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Answer :

Given, D, E and F are respectively the mid-points of sides AB, BC and CA of \(\triangle\) ABC.

In \(\triangle\) ABC,

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 = 1/2BC

FE || BC and FE || BD [BD = 1/2BC]

Since, opposite sides of parallelogram are equal and parallel

BDEF is parallelogram.

Similarly in \(\triangle\) FBD and \(\triangle\) DEF, we have

FB = DE (Opposite sides of parallelogram BDEF)

FD = FD (Common sides)

BD = FE (Opposite sides of parallelogram BDEF)

\(\triangle\) FBD \({\displaystyle \cong }\) \(\triangle\) DEF

Similarly, we can prove that

\(\triangle\) AFE \({\displaystyle \cong }\) \(\triangle\) DEF

\(\triangle\) EDC \({\displaystyle \cong }\) \(\triangle\) DEF

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

So,

Area(\(\triangle\) FBD) = Area(\(\triangle\) DEF) ……………………………(i)

Area(\(\triangle\) AFE) = Area(\(\triangle\) DEF) ……………………………….(ii)

and,

Area(\(\triangle\) EDC) = Area(\(\triangle\) DEF) ………………………….(iii)

Now,

Area(\(\triangle\) ABC) = Area(\(\triangle\) FBD) + Area(\(\triangle\) DEF) + Area(\(\triangle\) AFE) + Area(\(\triangle\) EDC) ………(iv)

Area(\(\triangle\) ABC) = Area(\(\triangle\) DEF) + Area(\(\triangle\) DEF) + Area(\(\triangle\) DEF) + Area(\(\triangle\) DEF)

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

Area(\(\triangle\) DEF) = (1/4)Area(\(\triangle\) ABC)

Area(\(\triangle\) DEF)/Area(\(\triangle\) ABC) = 1/4

Hence, Area(\(\triangle\) DEF):Area(\(\triangle\) ABC) = 1:4

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