Given, D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC.

In ΔABC,

F is the mid-point of AB (Already given)

E is the mid-point of AC (Already given)

So, by the mid-point theorem, we have,

FE || BC and FE = \( \dfrac{1}{2}\)BC

FE || BC and FE || BD [BD = \( \dfrac{1}{2}\)BC]

Since, opposite sides of parallelogram are equal and parallel

BDEF is parallelogram.

Similarly, in ΔFBD and ΔDEF, we have

FB = DE (Opposite sides of parallelogram BDEF)

FD = FD (Common sides)

BD = FE (Opposite sides of parallelogram BDEF)

∴ ΔFBD ≅ ΔDEF

Similarly, we can prove that

ΔAFE ≅ ΔDEF

ΔEDC ≅ ΔDEF

As we know, if triangles are congruent, then they are equal in area.

So,

Area(ΔFBD) = Area(ΔDEF) …………**(i)**

Area(ΔAFE) = Area(ΔDEF) ………**(ii)**

and,

Area(ΔEDC) = Area(ΔDEF) ……**(iii)**

Now,

Area(ΔABC) = Area(ΔFBD) + Area(ΔDEF)

+ Area(ΔAFE) + Area(ΔEDC) ………**(iv)**

Area(ΔABC) = Area(ΔDEF) + Area(ΔDEF)

+ Area(ΔDEF) + Area(ΔDEF)

From equation **(i)**, **(ii)** and **(iii)**,

⇒ Area(ΔDEF) = (\( \dfrac{1}{4}\))Area(ΔABC)

⇒ \( \dfrac{Area(ΔDEF)}{Area(ΔABC)}\) = \( \dfrac{1}{4}\)

Hence, Area(ΔDEF): Area(ΔABC) = 1:4

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