Find area of shaded regions in a triangle I am trying to solve this problem.

In $\triangle ABC$, $CD=3BD$ and $DE=AE$. Given that the area of $\triangle ABC$ is 14$cm^2$. Find the total area, in $cm^2$, of the shaded regions.
I divided the triangle into two parts and found the values of the area for both triangles.
How should I proceed and use the given to solve the problem?

 A: By applying a linear trasformation to the triangle you won't change the proprtion between the shaded area and the area of the entire triangle, so you can reduce the problem to computing the areas in a soecial case (one that makes the computation easy) and then apply the proportion to get the final result.
For example, suppose $|AB|=|BC|=4$ and that there is a right angle on $B$. Then you get that the shaded area is $24/7$, while the total area is $8$. You can then conclude that the answer to your problem is $\frac{3}{7}14=6$.
A: The area we are looking for is the sum of the area $CDE$ and the one $AEF$. Since $DE=AE$ we have $CDE=ACE\implies CDE=\frac 12 ACD=\frac 12\frac 34 ABC=\frac 38 14=\frac{21}4$. About $AEF$ we have: $DEF=AEF$ since $AE=DE$ and $BDF=\frac 13 CDF=\frac 13 (DEF+CDE)$ so $AEF=DEF=ABD-BDF-DEF=\frac 14 ABC-\frac 13 DEF-\frac 13 CDE-DEF=\frac 72 -\frac 74-\frac 43 DEF\iff \frac 73 DEF=\frac 73 AEF=\frac 72$
$\iff AEF=\frac 32\implies CDE+AEF=\frac {21}4+\frac 32=\frac {27}4$.
A: Let [$\cdot$] denote areas below.
$$  AE = DE \>\>\> \implies \>\>\> [DFC] = [AFC] $$
$$  CD = 3BD \> \>\> \implies \>\>\>[DFC]= 3[DFB]$$
Since [AFC] + [DFC] + [DFB] = [ABC], we get
$$[AFC]+[DFC]= [ACDF] =\frac 67  [ABC]$$
Thus, the shaded area is 
$$\frac 12 [AFD]+ \frac 12 [ACD] =\frac 12 [ACDF] =\frac 12 \frac 67  [ABC]=6cm^2$$
