In the given diagram, DE || BC and AD : DB = 2 : 3.

(a) Given,

DE ∥ BC

In triangle △ADE and △ABC,

∠ADE = ∠ABC (Corresponding angles are equal)

∠AED = ∠ACB (Corresponding angles are equal)

Therefore, △ADE ∼ △ABC (By AA similarity)

Given,

AD : DB = 2 : 3

Let AD = 2x and DB = 3x.

From figure,

AB = AD + DB = 2x + 3x = 5x.

We know that,

Corresponding sides of similar triangle are in proportion.

$\frac{DE}{BC} = \frac{AD}{AB} = \frac{2x}{5x} = \frac{2}{5}$.

DE : BC = 2 : 5.

Hence, proved ΔADE ~ ΔABC and DE : BC = 2 : 5.

(b) Given,

DE ∥ BC

In triangle ΔDFE and ΔCFB,

∠DFE = ∠CFB (Vertically opposite angles are equal)

∠DEF = ∠CBF (Alternate interior angles are equal)

ΔDFE ~ ΔCFB [By AA similarity]

Hence, proved that ΔDFE ~ ΔCFB.

(c) Let area of ΔCFB be x square units.

We know that,

The areas of two similar triangles are proportional to the squares of their corresponding sides.

$$\begin{aligned}\Rightarrow \frac{\text{Area of Δ DFE}}{\text{Area of Δ CFB}} = \Big(\frac{DE}{BC}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of Δ DFE}}{\text{Area of Δ CFB}} = \Big(\frac{2}{5}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of Δ DFE}}{\text{Area of Δ CFB}} = \frac{4}{25} \\[1em] \Rightarrow \frac{16}{x} = \frac{4}{25} \\[1em] \Rightarrow x = \frac{16 \times 25}{4} \\[1em] \Rightarrow x = \frac{400}{4} \\[1em] \Rightarrow x = 100.\end{aligned}$$

Hence, area of △CFB = 100 sq.units.