In the given figure, DE || BC and AE : EC = 5 : 4. Find :

(i) DE : BC

(ii) DO : DC

(iii) area of △DOE : area of △ DCE.

(i) Given,

DE || BC

In △ADE and △ABC,

∠ADE = ∠ABC [Corresponding angles are equal]

∠DAE = ∠BAC [Common angle]

△ADE ~ △ABC [By AA axiom]

Given,

AE : EC = 5 : 4

AE = 5x and EC = 4x

AC = AE + EC = 5x + 4x = 9x.

We know that,

Ratio of corresponding sides of two triangles are proportional to each other.

$\Rightarrow \dfrac{DE}{BC} = \dfrac{AE}{AC} \\[1em] \Rightarrow \dfrac{DE}{BC} = \dfrac{5x}{9x} = \dfrac{5}{9}.$

Hence, DE : BC = 5 : 9.

(ii) In △DOE and △BOC,

∠DOE = ∠BOC [Vertically opposite angles are equal]

∠ODE = ∠OCB [Alternate angles are equal]

△DOE ~ △BOC [By AA axiom]

Given,

DE : BC = 5 : 9

We know that,

Ratio of corresponding sides of two triangles are proportional to each other.

$\Rightarrow \dfrac{DO}{OC} = \dfrac{DE}{BC} \\[1em] \Rightarrow \dfrac{DO}{OC} = \dfrac{5}{9}.$

As,

DO : OC = 5 : 9

Let,

DO = 5y or OC = 9y

From figure,

DC = DO + OC = 5y + 9y = 14y.

DO : DC = 5y : 14y = 5 : 14.

Hence, DO : DC = 5 : 14.

(iii) We know that,

Ratio of areas of two triangles having equal heights is equal to the ratio of the corresponding bases.

$\Rightarrow \dfrac{\text{Area of △DOE}}{\text{Area of △DCE}} = \dfrac{DO}{DC} = \dfrac{5}{14}$

Hence, area of △DOE : area of △ DCE = 5 : 14.