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# 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 $$\cong$$ $$\triangle$$ DEF

Similarly, we can prove that
$$\triangle$$ AFE $$\cong$$ $$\triangle$$ DEF
$$\triangle$$ EDC $$\cong$$ $$\triangle$$ 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