In the given figure, if ∠ADE = ∠B, show that ∆ADE ∼ ∆ABC.

Given:

$\angle A D E=\angle A B C$ and $\angle A=\angle A$

Let DE be x cm

Therefore, by AA similarity theorem, $\triangle A D E \sim \triangle A B C$

$\Rightarrow \frac{A D}{A B}=\frac{D E}{B C}$

$\Rightarrow \frac{3.8}{3.6+2.1}=\frac{x}{4.2}$

$\Rightarrow x=\frac{3.8 \times 4.2}{5.7}=2.8$

Hence, DE = 2.8 cm