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# D, E and F are respectively the mid-points of the sides BC, CA and AB of a $$\triangle{ABC}$$. Show thati) BDEF is a parallelogram.ii) ar(DEF) = $$\frac{1}{4}$$ ar(ABC) iii) ar(BDEF) = $$\frac{1}{2}$$ ar(ABC) (i) In $$\triangle$$ ABC,
EF || BC and EF = $$\frac{1}{2}$$ BC (by mid point theorem)
also,
BD = $$\frac{1}{2}$$ BC (D is the mid point)
So, BD = EF
also,
BF and DE are parallel and equal to each other.
$$\therefore$$ the pair opposite sides are equal in length and parallel to each other.
$$\therefore$$ BDEF is a parallelogram.

(ii) Proceeding from the result of (i),

BDEF, DCEF, AFDE are parallelograms.
Diagonal of a parallelogram divides it into two triangles of equal area.
$$\therefore$$ ar(?BFD) = ar(?DEF) (For parallelogram BDEF) — (i)
also,
ar($$\triangle$$ AFE) = ar($$\triangle$$ DEF) (For parallelogram DCEF) — (ii)
ar($$\triangle$$ CDE) = ar($$\triangle$$ DEF) (For parallelogram AFDE) — (iii)

From (i), (ii) and (iii) $$ar(\triangle BFD) = ar(\triangle AFE) = ar(\triangle CDE) = ar(\triangle DEF)$$
$$\Rightarrow ar(\triangle BFD) +ar(\triangle AFE) +ar(\triangle CDE) +ar(\triangle DEF) = ar(\triangle ABC)$$
$$\Rightarrow 4 ar(\triangle DEF) = ar(\triangle ABC )$$
$$\Rightarrow$$ ar(DEF) = $$\frac{1}{4}$$ ar(ABC)

(iii) Area (parallelogram BDEF) = ar($$\triangle$$DEF) +ar($$\triangle$$ BDE)
$$\Rightarrow$$ ar(parallelogram BDEF) = ar($$\triangle$$ DEF) +ar($$\triangle$$ DEF)
$$\Rightarrow$$ ar(parallelogram BDEF) = 2× ar($$\triangle$$ DEF)
$$\Rightarrow$$ ar(parallelogram BDEF) = 2× $$\frac{1}{4}$$ ar($$\triangle$$ ABC)
$$\Rightarrow$$ ar(parallelogram BDEF) = $$\frac{1}{4}$$ ar($$\triangle$$ ABC)

Hence, proved.