Inscribed triangles and area 
As shown below, $\triangle GHJ$ is inscribed in $\triangle DEF$, which is inscribed in $\triangle ABC$. $GH\parallel AB$, $HJ\parallel BC$, $JG\parallel AC$, if $[ABC]=s$ and $[GHJ]=t$, find $[DEF]$. (These notations mean area)


Let $\begin{cases}\dfrac{BF}{FA}=x,\dfrac{AE}{EC}=z,\dfrac{CD}{DB}=y;\\[2pt]\dfrac{DH}{HF}=m,\dfrac{FG}{GE}=p,\dfrac{EJ}{JD}=n.\end{cases}$ Extend $HJ$ and intersect $AB$, $AC$. From $HJ\parallel BC$,
$$\frac{1+\dfrac x{1+m}}{\dfrac{\vphantom1xm}{1+m}}=\frac{z+\dfrac n{n+1}}{\dfrac1{1+n}}.$$
After a lot of work, $\dfrac{m+1}{n+1}=mx\cdot\dfrac{z+1}{x+1}$. Multiply the three cyclic equations to get$$xyzmnp=1.$$
I think this is useful, but I cannot solve out anything more.
 A: Using Ceva's Theorem or otherwise, we can prove the following lemma:

In the figure, if the $3$ sides of $\Delta ABC$ are respectively parallel to the $3$ sides of $\Delta A'B'C'$, then $AA'$, $BB'$ and $CC'$ are concurrent.

In the figure, let $[OBC]=s_1, [OHI]=t_1$, $[OCA]=s_2, [OIG]=t_2, [OAB]=s_3$ and $[OGH]=t_3$
Note that $s_1+s_2+s_3=s$ and $t_1+t_2+t_3=t$.
$(1)$ $\Delta ABC \sim \Delta GHI \implies \frac{HI}{BC}= \sqrt{\frac{t}{s}}$
$(2)$ $[HICB]=s_1-t_1 \implies [HID]=(s_1-t_1)\left(\frac{\sqrt t}{\sqrt s + \sqrt t}\right)$
$(3)$ Similarly $[GIE]=(s_2-t_2)\left(\frac{\sqrt t}{\sqrt s + \sqrt t}\right)$ and $[FGH]=(s_3-t_3)\left(\frac{\sqrt t}{\sqrt s + \sqrt t}\right)$
$(4)$ $\therefore$
\begin{align}
[DEF] & = [HID]+[GIE]+[FGH]+[GHI] \\
 & = (s_1-t_1)\left(\frac{\sqrt t}{\sqrt s + \sqrt t}\right)+(s_2-t_2)\left(\frac{\sqrt t}{\sqrt s + \sqrt t}\right)+(s_3-t_3)\left(\frac{\sqrt t}{\sqrt s + \sqrt t}\right)+t \\ 
 & = (s_1+s_2+s_3-t_1-t_2-t_3)\left(\frac{\sqrt t}{\sqrt s + \sqrt t}\right)+t \\
 & =  (s-t)\left(\frac{\sqrt t}{\sqrt s + \sqrt t}\right)+t\\
 & =  (\sqrt s- \sqrt t) \sqrt t +t\\ 
 & = \sqrt {st}
\end{align}
A: Let's put $\overrightarrow{AF}=\lambda\overrightarrow{AB}$, $\overrightarrow{BD}=\mu\overrightarrow{BC}$ and $\overrightarrow{CE}=\nu\overrightarrow{CA}$ where $(\lambda,\mu,\nu)\in\mathopen]0,1\mathclose[^3$.
Then let $x,y,z \in [0,1]$ be such that $\overrightarrow{EG}=x\overrightarrow{EF}$, $\overrightarrow{FH}=y\overrightarrow{FD}$ and $\overrightarrow{DJ}=z\overrightarrow{DE}$.
• We are going to express $x,y,z$ in terms of $\lambda,\mu,\nu$.
We deduce from these definitions that $\overrightarrow{GH}=\big(\color{blue}{\lambda(1-x)+(1-\lambda-\mu)y}\big)\overrightarrow{AB}
+ \big(\color{red}{(1-\nu)(1-x)-\mu y}\big)\overrightarrow{CA}$.
Since the lines $(AB)$ and $(GH)$ are parallel and because $\big(\overrightarrow{AB}\mathbin{,}\overrightarrow{CA}\big)$ is a base for the (vector) plane, the scalar before $\overrightarrow{CA}$ must be $0$, which can be written as $\color{red}{(1-\nu)x+\mu y=1-\nu}$.
The two conditions $(HJ)\parallel(BC)$ and $(GJ)\parallel(AC)$ provide similar relations between $x,y,z$ that lead to the linear system
$\begin{pmatrix}
\lambda&0&1-\mu\\
0&1-\lambda&\nu\\
1-\nu&\mu&0
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}
=\begin{pmatrix}
1-\mu\\
1-\lambda\\
1-\nu
\end{pmatrix}$.
Developed along the first column, its determinant is $\Delta=-\big(\lambda\mu\nu+(1-\lambda)(1-\mu)(1-\nu)\big)<0$,
and the unique solution is $\bbox[yellow,5px,border:1px solid]{(x,y,z)=(1-\mu,1-\nu,1-\lambda)}$

