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i) BDEF is a parallelogram.

ii) ar(DEF) = \(\frac{1}{4} \) ar(ABC)

iii) ar(BDEF) = \(\frac{1}{2} \) ar(ABC)

Answer :

(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.

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