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

Let , each side of the triangle be 2x.
So, AF=FB=BD=DC=CE=EA= x . EF=FD=DE=x ( as always EF= half of BC)
area of the parallelogram DEFB is $$ base\times height$$ = $$ (x)\times h_p$$
& area of the trapezium CAFD is $$\frac{1}{2}\times base\times height$$ = $$\frac{1}{2}\times (x+2x)\times h_t$$
Clearly , both the heights are to be measured from midpoint of sides to midpoints of the line joining the midpoints of the side. So , $$h_p = h_t$$ .
ratio of area of the parallelogram DEFB and area of the trapezium CAFD is
= $$\frac{(x)\times h_p}{\frac{1}{2}\times (x+2x)\times h_t}$$
= 2 : 3