Answer :

(i) In triangle ABC it is given that,


EF is parallel to BC

And,

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of
those sides is parallel to the third side and is half the length of the third side.

EF = BC   ( By using Mid – point theorem)

Also,

BD = BC (As D is the mid-point)

So,

BD = EF

BF and DE also parallel to each other

Therefore, the pair of opposite sides are equal and parallel in length.
Similarly BF = DE

Therefore,

BDEF is a parallelogram

(ii) Diagonal of a parallelogram divides it into two equal area

Therefore,

In parallelogram BDEF
DF is the diagonal

⇒ Area of triangle BFD = Area of triangle DEF  ......(1)
In parallelogram DCEF

ED is the diagonal 

⇒  Area of triangle AFE = Area of triangle DEF   .....(2)

In parallelogram AFDE
EF is the diagonal.

⇒ Area of triangle CDE = Area of triangle DEF  ......(3)

So from (1), (2) and (3)

Area of triangle BFD = Area of triangle AFE = Area of triangle CDE = Area of triangle DEF

Area of triangle ABC = Area of triangle BFD + Area of triangle AFE + Area of triangle CDE + Area of triangle DEF

4 (Area of triangle DEF) = Area of triangle ABC

Area of triangle DEF = × Area of triangle

(iii) Area of parallelogram BDEF = Area of triangle DEF + Area of triangle BDE

From part (ii)

Area of parallelogram BDEF = Area of triangle DEF + Area of triangle DEF


                                      = 2 × Area of triangle DEF


                                    = 2 ×  × Area of triangle ABC


                                  = × Area of triangle ABC

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