Finding the area of a triangle

Let $BD=x$ . Note that $AF=AG$ , $FD=GE$ and $BD=CE=x$ . Since $FE || DC$ , we have:

$\begin{aligned} \frac {EC}{AE} & = \frac {FD}{AF} \\ \frac {BD}{AG+GE} & = \frac {10}8 \\ \frac {BD}{AF+FD} & = \frac {10}8 \\ \frac x {8+10} & = \frac {10}8 \\ \implies x & = \frac {18 \times 10}8 = 22.5 \\ \implies AB & = AF+FD+BD \\ & = 8 + 10 + 22.5 \\ & = 40.5 \end{aligned}$

Since $AB=AC=BC=40.5$ , $triangle ABC$ is equilateral and its area is $\dfrac 12 \times 40.5^2 \times \dfrac {\sqrt 3}2 \approx \boxed{710.24908}$

Consider $\triangle ADC$

$\dfrac{AF}{FD}=\dfrac{AE}{EC}$

$\dfrac{8}{10}=\dfrac{AE}{EC}$

$\dfrac{4}{5}=\dfrac{AE}{EC}$ $\color{#D61F06}(1)$

Consider $\triangle ABC$

$\dfrac{AD}{DB}=\dfrac{AE}{EC}$

$\dfrac{18}{DB}=\dfrac{AE}{EC}$ $\color{#D61F06}(2)$

Equate $\color{#D61F06}(1)$ and $\color{#D61F06}(2)$

$\dfrac{4}{5}=\dfrac{18}{DB}$

$4(DB)=(5)(18)$

$DB=22.5$

It follows that

$AB=8+10+22.5=40.5$

Therefore, $\triangle ABC$ is an equilateral triangle.

The formula for the area of an equilateral triangle is given by $A=\dfrac{s^2\sqrt{3}}{4}$ where $s$ = side length

Substituting, we obtain

$A=\dfrac{40.5^2\sqrt{3}}{4}=$ $\color{#3D99F6}\boxed{710.24908}$ $\boxed{\text{answer}}$