Given D, E and F are the mid-points of sides BC, CA and AB respectively of ΔABC.

Then DE || AB, DE || FA … (1)

And DF || CA, DF || AE … (2)

From (1) and (2), we get AFDE is a parallelogram.

Similarly, BDEF is a parallelogram.

In ΔADE and ΔABC,

⇒ ∠FDE = ∠A [Opposite angles of ||gm AFDE]

⇒ ∠DEF = ∠B [Opposite angles of ||gm BDEF]

∴ By AA similarity criterion, ΔABC ~ ΔDEF.

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ ar (ΔDEF): ar (ΔABC) = 1: 4