Morley Theorem variant: Trisecting a triangle's sides instead of its angles 
What happens when the Morley process is followed, but modified so as to use trisection of the opposite side instead of trisection of the vertex angle, for the three vertices? 

This Go Geometry page claims that the area of the resulting triangle is (exactly) 25 times smaller than the area of the original triangle. Three proofs are given. 

It is the first of the three proofs that I am interested in, but am having difficulty following it, starting at the very first line. This is the proof as given (including lack of formatting):

  AH/HB" * 1/3 * 2/1 = 1; hence AH/HB" = 3/2
  BD/DB" * 1/3 * 2/1 = 1; hence BD/DB" = 3/2
  Or HD is parallel to AC & HD/BB" = AH/(AH+HB")= 3/5 or HD/AC = 3/15 = 1/5
  Similarly HF is parallel to BC and HF/BC = 1/5
  and FD is parallel to AB and FD/AB = 1/5
  Thus, Tr.HFD is similar to Tr.ABC and therefore their areas are proportional to the squares of their sides, so Tr. HFD/ Tr. ABC =(1/5)^2 = 1/25 or S1 = S/25
From where, in the first line of the first proof, do the $1/3$ and $2/1$ come from? And, is the rest of the proof in fact free of typos? (I’m thinking that, in the second line of the proof, $DB’’$ should be $DB’$.)

 A: The posted problem is beautiful, the idea to show $DE\|BC$ (with notations as in the picture below) taken from the link is a quick way to proceed, but the presentation in the offered link is ugly and buggy, to be mild with the circumstances. So let us restate.

In the picture

we start with the triangle $\Delta ABC$, then take division points $A',A''$, and $B', B''$, and $C',C''$ (cyclically in this order on the sides) so that they divide in three equal parts the sides $BC$, and $CA$, and respectively $AB$.
Construct intersection $D,E,F$ of cevians as in the picture.

The problem asks to find the proportion
$$
\frac
{
\operatorname{Area}(\Delta DEF)
}
{
\operatorname{Area}(\Delta ABC)
}
\ .
$$

Two of the cevians $AA'$ and $AA''$, were drawn in a darker color. We show where is the positions of the points $E,F$ on them. The same argument applies than similarly for the other cevians from $B$, and from $C$.
We apply the theorem of Menelaos in the triangle $\Delta AA'C$ with respect to the transversal line $BFB''$. Note that there is a sign $-1$ in some fractions because segments on the same are considered to be oriented (w.r.t. some fixed orientation for each line).
We get:
$$
\begin{aligned}
1 &= 
\frac{FA}{FA'}\cdot
\frac{BA'}{BC}\cdot
\frac{B''C}{B''A}\ ,&&\text{ i.e.}
\\
1 &= 
\frac{FA}{FA'}\cdot
\frac 13\cdot
\frac{-2}{1}\ .
&&\text{ This gives:}
\\
\frac{AF}{FA'}
&=
\frac 32\ .
&&\text{ Derived proportions:}
\\
\frac{AF}{AA'}
&=
\frac{AF}{AF+FA'}
=\frac{3}{3+2}=\frac 35\ .
&&\text{ Similarly:}
\\
\frac{AE}{AA''}
&=
\frac 35\ .
\end{aligned}
$$
The equality of proportions implies $EF\| BC$.
Similarly $FD\| CA$, and $DE\| AB$.
To see the needed proportion of areas (of the two similar triangles), we need only to compute the proportion of corresponding sides. One more line:
$$
\frac{FE}{BC}
=
\frac{FE}{A'A''}
\cdot
\frac{A'A''}{BC}
=
\frac 35\cdot\frac 13
=\frac 15\ .
$$
