In the following figure, point D divides AB in the ratio 3 : 5. Find:

(i) $\dfrac{\text{AE}}{\text{EC}}$

(ii) $\dfrac{\text{AD}}{\text{AB}}$

(iii) $\dfrac{\text{AE}}{\text{AC}}$

Also if,

(iv) DE = 2.4 cm, find the length of BC.

(v) BC = 4.8 cm, find the length of DE.

(i) Given,

$\dfrac{AD}{DB} = \dfrac{3}{5}$ and DE || BC.

By basic proportionality theorem we have :

A line drawn parallel to one side of triangle divides the other two sides proportionally.

$\therefore \dfrac{AE}{EC} = \dfrac{AD}{DB} = \dfrac{3}{5}.$

Hence, AE : EC = 3 : 5.

(ii) Given,

$\dfrac{AD}{DB} = \dfrac{3}{5}$

Let AD = 3x and DB = 5x.

AB = AD + DB = 3x + 5x = 8x.

$\dfrac{AD}{AB} = \dfrac{3x}{8x} = \dfrac{3}{8}$ = 3 : 8.

Hence, AD : AB = 3 : 8.

(iii) Given,

$\Rightarrow \dfrac{AE}{EC} = \dfrac{3}{5} \\[1em] \Rightarrow \dfrac{EC}{AE} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{EC}{AE} + 1 = \dfrac{5}{3} + 1 \\[1em] \Rightarrow \dfrac{EC + AE}{AE} = \dfrac{5 + 3}{3} \\[1em] \Rightarrow \dfrac{AC}{AE} = \dfrac{8}{3} \\[1em] \Rightarrow \dfrac{AE}{AC} = \dfrac{3}{8}.$

Hence, AE : AC = 3 : 8.

(iv) In ∆ADE and ∆ABC,

∠ADE = ∠ABC [As DE || BC, Corresponding angles are equal.]

∠A = ∠A [Common angles]

Hence, ∆ADE ~ ∆ABC by AA criterion for similarity.

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{AD}{AB} = \dfrac{DE}{BC} \\[1em] \Rightarrow \dfrac{3}{8} = \dfrac{2.4}{BC} \\[1em] \Rightarrow BC = \dfrac{8 \times 2.4}{3} \\[1em] \Rightarrow BC = 6.4 \text{ cm}.$

Hence, BC = 6.4 cm.

(v) Since, ∆ADE ~ ∆ABC by AA criterion for similarity

So, we have

$\Rightarrow \dfrac{AD}{AB} = \dfrac{DE}{BC} \\[1em] \Rightarrow \dfrac{3}{8} = \dfrac{DE}{4.8} \\[1em] \Rightarrow DE = \dfrac{3 \times 4.8}{8} \\[1em] \Rightarrow DE = 1.8 \text{ cm}.$

Hence, DE = 1.8 cm.