Ratio of area of triangle and hexagon I am looking for the proof of the following claim:

Claim. If the sides of the triangle are  partitioned into $n$ equal segments for $n$ an even integer and each division point adjacent to the corresponding central division point  is connected to the opposite vertex, a central hexagon which has an area of $\frac{32}{(3n-2)(3n+2)}$ of the area of the original triangle  is still formed.


GeoGebra applet that demonstrates this claim can be found here.
I tried to adapt the proof of the Marion's theorem given on this page. The problem I encountered is the following: How to find a trilinear coordinates of the division points : $D,E,F,G,H,I$ . The only formula I am aware of is the formula for the first n-multisection.
 A: We can use the Routh-Steiner theorem, which gives a method to compute the area of a sub-triangle bounded by three cevians.
The three medians of $ABC$ (meeting at the centroid $X$) divide the central hexagon into six triangles.  The triangle $LMX$ is bounded by cevians defined by side ratios $x=y=1,z=(n+2)/(n-2)$.
By Routh-Steiner
$$
\begin{align}
\cfrac{|LMX|}{|ABC|} &=
 \cfrac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)} \\
&= \cfrac{\cfrac{16}{(n-2)^2}}{(3)\left(\cfrac{3n+2}{n-2}\right)\left(\cfrac{3n-2}{n-2}\right)} \\
&= \cfrac{16}{3(3n+2)(3n-2)}
\end{align}
$$   The other five triangles are built using the same ratios, so have the same area as $LMX$.  Thus the area of the hexagon as a fraction of the large triangle is
$$
\cfrac{|JKLMNO|}{|ABC|}=6 \left(\cfrac{16}{3(3n+2)(3n-2)}\right)=\cfrac{32}{(3n+2)(3n-2)}.
$$
Note that the construction and formula also work for non-integer values of $n.$  Here, $n=AC/HR.$
A: I will provide a picture and go into detail, but need my notations first - as in the following picture:

Let $\Delta ABC$ be the given triangle. Let $A'$, $B'$, $C'$ be the mid points of the sides respectively opposed to $A$, $B$, $C$. We denote by $A_-$ and $A_+$ the two division points ($\ne A'$) to the left and to the right of $A'$ on the side $BC$, then use similar notations for the other sides,
so that we have
$$
\begin{aligned}
A_-A'&=A'A_+=xBC=xa\ ,\\
B_-B'&=B'B_+=xCA=xb\ ,\\
C_-C'&=C'C_+=xAB=xc\ ,
\end{aligned}
$$
where $x$ is one and the same proportion. We can let it be general,
in the OP it is $\displaystyle \color{blue}{\frac1n}$.
Let $A^*$ be the intersection of the cevians $BB_+$, $CC_-$, and the median $AA'$.
Let $A^!$ be the intersection of the cevians $BB_-$, $CC_+$, and the median $AA'$.
Similarly we construct $B^*$, $B^!$; $C^*$, $C^!$.
The OP is immediately solved if we answer the following questions:

*

*Which is the position of $A^*$ on the segment $GA$?

*Which is the position of $A^!$ on the segment $GA'$?
(Try to get it on your own.) Applying Menelaus in $\Delta AGB'$ w.r.t. the line $BA^*B+$, and w.r.t. the line $BA^!B_-$ we have:
$$
\begin{aligned}
1 &=
\frac {A^*G}{A^*A} 
\cdot 
\frac {B_+A}{B_+B'} 
\cdot 
\frac {BB'}{BG} 
=
\frac {A^*G}{A^*A} 
\cdot 
\frac {\frac 12-x}{x} 
\cdot 
\frac 32\ ,
&\text{ so }\qquad
\frac {GA^*}{GA} 
&=
\frac {4x}{3-2x}\ ,
\\
1 &=
\frac {A^!G}{A^!A} 
\cdot 
\frac {B_-A}{B_-B'} 
\cdot 
\frac {BB'}{BG} 
=
\frac {A^!G}{A^!A} 
\cdot 
\frac {\frac 12+x}{x} 
\cdot 
\frac 32\ ,
&\text{ so }\qquad
\frac {GA^!}{GA} 
&=
\frac {4x}{3+2x}\ ,
\\
&
&\text{ so }\qquad
\frac {GA^!}{GA'} 
&=
\frac {8x}{3+2x}\ .
\end{aligned}
$$
Similar computations can be done for $B^*$, $B^!$, and $C^*$, $C^!$.
This gives the proportions of areas:
$$
\frac{[\color{magenta}G\color{blue}{A^*B^!}]}{[\color{magenta}{G}AB']}
=
\frac {\color{magenta}{G}\color{blue}{A^*}}{\color{magenta}{G}A}\cdot
\frac {\color{magenta}{G}\color{blue}{B^!}}{\color{magenta}{G}B'}
=
\frac {4x}{3-2x}\cdot
\frac {8x}{3+2x}
=
\color{blue}{
\frac{32}{(3n-2)(3n+2)}}
\ .
$$
This is also the final answer since the medians cut the triangle $\Delta ABC$ in six smaller triangles of same area. This matches the claimed formula.
$\square$

Here is also explicitly the situation $n=4$:

We have placed

*

*on $GA$, $GB$, $GC$ points to illustrate $GA^*:GA=2:5$ (and similar proportions for the other two segments),
and

*on $GA'$, $GB'$, $GC'$ points to illustrate
$GA^!:GA'=4:7$ (and similar proportions for the other two segments).

