geometry - One of the diagonals in a hexagon cuts of a triangle of area $leq 1/6^{th}$ of the hexagon

Problem: Show that, in a convex hexagon, there exists a diagonal which cuts off a triangle of area not more than one-sixth of the hexagon.

My attempt: Suppose we have a hexagon $ABCD$. There are two possible cases: either the main diagonals are concurrent, or they are not.

If the main diagonals $AD, BE, CF$ concur at a point $G$, then the main diagonals cut the hexagon into $6$ triangles, atleast one of which has area $\leq \frac 16 [ABCDEF]$ Suppose one such triangle is $DEG$. Thus one of the triangles $DEF$ or $DEC$ has area $\leq[DEG]$, and we are done.

But suppose the main diagonals are not concurrent, i.e., they form a triangle $PQR$. How can I prove the statement in this case?

Answer

Consider the six triangles ABQ, BCQ, CDR, DER, EFP, FAP. They are disjoint and cover an area smaller than the entire hexagon. And each of them reaches up to a diagonal that doesn't touch their base.