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

(i) Determine the ratios $\dfrac{AD}{AB} \text{ and } \dfrac{DE}{BC}$.

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

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

(i) Given,

$\dfrac{AD}{DB} = \dfrac{3}{2}$

Let AD = 3x and BD = 2x.

From figure,

AB = AD + DB = 3x + 2x = 5x.

$\dfrac{AD}{AB} = \dfrac{3x}{5x} = \dfrac{3}{5}$.

In △ADE and △ABC,

∠A = ∠A [Common]

∠ADE = ∠ABC [Corresponding angles are equal]

∴ △ADE ~ △ABC [By AA]

Since, corresponding sides of similar triangle are proportional to each other.

$\therefore \dfrac{DE}{BC} = \dfrac{AD}{AB} = \dfrac{3}{5}$ ………..(1)

Hence, $\dfrac{AD}{AB} = \dfrac{DE}{BC} = \dfrac{3}{5}$.

(ii) In △DEF and △CBF,

∠FDE = ∠FCB (Alternate angles are equal)

∠DFE = ∠BFC (Vertically opposite angles are equal)

∴ △DEF ~ △CBF [By AA]

Since, corresponding sides of similar triangle are proportional to each other.

$\therefore \dfrac{EF}{FB} = \dfrac{DE}{BC} = \dfrac{3}{5}$

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

(iii) We know that,

The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

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

Hence, ratio of the areas of △DEF and △BFC = 9 : 25.