Solution:-

(i) From the question it is given that, DE || BC

We have to prove that, ∆ADE and ∆ABC are similar

∠A = ∠A … [common angle for both triangles]

∠ADE = ∠ABC … [because corresponding angles are equal]

Therefore, ∆ADE ~ ∆ABC … [AA axiom]

(ii) From (i) we proved that, ∆ADE ~ ∆ABC

Then, AD/AB = AB/AC = DE/BC

So, AD/(AD + BD) = DE/BC

(½ BD)/ ((½BD) + BD) = DE/4.5

(½ BD)/ ((3/2)BD) = DE/4.5

½ × (2/3) = DE/4.5

1/3 = DE/4.5

Therefore, DE = 4.5/3

DE = 1.5 cm

(iii) From the question it is given that, area of ∆ABC = 18 cm²

Then, area of ∆ADE/area of ∆ABC = DE²/BC²

area of ∆ADE/18 = (DE/BC)²

area of ∆ADE/18 = (AD/AB)²

area of ∆ADE/18 = (1/3)² = 1/9

area of ∆ADE = 18 × 1/9

area of ∆ADE = 2

So, area of trapezium DBCE = area of ∆ABC – area of ∆ADE

= 18 – 2

= 16 cm²