Given:ABC is a Triangle in which the midpoints of sides BC ,CA and AB are D, E and F.

To show:(i) BDEF is a parallelogram. (ii) ar (DEF) = 1/4ar (ABC)(iii) ar (BDEF) =1/2 ar (ABC)

Proof: i)Since E and F are the midpoints of AC and AB.

BC||FE & FE= ½ BC= BD

(By mid point theorem)

BD || FE & BD= FE

Similarly, BF||DE & BF= DE

Hence, BDEF is a parallelogram

.[A pair of opposite sides are equal and parallel]

(ii)Similarly, we can prove that FDCE & AFDE are also parallelograms.

Now, BDEF is a parallelogram so its diagonal FD divides its into two Triangles of equal areas.

∴ ar(ΔBDF) = ar(ΔDEF) — (i)

In Parallelogram AFDE

ar(ΔAFE) = ar(ΔDEF) (EF is a diagonal) — (ii)

In Parallelogram FDCE

ar(ΔCDE) = ar(ΔDEF) (DE is a diagonal) — (iii)

From (i), (ii) and (iii)

ar(ΔBDF) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF).....(iv)

ar(ΔBDF) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = ar(ΔABC)

4ar(ΔDEF) = ar(ΔABC)(From eq iv)

ar(∆DEF) = 1/4 ar(∆ABC)........(v)

(iii)Area (parallelogram BDEF) = ar(ΔDEF) + ar(ΔBDF)ar(parallelogram BDEF) = ar(ΔDEF) + ar(ΔDEF)

ar(parallelogram BDEF) = 2×ar(ΔDEF)(From eq iv)

ar(parallelogram BDEF) = 2× 1/4

ar(ΔABC)(From eq v)

ar(parallelogram BDEF) = 1/2 ar(ΔABC