AOB)\over area(\triangle COD)$$ = $${AB}^2\over {DC}^2$$

$$\implies$$  $$area(\triangle AOB)\over area(\triangle COD)$$ = = $${2DC}^2\over {DC}^2$$ = $$4\over 1$$

Hence, Area of triangle AOB : Area of triangle COD = 4 : 1