Proof that each of the three cevians is divided in the ratio $1:3:3$ Points $D,E,F$ are the first trisection points of $BC,CA,AB$ respectively. Let $[ABC]$ denotes the area of triangle $ABC$. If $[ABC]=1$, find $[GHI]$, the area of the shaded triangle. (the two images are from "The Art and Craft of Problem Solving" -Paul Zeitz)


attempt:
Let us look at the image below first

We know $AF:FB = 1:2$ ,this implies $[FIA]=x \implies [FBI]=2x$. Similarly $[DIB]=2y \implies [CID] = 4y$
Because $[CAF]:[CFB]=1:2$ and $[CFB]=6y+2x$, then $[CAF]=3y+x$ and $[CIA]=3y$.
We know the ratio $AF:AB = 1:3$, so ratio $[CAF]:[CFB]=1:2$. 
Why at some point we can conclude that each of the three cevians $AD, EB, CF$, are divided by the ratio $1:3:3$?
 A: You have stated in the question that $[CIA]=3y$. Thus, using this and $[CID] = 4y$ gives that $AI:ID = 3:4$. Similarly, $BG:GE = CH:HF = 3:4$.
Also, $[CAD] = [CIA] + [CID] = 3y + 4y = 7y$. In addition, using your diagram's area values, $[BAD] = x + 2x + 2y = 3x + 2y$. Since $[CAD]:[BAD] = 2:1$, we get
$$\frac{7y}{3x + 2y} = \frac{2}{1} \; \; \to \; \; 7y = 6x + 4y \; \; \to \; \; y = 2x \tag{1}\label{eq1}$$
Using $[FIA] = x$ and $[CIA]=3y$ (that you've determined) gives that $FI:IC = x:3y = x:6x = 1:6$. Also, I earlier showed $CH:HF = 3:4$, so since $1 + 6 = 3 + 4$, the base values used in both ratios are the same, and thus $FI:IH:HC = 1:3:3$.
It can be similarly proven (left to the reader) that the other $2$ cevians are divided by the same ratios.
A: The use of areas and ratio of areas suggest that barycentric coordinates (also named areal coordinates) could be a rather systematic way to consider this issue.
Let us show in particular (it is equivalent to the (1:3:3) proportionnality) that 
$$G=\tfrac47B+\tfrac37E \ \ \ (a) \ \ \ \ \text{and} \ \ \ \ H=\tfrac17B+\tfrac67E \ \ \ (b) \tag{1}$$
Normalized barycentric coordinates of $G$ and $H$ are shown (see Edit 1 below) to be :
$$G=[\tfrac17,\tfrac47,\tfrac27] \ \text{and} \ H=[\tfrac27,\tfrac17,\tfrac47] $$
As $B = [0,1,0]$ and $E = [\tfrac13,0,\tfrac23]$, checking (a) and (b) is immediate. 

Edit 1 : Barycentric equations of 
$$\begin{cases}\text{Line AD} &:& b-2c&=&0& B.C. : \{0,1,-2\}\\
\text{Line BE} &:& c-2a&=&0& B.C. : \{-2,0,1\}\\
\text{Line CF} &:& a-2b&=&0& B.C. : \{1,-2,0\}\end{cases}\tag{2}$$
where $B.C.$ denote (unnormalized) barycentric coordinates of the corresponding lines.
As is well known the barycentric coordinates of the crossing point of two lines are obtained as the cross product of their  B.C. 
Here, for example, the barycentric coordinates of $G=AD \cap BE$ are obtained by the cross product  (see (2)) :$$ \{0,1,-2\} \times \{-2,0,1\}= (1,4,2)$$
which, under its normalized form, gives $[\tfrac17,\tfrac47,\tfrac27]$ for the barycentric coordinates of $G$, as awaited.

