Edit: What was I thinking? Was I even already fully awake when I posted this? Kyaaaah! *facepalm*

^ What? Drawing all the diagonals from a vertex cuts the hexagon to four triangles only. (FYI: TheChainheartMachine originally posted a solution along the lines of "diagonals from a vertex cut a hexagon to six triangles", that is now edited out.)

First suppose that the main diagonals pass through one point and use pigeon hole . Domi

Amir Hossein wrote: Prove that in every convex hexagon of area $S$ one can draw a diagonal that cuts off a triangle of area not exceeding $\frac{1}{6}S.$ It's obvious if the primary diagonals pass through one point, say O. Let the hexagon be $ABCDEI$. By primary diagonals, I mean diagonals that do not directly cut off a triangle, like AD,BE and CI. Let $\triangle EOD$ be the triangle with the required property. Now one of the triangles $\triangle ECD$ or $\triangle EID$ has area less that $\triangle EOD$ and we are done. Now, let the primary diagonals make a triangle FGH, with F&G on IC, H on BE & AD. Look at the triangles - $\triangle ABF,BFC,GDC,AHI,EGD,EHI$ From the preceding principle there exists one triangle satisfying the given requirement.