If D, E and F are the mid points of BC, CA and AB respectively of the ΔABC. The ratio of area of the parallelogram DEFB and area, of the trapezium CAFD is:

We know when a new triangle is formed by using mid points of big triangle.
⇒ In this case Area of 4 triangle is same

⇒ i.e. Area of ΔAFE = ΔFBD
= ΔFDE = ΔDEC = 1
⇒ Parallelogram
DEFB = ΔBFD + ΔDFE = 1 + 1
⇒ Area of Parallelogram
DEFB = 2 . . . . . . (i)
⇒ Again trapezium CAFD
= ΔAFE + ΔFED + ΔDCE = 1 + 1 + 1
Area of Trapezium
CAFD = 3 . . . . . . (ii)
Required Ratio will be = 2 : 3