Find the unknown area by different methods this problem has many ways of calculating, could provide some solutions, particularly using proportions

EDIT : A it's not midpoint
If we view $AEB$ as having base $AE$ and triangle $FEB$ as have base $FE$, then the both have the same height--- the altitude of line $AF$ to point $B$.

Since triangles $AEB$ and $FEB$ both have the same height then their bases are in proportion to their areas.  So $EF = \frac 32 AE$.

Triangle $DEF$ is similar $AEB$ and the sides are that $EF$ corresponds to $AE$ and so the sides are $\frac 32$ as long.  So the area of the $DEF$ (white) area is $2cm^2 *(\frac 32)^2 = 4.5 cm^2$.

So the area of the white plus the yellow is $4.5cm^2 + 3cm^2 = 7.5 cm^2$.

As this is a rectangle triangles $DCB$ and $DFB$ are congruent and have the same area.

And so the area of the blue plus the green is $7.5 cm^2$.  So the area of the blue is $7.5 cm^2 - 2cm^2 = 5.5 cm^2$
Area of the triangle $ABF=5$.

And $A$ is the midpoint of the side $CB$.

So the area of rectangle $BCDF$= 4 × area  of triangle$ ABF$
$=4×5= 20$.

Thus area of the triangle $BDC$ is half of the area of the rectangle =$10$.

Thus area of the blue reigion= $10-2=8$.