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

Solution 1:

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*}

Solution 2:

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 $h_b = 2 h_d = h_a + h_c$