In the figure (ii) given below, DE || BC and AD : DB = 1 : 2, find the ratio of the areas of △ADE and trapezium DBCE.

Considering △ADE and △ABC,

∠ A = ∠ A (Common angles)
∠ ADE = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △ADE ~ △ABC.

Given AD : DB = 1 : 2.

$\Rightarrow \dfrac{AD}{AB - AD} = \dfrac{1}{2} \\[1em] \Rightarrow 2AD = AB - AD \\[1em] \Rightarrow 2AD + AD = AB \\[1em] \Rightarrow 3AD = AB \\[1em] \Rightarrow \dfrac{AD}{AB} = \dfrac{1}{3} \\[1em] \Rightarrow AD : AB = 1 : 3.$

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ADE}}{\text{Area of △ABC}} = \dfrac{AD^2}{AB^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ADE}}{\text{Area of △ADE + Area of ⏢DBCE}} = \dfrac{1^2}{3^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ADE}}{\text{Area of △ADE + Area of ⏢DBCE}} = \dfrac{1}{9} \\[1em] \Rightarrow 9 \times \text{Area of △ADE} = \text{Area of △ADE + Area of ⏢DBCE} \\[1em] \Rightarrow 8 \text{ Area of △ADE} = \text{ Area of ⏢DBCE } \\[1em] \Rightarrow \dfrac{\text{Area of △ADE}}{\text{ Area of ⏢DBCE }} = \dfrac{1}{8} = 1 : 8.$

Hence, the ratio of the areas of △ADE and trapezium DBCE is 1 : 8.