Find the area of ​the shaded region in the triangle below? For reference: In figure $G$ is the centroid of the triangle $ABC$; if the area of ​​the $FGC$ triangle is $9m^2$, the area of ​​the FGB triangle is $16m^2$
Calculate the area of ​​the shaded region. (Answer:$7m^2$)
If possible by geometry
My progress:

$S_{FGC} = \frac{b.h_1}{2} = \frac{FG.h_1}{2}\implies FG = \frac{18}{h_1}\\
S_{FGB}=\frac{b.h_2}{2} = \frac{FG.h_2}{2} \implies FG = \frac{32}{h_2}\\
\therefore \frac{18}{h_1} = \frac{32}{h_2}\implies \frac{h_1}{h_2} = \frac{32}{18}=\frac{16}{9}\\
S_{ABG} = S_{BCG} = S_{ACG}$
...??? I'm not able to develop this
 A: With the usual conventions:
\begin{aligned}
\overrightarrow{GF}\times\overrightarrow{GB} &= (0,0,32)\\
\overrightarrow{GF}\times\overrightarrow{GC} &= (0,0,-18)\\
\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC} &= \overrightarrow{0}
\end{aligned}
The area you search:
\begin{multline*}
\frac{1}{2}\big\lvert\overrightarrow{GF}\times\overrightarrow{GA}\big\rvert=\frac{1}{2}\Big\lvert\overrightarrow{GF}\times\big(-\overrightarrow{GB}-\overrightarrow{GC}\big)\Big\rvert
=\frac{1}{2}\big\lvert-\overrightarrow{GF}\times\overrightarrow{GB}-\overrightarrow{GF}\times\overrightarrow{GC}\big\rvert=\frac{1}{2}\big\lvert(0,0,-14)\big\rvert = 7
\end{multline*}
A: This can be directly deduced from a known property that establishes the relationship between altitudes from vertices of a triangle to a secant passing through its centroid.

In the given diagram, if we draw altitudes from vertices $A, B$ and $C$ to the line through $FG$ where $G$ is the centroid. we have the below relationship:
$h_b = h_a + h_c$, where $h_a$ is altitude from $A$ to $FG$, $h_b$ is altitude from $B$ to $FG$ and $h_c$ is altitude from $C$ to $FG$ extend.
So,
$S_{\triangle FAG} = \frac 12 \cdot FG \cdot h_a = \frac 12 \cdot FG \cdot (h_b - h_c) $
$ = S_{\triangle FBG} - S_{\triangle FCG} = 16 - 9 = 7$

If you want to show $h_b = h_a + h_c$, note that
$h_d = \frac 12 (h_a + h_c)$ as $D$ is the midpoint of $AC$.
Now as $BG = 2 DG$, we have $h_b = 2 h_d = h_a + h_c$
A: Let the distance between vertex $X$ and line $FG$ be $h_X$ ($X=A,B,C$), then $S_{FGX}=\dfrac{1}{2}FG \cdot h_X$. I will prove that in your case $h_B=h_A+h_C$, thus $S_{FGB}=S_{FGA}+S_{FGC}$.
Since $G$ is the centroid, we have $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}$. Project the equation to the right-hand orthogonal direction $\overrightarrow{n}$ of $\overrightarrow{GF}$ we get $(-h_B+h_A+h_C)\overrightarrow{n}=\overrightarrow{0}$, and the proof is over.
A: Here is yet an other solution using only the simple properties of areas, and the position of the centroid $G$ on the medians. Let $A'$, $B'$, $C'$ be the mid points of the sides $BC$, $CA$, $AB$. Let $X$ be the intersection of the line $GF$ with the line $AB$. In a picture:

Then we have three equal proportions of areas:
$$
\frac{[FGA]}{[XGA]} =
\frac{[FGB]}{[XGB]} =
\frac{[FGC]}{[XGC]} =
\frac{FG}{FX} \ ,
$$
so the relation between areas $[FGB]=[FGA]+[FGC]$
is equivalent to the same relation obtained by replacing $F$ by $X$. So let us show this relation for $X$, using the areas $[GAC']$ and $[GC'B]$ as intermediates. The following computation applies in case $X$ is between $C'$ and $A$.
$$
\begin{aligned}{}
[GXB]-[GAX]
&=
[GXC'] + [GC'B] -[GAX]\\
&=
[GXC'] + [GAC'] -[GAX]\\
&=
2[GXC']=2[XC'G]\\
&= [XGC]\ .
\end{aligned}
$$
It is now left to the reader to consider the cases when $X$ is between $C'$ and $B$, respectively not on the segment $AB$ - if they can be realized.
A: $\begin{array}{} A=(p,q) & B=(a,0) & C=(0,0) & F=(x,y) \end{array}$
$S_{FGC}=|\frac{1}{2}\left| \begin{array}{} x & y & 1 \\ x_{G} & y_{G} & 1 \\ 0 & 0 & i \\ \end{array} \right| |=9 \\ S_{FGB}=|\frac{1}{2}\left| \begin{array}{} x & y & 1 \\ x_{G} & y_{G} & 1 \\ a & 0 & i \\ \end{array} \right| |=16$
$\begin{array}{} |x·y_{G}-x_{G}·y|=18 &  & |x·y_{G}-x_{G}·y+a·y-a·y_{G}|=32  \end{array}$
Solving the modular equations we get four solutions. Replace $x_{G}=\frac{a+p}{3}$ and $y_{G}=\frac{q}{3}$
$\begin{array}{} \left( x=\frac{aq(a+p)±6(2a-25p)}{3a},y=\frac{aq∓150}{3a}  \right)  & ⇒ & S_{AFG}=7 \\ \left( x=\frac{aq(a+p)±6(34a+7p)}{3a},y=\frac{aq±42}{3a}  \right)  & ⇒ & S_{AFG}=25 \\ \end{array}$