• Now the coefficient before $\overrightarrow{AB}$ above is $\color{blue}{\lambda(1-x)+(1-\lambda-\mu)y}$, that is (by developping)
$\lambda\mu+\mu\nu+\nu\lambda-\lambda-\mu-\nu+1=-\Delta$, and this is thus the ratio of the lengths $\frac{GH}{AB}\cdot$
Since $\triangle ABC$ and $\triangle GHJ$ are similar, the ratio of the areas is $\frac ts=\Big(\frac{GH}{AB}\Big)^2=\Delta^2$.
• On the other hand $\mathscr{A}(DEF)=s-\big( \mathscr{A}(DCE) + \mathscr{A}(EAF) + \mathscr{A}(FBD)\big)$.
Also, since $\overrightarrow{CD}=(1-\mu)\overrightarrow{CB}$ and $\overrightarrow{CE}=\nu\overrightarrow{CA}$, it follows that $\mathscr{A}(DCE)=\nu(1-\mu)\mathscr{A}(ABC)=\nu(1-\mu)s$.
Similarly $\mathscr{A}(EAF)=\lambda(1-\nu)s$ and $\mathscr{A}(FBD)=\mu(1-\lambda)s$.
Finally grouping these results we obtain $\mathscr{A}(DEF)=s-\big(\nu(1-\mu)+\lambda(1-\nu)+\mu(1-\lambda)\big)s
=(-\Delta) s=s\sqrt{\frac ts}=\sqrt{st}$.
• Remark 1.
In the example above, we choose $\lambda=\nu=\frac13$ and $\mu=\frac23$; we obtain $-\Delta=\lambda\mu+\mu\nu+\nu\lambda - \lambda-\mu-\nu+1=\frac29+\frac29+\frac19-\frac13-\frac23-\frac13+1=\frac29\cdot$
• Remark 2.
The calculations above provide several relations such as, for example, $\frac{FG}{FE}=1-x=\mu=\frac{BD}{BC}$, etc.
A: The angles of triangles $\triangle ABC$ and $\triangle GHJ$ are congruent, so the triangles are similar. Since corresponding sides of the triangles are parallel, the triangles are homeothetic. Let $O$ be the center of the homeothety.

Let $r = \dfrac{\lvert GH\rvert}{\lvert AB\rvert}.$
The triangles $\triangle FGH$ and $\triangle OGH$ have a combined area equal to the area of a triangle $\triangle XGH$ with the same shared base $GH$ and an altitude equal to the sum of the altitudes of $\triangle FGH$ and $\triangle OGH$ with respect to base $GH.$ But then the altitude of $\triangle XGH$ is equal to the altitude of $\triangle OAB.$ Hence the area of the quadrilateral $FGOH$ is
$$ [FGOH] = [\triangle FGH] + [\triangle OGH] = [\triangle XGH] = r [\triangle OAB].$$
Applying similar calculations for the quadrilaterals $DHOJ$ and $EGOJ,$
we have
\begin{align}
[DHOJ] &= r [\triangle OBC], \\
[EGOJ] &= r [\triangle OAC].
\end{align}
Since $FGOH,$ $DHOJ,$ and $EGOJ$ are a partition of $\triangle DEF,$
while $\triangle OAB,$ $\triangle OBC,$ and $\triangle OAC$ are a partition of
$\triangle ABC,$ we have
$$ [\triangle DEF] = r [\triangle ABC] .$$
Meanwhile by the similarity of $\triangle GHJ$ and $\triangle ABC$ and the fact that
$\lvert GH\rvert = r \lvert AB\rvert,$
$$ [\triangle GHJ] = r^2 [\triangle ABC] .$$
It follows that the area of $\triangle DEF$ is the geometric mean of
$[\triangle ABC] = s$ and $[\triangle GHJ] = t,$ that is,
$$ [\triangle DEF] = \sqrt{st}. $$
