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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 $DB’$.)

Blue
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    The phrase "25 times smaller" is sloppy and confusing. (See *Twice as less* by Elanor Orr.) – David G. Stork Jul 26 '19 at 18:06
  • @David G. Stork: Consistency is the hobgoblin of little minds. What actually applies here is FLT logic. That is, for n < 3, ‘n times less’ is, as you say, sloppy and confusing, but for n > 2, it is entirely appropriate. –  Jul 26 '19 at 20:38
  • “Resulting triangle”? There are lots of resulting triangles, along with quadrilaterals and a central hexagon. – John Bentin Jul 26 '19 at 20:46
  • @EulerSpoiler: I would never object to "two times the size" or anything like that. Never have, never will. But "25 times smaller" (as you still post) remains sloppy and frankly incorrect. And this of course has nothing whatsoever to do with "consistency." Frankly, I still don't think you get it. (Over and out.) – David G. Stork Jul 26 '19 at 21:25
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    FYI: Mathematicians Bényi and Ćurgus recently described a generalization of this result, allowing the points $A'$, $A''$, etc, to be anywhere along the side-lines of the triangle. (Their discussion is more-closely tied to Routh's Theorem, hence the title of their paper "A generalization of Routh's Triangle Theorem" (PDF).) They give a formula for the area of the "resulting triangle" in terms of the Ceva-like ratios determined by $A'$, $A''$, etc. – Blue Jul 27 '19 at 02:37
  • @David G. Stork: For a cry-baby spiel like that, you should say “Oeuvre and opera.”. –  Jul 27 '19 at 08:25

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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 Area of a triangle in the triangle with obtained by a geometric design where sides are divided in three equal parts 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: $$ \frac{FE}{BC} = \frac{FE}{A'A''} \cdot \frac{A'A''}{BC} = \frac 35\cdot\frac 13 =\frac 15\ . $$

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