D, E and F are respectively the mid-points of the sides BC, CA and AB of a \(\triangle{ABC}\). Show that
i) 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.