In the adjoining figure, ABC is a triangle. DE is parallel to BC and $\dfrac{AD}{DB} = \dfrac{3}{2}.$

(i) Determine the ratio $\dfrac{AD}{AB}, \dfrac{DE}{BC}.$

(ii) Prove that △DEF is similar to △CBF. Hence, find $\dfrac{EF}{FB}.$

(iii) What is the ratio of the areas of △DEF and △CBF ?

(i) We need to find $\dfrac{AD}{AB}$,

Given,

$\dfrac{AD}{DB} = \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{AD}{AB - AD} = \dfrac{3}{2} \\[1em] \Rightarrow 2AD = 3(AB - AD) \\[1em] \Rightarrow 2AD = 3AB - 3AD \\[1em] \Rightarrow 5AD = 3AB \\[1em] \Rightarrow \dfrac{AD}{AB} = \dfrac{3}{5}.$

Hence, the ratio $\dfrac{AD}{AB} = \dfrac{3}{5}$.

Considering △ADE and △ABC,

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

Hence, by AA axiom △ADE ~ △ABC.

Since triangles ADE and ABC are similar so the ratio of corresponding sides will be equal.

$\therefore \dfrac{DE}{BC} = \dfrac{AD}{AB} \\[1em] \Rightarrow \dfrac{DE}{BC} = \dfrac{3}{5}.$

Hence, the ratio $\dfrac{DE}{BC} = \dfrac{3}{5}$.

(ii) Considering △DEF and △CBF,

∠ DFE = ∠ BFC (Vertically opposite angles)
∠ EDF = ∠ FCB (Alternate angles are equal)

Hence, by AA axiom △DEF ~ △CBF.

Since triangles are similar hence the ratio of the corresponding sides will be equal,

$\therefore \dfrac{EF}{FB} = \dfrac{DE}{BC} \\[1em] \Rightarrow \dfrac{EF}{FB} = \dfrac{3}{5}.$

Hence, the value of $\dfrac{EF}{FB} = \dfrac{3}{5}.$

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

$\therefore \dfrac{\text{Area of △DEF}}{\text{Area of △CBF}} = \dfrac{EF^2}{FB^2} \\[1em] \Rightarrow \dfrac{\text{Area of △DEF}}{\text{Area of △CBF}} = \dfrac{3^2}{5^2} \\[1em] \Rightarrow \dfrac{\text{Area of △DEF}}{\text{Area of △CBF}} = \dfrac{9}{25}.$

Hence, the ratio of the area of △DEF : area of △CBF = 9 : 25.