In the given figure, DE || BC.

(i) Prove that △ADE and △ABC are similar.

(ii) Given that AD = $\dfrac{1}{2}$BD, calculate DE, if BC = 4.5 cm.

(iii) If area of △ABC = 18 cm², find area of trapezium DBCE.

(i) Considering △ADE and △ABC,

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

Hence, by AA axiom △ADE ~ △ABC.

(ii) Given AD = $\dfrac{1}{2}$BD

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

Since triangles ADE and ABC are similar so, ratio of their corresponding sides will be equal

$\therefore \dfrac{AD}{AB} = \dfrac{DE}{BC} \\[1em] \Rightarrow \dfrac{1}{3} = \dfrac{DE}{4.5}\\[1em] \Rightarrow DE = \dfrac{4.5}{3} \\[1em] \Rightarrow DE = 1.5$

Hence, the length of DE = 1.5 cm.

(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.

$\therefore \dfrac{\text{Area of △ADE}}{\text{Area of △ABC}} = \dfrac{AD^2}{AB^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ADE}}{\text{Area of △ABC}} = \dfrac{1^2}{3^2} \\[1em] \Rightarrow \text{Area of △ADE} = \dfrac{1}{9} \times \text{ Area of △ABC} \\[1em] \Rightarrow \text{Area of △ADE} = \dfrac{1}{9} \times 18 \\[1em] \Rightarrow \text{Area of △ADE} = 2 \text{ cm}^2$

Area of trapezium DBCE = Area of △ABC - Area of △ADE = (18 - 2) cm² = 16 cm².

Hence, the area of trapezium DBCE = 16 cm².