In the given figure, DE ∥ BC.

(i) Prove that ΔADE ∼ ΔABC.

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

(iii) Find $\dfrac{\text{ar(ΔADE)}}{\text{ar(ΔABC)}}$.

(iv) Find $\dfrac{\text{ar(ΔADE)}}{\text{ar(trap. BCED)}}$.

(i) Considering ΔADE and ΔABC,

∠BAC = ∠DAE [Common angles]

∠ADE = ∠ABC [Corresponding angles are equal]

∴ ΔADE ∼ ΔABC (By A.A. axiom)

Hence, proved that ΔADE ∼ ΔABC.

(ii) Given,

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

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

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

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 square of the ratio of their corresponding sides.

$\Rightarrow \dfrac{\text{ar(ΔADE)}}{\text{ar(ΔABC)}} = \dfrac{DE^2}{BC^2} \\[1em] = \Big(\dfrac{DE}{BC}\Big)^2 \\[1em] = \Big(\dfrac{1}{3}\Big)^2 \\[1em] = \dfrac{1}{9}.$

Hence, $\dfrac{\text{ar(ΔADE)}}{\text{ar(ΔABC)}} = \dfrac{1}{9}$.

(iv) From figure,

Area of trap.(BCED) = area(Δ ABC) - area(Δ ADE)

$\Rightarrow \dfrac{\text{ar(ΔABC)}}{\text{ar(ΔADE)}} = \dfrac{9}{1} \\[1em] \Rightarrow \text{ar(ΔABC)} = 9[\text{ar(ΔADE)}]\\[1em] \Rightarrow \text{ar(trap. BCED)} = 9[\text{ar(ΔADE)}] - \text{ar(ΔADE)} \\[1em] \Rightarrow \text{ar(trap. BCED)} = 8[\text{ar(ΔADE)}] \\[1em] \Rightarrow \dfrac{\text{ar(ΔADE)}}{\text{ar(trap. BCED)}} = \dfrac{1}{8}.$

Hence, $\dfrac{\text{ar(ΔADE)}}{\text{ar(trap. BCED)}} = \dfrac{1}{8}$.