In the given figure, AE : EC = 2 : 3 and BC = 20 cm then DE is equal to :

1. 10 cm

2. 12 cm

3. 8 cm

4. 16 cm

From figure,

In △ ADE and △ ABC,

⇒ ∠DAE = ∠BAC (Common angle)

⇒ ∠ADE = ∠ABC (Corresponding angles are equal)

∴ △ ADE ~ △ ABC (By A.A. postulate)

Given,

AE : EC = 2 : 3

Let AE = 2x and EC = 3x.

From figure,

AC = AE + EC = 2x + 3x = 5x.

We know that,

Corresponding sides of similar triangle are in proportion.

$\therefore \dfrac{DE}{BC} = \dfrac{AE}{AC} \\[1em] \Rightarrow \dfrac{DE}{20} = \dfrac{2x}{5x} \\[1em] \Rightarrow DE = 20 \times \dfrac{2}{5} \\[1em] \Rightarrow DE = \dfrac{40}{5} = 8\text{ cm}.$

Hence, Option 3 is the correct option.