The line through Centroid $G$ divides the triangle $ABC$ into two figures. I need help with formal proof that the ratio between the areas of triangle ABC and the one that was sectioned is between $2$ and $2.25$
$r∩ΔABC=\left\{ D,E \right\}$
distance between point and line
$dCr=dAr+dBr$
$\begin{array}{} \text{similar triangles:} & ΔAEJ∼ΔCEI & ΔBDK∼ΔCDI \end{array}$
$\begin{array}{} \frac{dAr}{dCr}=\frac{AE}{CE} & \frac{dAr+dCr}{dCr}=\frac{AE+CE}{CE}=\frac{b}{CE} & CE=\frac{b·dCr}{dAr+dCr} \end{array}$
$\begin{array}{} \frac{dBr}{dCr}=\frac{BD}{CD} & \frac{dBr+dCr}{dCr}=\frac{BD+CD}{CD}=\frac{a}{CD} & CD=\frac{a·dCr}{dBr+dCr} \end{array}$
$ratioC=\frac{Δ(ABC)}{Δ(CDE)}$
$\begin{array}{} Δ(ABC)=\frac{1}{2}·a·b·sin(\hat{C}) & Δ(CDE)=\frac{1}{2}·CE·CD·sin(\hat{C}) \end{array}$
$\begin{array}{} rC=\frac{\left( dAr+dCr \right)\left( dBr+dCr \right) }{dCr^2} & rC=\frac{\left( 2·dAr+dBr \right)\left( dAr+2·dBr \right) }{\left( dAr+dBr \right)^2 } \end{array}$
$\begin{array}{} if & dBr=0 & rC=\frac{2·dAr·dAr}{dAr^2}=2\\ if & dAr=dBr & rC=\frac{3·dAr·3·dAr}{\left( 2·dAr \right)^2 }=\frac{9}{4}=2.25\end{array}$
The Euler line is a particular case of the line $r$ passing through the centroid
$rA=|\frac{\left| \begin{array}{} -m & l & 0 & 0 \\ x(A) & y(A) & 1 & 0 \\ x(B) & y(B) & 1 & 0\\x(C) & y(C) & 1 & 1 \\ \end{array} \right|·\left| \begin{array}{} -m & l & 0 & 0 \\ x(A) & y(A) & 1 & 0 \\ x(B) & y(B) & 1 & 1\\x(C) & y(C) & 1 & 0 \\ \end{array} \right|}{(l·x(A)+m·y(A)+n)^{2}}|$
$rB=|\frac{\left| \begin{array}{} -m & l & 0 & 0 \\ x(A) & y(A) & 1 & 0 \\ x(B) & y(B) & 1 & 0\\x(C) & y(C) & 1 & 1 \\ \end{array} \right|·\left| \begin{array}{} -m & l & 0 & 0 \\ x(A) & y(A) & 1 & 1 \\ x(B) & y(B) & 1 & 0\\x(C) & y(C) & 1 & 0 \\ \end{array} \right|}{(l·x(B)+m·y(B)+n)^{2}}|$
$rC=|\frac{\left| \begin{array}{} -m & l & 0 & 0 \\ x(A) & y(A) & 1 & 0 \\ x(B) & y(B) & 1 & 1\\x(C) & y(C) & 1 & 0 \\ \end{array} \right|·\left| \begin{array}{} -m & l & 0 & 0 \\ x(A) & y(A) & 1 & 1 \\ x(B) & y(B) & 1 & 0\\x(C) & y(C) & 1 & 0 \\ \end{array} \right|}{(l·x(C)+m·y(C)+n)^{2}}|$
$\begin{array}{} \text{Euler line:} & l·x+m·y+n=0 \end{array}$
The ratio $r$ is between $2$ and $2.25$ as shown by “Geogebra” ($r = 2$ when the Euler line passes through one of the vertices of the triangle ABC and $r = 2.25$ when the Euler line is parallel to one side of the triangle $ABC$. The parallelism condition is given by:
$\begin{array}{} \text{Euler line parallel to the side a} & \text{Euler line parallel to the side b} & \text{Euler line parallel to the side c} \\ \frac{b^2+c^2}{a^2}+\left( \frac{b^2-c^2}{a^2} \right)^2=2 & \frac{a^2+c^2}{b^2}+\left( \frac{a^2-c^2}{b^2} \right)^2=2 & \frac{a^2+b^2}{c^2}+\left( \frac{a^2-b^2}{c^2} \right)^2=2 \\ \end{array}$
If we consider the sectioned areas (triangle and quadrilateral) the ratio $r'$ will be defined as: $r'=\frac{1}{r-1}$ $(0.8\le r'\le1)$
If you are so inclined to fill out the details, then you can solve it by a simple Menelaus. Let $ED$ intersect $AB$ at $F.$ If the two are parallel, then the ratio is precisely $\dfrac 94$, so we can assume they must intersect. Then, draw the median $CM$, passing through $G$, and write the Menelaus' theorem for the triangles $\triangle CMB$ and $\triangle CMA.$ After manipulating the ratios obtained from them, you will find that:
$$\dfrac{CB}{CD}+\dfrac{CA}{CE} = 3.$$
Call these two ratios $x$ and $y$, then by your assumption, $x,y\in[1,2]$ and now we know that $x+y = 3.$ Now AM-GM immediately yields that $xy\leq\dfrac 94.$
For the other direction:
$$xy-2 = x(3-x) - 2 = (x-1)(2-x)\geq 0.$$ Finally, notice that the ratio of area which are you trying to bound is precisely: $$\dfrac{\triangle CAB}{\triangle CED} = \dfrac{CA\cdot CB}{CE\cdot CD} = xy$$ and we are done.