Solution:-

From the question it is given that,

ABCD is a parallelogram. E is mid-point of BC.

DE meets the diagonal AC at O.

(i) Now consider the ∆AOD and ∆EDC,

∠AOD = ∠EOC … [because Vertically opposite angles are equal]

∠OAD = ∠OCB … [because alternate angles are equal]

Therefore, ∆AOD ~ ∆EOC

Then, OA/OC = DO/OE = AD/EC = 2EC/EC

OA/OC = DO/OE = 2/1

Therefore, OA: OC = 2: 1

(ii) From (i) we proved that, ∆AOD ~ ∆EOC

So, area of ∆OEC/area of ∆AOD = OE²/DO²

area of ∆OEC/area of ∆AOD = 1²/2²

area of ∆OEC/area of ∆AOD = ¼

Therefore, area of ∆OEC: area of ∆AOD is 1: 4.