Edit 2 : Back to areas
There is a noteworthy application  of barycentric coordinates : they provide easy calculations of areas. For example, the ratio (area of triangle GHI)/(area of reference triangle ABC) is equal to the determinant :
$$\det(G,H,I)=\begin{vmatrix}\tfrac17&\tfrac27&\tfrac47\\
\tfrac47&\tfrac17&\tfrac27\\
\tfrac27&\tfrac47&\tfrac17\end{vmatrix}=\frac17.$$
Connected... and interesting : https://en.wikipedia.org/wiki/One-seventh_area_triangle
A: We may and do assume that the area $[ABC]$ is one.
Following the arguments of the OP, we already have the following information regarding the areas cut / separated from $[ABC]$ by two cevians, $AF$ and $BD$:

In particular we have $2:1=CD:DB=\color{red}{7y}:\color{blue}{3x+2y}$,  which implies $7y=2(3x+2y)$, i.e. $y=2x$. Then we can immediately express the piece $x$ in terms of the total area 
$$
1=[ABC]=\color{red}{7y}+\color{blue}{3x+2y}=14x+3x+4x=21x
\ ,
$$
so $x=[AIF]=1/21$, and $y=2/21$.
By the same argument, and/or by cylcic shift in the notation, we have 
$$
[AIF]=[BGD]=[CHE]=\frac 1{21}\ .
$$
We also have so far $3y=[AIC]=6/21$. We remove the area of the triangle $\Delta CHE$ from this quantity, thus getting by symmetry the following quadrilateral areas:
$$
[AIHE]=[BGIF]=[CHGD]=\frac6{21}-\frac 1{21}=\frac 5{21}\ .
$$
For the triangle in the middle, the remained area is
$$
[GHI]=\frac{21-1-1-1-5-5-5}{21}=\frac 3{21}=\frac 17\ .
$$
In a picture:


Note: The above computed the area of the "inner" triangle $\Delta GHI$ without knowing the proportion $AI:IG:GD$. A posteriori, if we really need it, we can for instance compute $AI:ID=[CAI]:[CID]=3y:4y=3:4$, $AG:GD=[AGC]:[GDC]=6:1$.
Note: In such cases much of the effort is reduced by using directly the Theorem of Menelaus. For instance, applying it for $\Delta ABD$ with the line intersecting the sides in $F,C,I$, we can write immediately:
$$
\frac{FA}{FB}\cdot\frac{CB}{CD}\cdot\frac{ID}{IA}=1\ .
$$
So $AI:ID=3:4$.
A: 
Note that $[ACF] = [BAD] =[CBE] = \frac 13 [ABC]$ and observe that 
$$[GHI] = [ABC] - \left( [ACF] + [BAD] + [CBE] \right)+ S = S$$
$S$ is the sum of the shaded (purple) areas, which is equal to the area to be evaluated. Next, we evaluate one of the shaded areas $[CEH]$. Since,
$$\frac{[CEH]}{[FEH]}= \frac{[CEB]}{[FEB]}
= \frac{\frac 13 [ABC]}{\frac 23 [AEB]}= \frac{\frac 13 [ABC]}{\frac 23 \frac 23[ABC]}
=\frac 34 $$
we have,
$$[CEH] = \frac{3}{7} [CEF] = \frac 37\frac 13[CAF]
= \frac 37\frac 13 \frac 13[ABC]
=\frac{1}{21}$$
Similarly, the other two shaded areas $[AFI] = [BDG] = \frac{1}{21}$. Therefore, 
$$[GHI] = \frac{1}{21}+\frac{1}{21}+\frac{1}{21}=\frac 17$$
A: Here's an approach using the Extended Ceva's Theorem that I presented in an earlier answer way back in 2013.


Extended Ceva's Theorem. Consider points $D_1$, $D_2$, $E_1$, $E_2$, $F_1$, $F_2$ on the (extended) edges of $\triangle ABC$, with $D_i$, $E_i$, $F_i$ on the (extended) edge opposite vertex $A$, $B$, $C$, respectively.

