Question

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

1. Prove that: ΔADE ~ ΔABC and hence find DE : BC.
2. Prove: ΔDFE ~ ΔCFB
3. Given, area of ΔDFE = 16 square units, find the area of ΔCFB.

Answer 1

(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 the figure,

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

We know that,

Corresponding sides of similar triangle are in proportion

`(DE)/(BC) = (AD)/(AB) = (2x)/(5x)`

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]

(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.

⇒ `"Area of Δ DFE"/"Area of Δ CFB" = ((DE)/(BC))^2`

⇒ `"Area of Δ DFE"/"Area of Δ CFB" = (2/5)^2`

⇒ `"Area of Δ DFE"/"Area of Δ CFB" = 16/ x`

⇒ `4/25 = 16/ x`

⇒ x = `(16 xx 25)/4`

⇒ `x = 400/4`

⇒ x = 100

⇒ Area of Δ CFB