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# Diagonals of a 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.

Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O. In $$\triangle$$ AOB and $$\triangle$$ COD, we have
$$\angle$$ 1 = $$\angle$$ 2 (Alternate angles)
$$\angle$$ 3 = $$\angle$$ 4 (Alternate angles)
$$\angle$$ 5 = $$\angle$$ 6 (Vertically opposite angle)
$$\triangle$$ AOB ~ $$\triangle$$ COD [AAA similarity criterion]

As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,

$$\frac{Area of ( \triangle AOB)}{Area of ( \triangle COD) } = \frac{AB^2}{CD^2}$$
= $$\frac{(2CD)^2}{CD^2}$$ [ AB = CD]
$$\frac{Area of (\triangle AOB)}{Area of (\triangle COD)}$$
= $$\frac{4CD^2}{CD} = \frac{4}{1}$$

Hence, the required ratio of the area of $$\triangle$$ AOB and $$\triangle$$ COD = 4:1