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Given: In ∆ABC, D, E and F are midpoints of sides AB, BC and CA respectively.
BC=EC
Recall that the line joining the midpoints of two sides of traingle is parallel to third side and half of it.
Therefore, we have:
DF= d½BC
=> DF/BC = 1/2.......(1)
DE/AC = 1/2........(2)
EF/AB = 1/2.......(3)
From(1), (2), and (3) we have,
two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.
Therefore, ∆ABC~ ∆ EDF [by SSS similarity theorem]
Hence, area of ∆DEF: area of ∆ABC = 1:4