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 $1:3:3$?


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, 
$$[GHI] = \frac{1}{21}+\frac{1}{21}+\frac{1}{21}=\frac 17$$
