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

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