find ratio of area of triangle $\Delta{AFG}$ to area of $\Delta {ABC}$ In the figure below, $BD=DE=EC$, $F$ divides $AD$ so that $FA:FD=1:2$ and $G$ divides $AE$ so that $GA:GE=2:1$.
Find ratio of area of triangle $\Delta{AFG}$ to area of $\Delta {ABC}$

My Try:
I noticed that $G$ is centroid of $\Delta{ADC}$ and
$$Ar(ABD)=Ar(ADE)=Ar(AEC)$$
any clue?
 A: We have that
$$|AFG|\stackrel{|AD|=3|AF|}{=}\frac{1}{3}\cdot |AGD|\stackrel{3|AG|=2|AE|}{=}\frac{1}{3}\cdot\frac{2}{3}\cdot |ADE|\stackrel{|BC|=3|DE|}{=}\frac{1}{3}\cdot\frac{2}{3}\cdot\frac{1}{3}\cdot |ABC|.$$
A: HINT:

Lemma: Prove that $S_{ABC}=\dfrac{1}{2}AB\times AC\times\sin{BAC}$ and apply similarly for other angles

I will use the lemma above, but I will not prove it here.
Firstly, notice that $S_{ADE}$ has the same altitude as $S_{ABC}$, but $\dfrac{DE}{BC}=\dfrac{1}{3}$, so using the area formula $\dfrac{\text{base}\times\text{height}}{2}$, we will have $\dfrac{S_{ADE}}{S_{ABC}}=\dfrac{1}{3}$ (this step have not used the lemma yet).
$\Rightarrow S_{ADE}=\dfrac{1}{3}S_{ABC}$
Secondly, using the lemma above, we can prove that $S_{ADE}=\dfrac{1}{2}AD\times AE\times\sin{DAE}$ and $S_{AFG}=\dfrac{1}{2}AF\times AG\times\sin{DAE}$
$\Rightarrow \dfrac{S_{AFG}}{S_{ADE}}=\dfrac{\frac{1}{2}AF\times AG\times\sin{DAE}}{\frac{1}{2}AD\times AE\times\sin{DAE}}=\dfrac{AF}{AD}\times\dfrac{AG}{AE}$
Hope these hints are enough.
