In the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meet AB (produced) at F. Prove that

(i) DO : OE = 2 : 1

(ii) area of △OEC : area of △OAD = 1 : 4

(i) Given E is the mid-point of BC,

∴ 2EC = BC

Since ABCD is a parallelogram so, BC = AD or, 2EC = AD.

Considering △AOD and △EDC,

∠AOD = ∠EOC (Vertically opposite angles are equal)

∠OAD = ∠OCB (Alternate angles are equal)

Hence by AA axiom △AOD ~ △EOC. Since triangles are similar so the ratio of their corresponding sides are equal.

$\therefore \dfrac{DO}{OE} = \dfrac{AD}{EC} = \dfrac{2EC}{EC} \\[1em] = \dfrac{2}{1} \\[1em]$

Hence, proved that DO : OE = 2 : 1.

(ii) From (i) we have proved that △AOD ~ △EOC.

$\therefore \dfrac{\text{Area of △OEC}}{\text{Area of △AOD}} = \dfrac{OE^2}{DO^2} \\[1em] = \dfrac{1^2}{2^2} \\[1em] = \dfrac{1}{4} \\[1em] = 1 : 4.$

Hence, proved that area of △OEC : area of △OAD = 1 : 4.