Notes
four equilateral triangles inside a hexagon solution

Solution to the four equilateral triangles inside a hexagon Puzzle

Solution by Symmetry

The solution to this is perhaps easier to see if some of the lines are removed, as in the above diagram. The angles at the centre of the white regions are all $60^\circ$. As the hexagon is invariant under rotating by $60^\circ$, this means that when rotating anticlockwise by $60^\circ$ then $O B$ maps to $O D$, $O E$ to $O G$, and $O I$ to $O K$.

Therefore, rotating triangle $O L K$ by $60^\circ$ clockwise brings $O K$ along $O I$, creating a new region $O G H J$. Rotating this region by $60^\circ$ clockwise brings $O G to O E$, creating a new region $O D F H O$. Lastly, rotating this region by $60^\circ$ brings $O D$ to $O B$ with the final region $O A C F O$.

This is $\frac{1}{3}$ of the area of the hexagon.