Notes
crossed hexagon solution

Crossed Hexagon

Solution by Similar Triangles and Vertically Opposite Angles

In triangles $B G D$ and $A G E$, the sides $B D$ and $A E$ are parallel, and the angles $B \hat{G} D$ and $E \hat{G} A$ are vertically opposite and so equal. Therefore, triangles $B G D$ and $A G E$ are similar. Since $A E$ has twice the length of $B D$, the length scale factor is $2$ and the area scale factor is $4$. The length of $F G$ is twice that of $C G$, so $C G$ has length one third of that of $C F$. Triangle $B F D$ is an equilateral triangle with area one sixth of the area of the whole hexagon. So the area of $B G D$ is $\frac{1}{18}$th of the area of the whole hexagon. Triangle $A G E$ then has area four times that, so has area $\frac{4}{18}$ths of the whole hexagon. Therefore, the area of the shaded region is $\frac{10}{18} = \frac{5}{9}$ths of the hexagon.

Solution by Crossed Trapezium

The quadrilateral $A B D E$ with its diagonal lines is a crossed trapezium. In such, the areas of the four triangles are in the ratio $1 : s : s : s^2$ where $s$ is the ratio of the parallel sides. Therefore the ratio of the area of the shaded region to the whole trapezium is $\frac{1 + s^2}{1 + 2 s + s^2}$. As $s = 2$, this is $\frac{5}{9}$. Since the other half is a copy of this trapezium, that ratio then holds for the whole hexagon.