Lines $\overleftrightarrow{D_1E_2}$, $\overleftrightarrow{E_1F_2}$, $\overleftrightarrow{F_1D_2}$ concur if and only if
$$\begin{align}
1 &= \frac{|BD_1|}{|D_1C|} \frac{|CE_1|}{|E_1A|} \frac{|AF_1|}{|F_1B|} 
+ \frac{|D_2C|}{|BD_2|} \frac{|E_2A|}{|CE_2|} \frac{|F_2B|}{|AF_2|} \\[6pt]
&+ \frac{|BD_1|}{|D_1C|} \frac{|D_2C|}{|BD_2|}
+ \frac{|CE_1|}{|E_1A|} \frac{|E_2A|}{|CE_2|}
+ \frac{|AF_1|}{|F_1B|} \frac{|F_2B|}{|AF_2|} \qquad\qquad \tag{$\star$}
\end{align}$$

Note: The above uses signed lengths, with $\overrightarrow{AB}$, $\overrightarrow{BC}$, $\overrightarrow{CA}$ indicating the direction of a positively-signed segment on each of the triangle's sides.

For the problem at hand, I'll use two of the given trisecting cevians at a time, along with a line through their intersection, parallel to a side of the triangle.
First, we consider trisecting cevians $\overline{BE}$ and $\overline{CF}$ with intersection $X$, with $\overline{PQ}$ through $X$, parallel to $\overline{AB}$. To match the elements of $(\star)$, we make the following assignments:
$$D_1 = P \qquad D_2 = C \qquad E_1 = E \qquad E_2 = Q \qquad F_1 = F \qquad F_2 = B \tag{1}$$

Then $(\star)$ becomes
$$\begin{align}
1 &= \frac{|BP|}{|PC|} \frac{|CE|}{|EA|} \frac{|AF|}{|FB|} 
+ \frac{|CC|}{|BC|} \frac{|QA|}{|CQ|} \frac{|BB|}{|AB|} + \frac{|BP|}{|PC|} \frac{|CC|}{|BC|}
+ \frac{|CE|}{|EA|} \frac{|QA|}{|CQ|}
+ \frac{|AF|}{|FB|} \frac{|BB|}{|AB|} \tag{2} \\[6pt]
&= \frac{|BP|}{|PC|}\cdot \frac12\cdot \frac12 + 0 + 0  
+ \frac12 \cdot \frac{|QA|}{|CQ|} + 0 \tag{3}
\end{align}$$
Now, defining
$$\lambda := \frac{|CX|}{|XF|} = \frac{|CP|}{|PB|} = \frac{|CQ|}{|QA|} \tag{4}$$
we have
$$1 = \frac1{\lambda} \cdot \frac14 + \frac12 \cdot \frac1{\lambda} \quad\to\quad \lambda = \frac34 \tag{5}$$

On the other hand, we can take $\overline{RS}$ through $X$, parallel to $\overline{AC}$. We make new assignments with respect to $(\star)$:
$$D_1 = B \qquad D_2 = R \qquad E_1 = C \qquad E_2 = E \qquad F_1 = S \qquad F_2 = F \tag{6}$$

This time, $(\star)$ becomes
$$\begin{align}
1 &= \frac{|BB|}{|BC|} \frac{|CC|}{|CA|} \frac{|AS|}{|SB|} 
+ \frac{|RC|}{|BR|} \frac{|EA|}{|CE|} \frac{|FB|}{|AF|} + \frac{|BB|}{|BC|} \frac{|RC|}{|BR|}
+ \frac{|CC|}{|CA|} \frac{|EA|}{|CE|}
+ \frac{|AS|}{|SB|} \frac{|FB|}{|AF|} \tag{7} \\[6pt]
&= 0 + \frac{|RC|}{|BR|} \cdot 2 \cdot 2 + 0 + 0
+ \frac{|AS|}{|SB|} \cdot 2 \tag{8}
\end{align}$$
Defining
$$\mu := \frac{|BX|}{|XE|} = \frac{|BR|}{|RC|} = \frac{|BS|}{|SA|} \tag{9}$$
we have ...
$$ 1 = \frac1{\mu} \cdot 4 + \frac1{\mu} \cdot 2 \qquad\to\qquad \mu = 6 \tag{10}$$

The result follows from $(5)$ and $(10)$, as well as the symmetry in how the ratios apply across all of the cevians. $\square$
