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

(4:1) Since $AB \parallel CD,$
$\therefore \angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$ (Alternate interior angles).
In $\triangle AOB$ and $\triangle COD,$
$\angle AOB = \angle COD$ (Vertically opposite angles).
$\angle OAB = \angle OCD$ (Alternate interior angles).
$\angle OBA = \angle ODC$ (Alternate interior angles).
$\therefore \triangle AOB \sim \triangle COD$ (By $AAA$ similarity criterion).
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\triangle AOB)}{\operatorname{ar}(\triangle COD)} = \left(\frac{AB}{CD}\right)^2$.
Given $AB = 2 CD,$
$\therefore \frac{\operatorname{ar}(\triangle AOB)}{\operatorname{ar}(\triangle COD)} = \left(\frac{2 CD}{CD}\right)^2 = \left(\frac{2}{1}\right)^2 = \frac{4}{1} = 4:1$.
Thus,the ratio of the areas of triangles $AOB$ and $COD$ is $4:1$.