D, E and F are the mid-points of the sides BC, CA and AB respectively of triangle ABC. Prove that :

(i) BDEF is a parallelogram.

(ii) area of BDEF is half the area of △ABC.

(i) Given: ABC is a triangle. D, E and F are the mid-points of the sides BC, CA and AB respectively.

To prove: BDEF is a parallelogram.

Construction: Join DE, EF and FD.

Proof: D, E and F are mid-points of sides BC, AC and AB respectively. Using mid-point theorem,

So, DE ∥ BA and DE = $\dfrac{1}{2}$ BA

⇒ DE ∥ BF and DE = BF

And, EF ∥ BC and EF = $\dfrac{1}{2}$ BC

⇒ EF ∥ BD and EF = BD.

Hence, BDEF is a parallelogram.

(ii) To prove: Ar.(∥gm BDEF) = $\dfrac{1}{2}$ Ar.(Δ ABC)

Proof: From (i), BDEF is a parallelogram.

D, E and F are mid-points of sides BC, AC and AB respectively. Using mid-point theorem,

So, FD ∥ AC and FD = $\dfrac{1}{2}$ AC

⇒ FD ∥ EC and FD = EC

And, EF ∥ BC and EF = $\dfrac{1}{2}$ BC

⇒ EF ∥ DC and EF = DC

Hence, DCEF is a parallelogram.

And, DE ∥ BA and DE = $\dfrac{1}{2}$ BA

⇒ DE ∥ AF and DE = AF

And, DF ∥ AC and DF = $\dfrac{1}{2}$ AC

⇒ DF ∥ AE and DF = AE

Hence, AFDE is a parallelogram.

That means, AFDE, DCEF and BDEF all are parallelograms.

Now, DF is a diagonal of parallelogram BDEF.

Therefore, Ar.(Δ BDF) = Ar.(Δ DEF) …………….(1)

DE is a diagonal of parallelogram DCEF.

So, Ar.(Δ DCE) = Ar.(Δ DEF) …………….(2)

FE is a diagonal of parallelogram AFDE.

So, Ar.(Δ AFE) = Ar.(Δ DEF) …………….(3)

From (1), (2) and (3), we have

Ar.(Δ BDF) = Ar.(Δ DCE) = Ar.(Δ AFE) = Ar.(Δ DEF)

But, Ar.(Δ BDF) + Ar.(Δ DCE) + Ar.(Δ AFE) + Ar.(Δ DEF) = Ar.(Δ ABC)

So, 4 x Ar.(Δ DEF) = Ar.(Δ ABC)

⇒ Ar.(Δ DEF) = $\dfrac{1}{4}$ Ar.(Δ ABC)

We know that the diagonal of a parallelogram divides it into two triangles of equal areas.

⇒ Ar.(∥gm BDEF) = 2Ar.(Δ DEF)

⇒ Ar.(∥gm BDEF) = 2 x $\dfrac{1}{4}$ Ar.(Δ ABC)

⇒ Ar.(∥gm BDEF) = $\dfrac{1}{2}$ Ar.(Δ ABC)

Hence, area of BDEF is half the area of △ABC.