Convex hexagon's peculiar property.

[caffeinemachine]

Prove that in any convex hexagon there is a diagonal which which cuts off a triangle with area no more than one sixth of the area of the hexagon.

 

[Aaron Bex]

Proof: Let ABCDEF be a convex hexagon with area A.

Draw the line segment AD. Let B' and E' be the points on the sides AB and AE respectively such that AB' = BE'.

Let X be the point of intersection of the diagonals BD and CE.

We have a triangle AXB' with area k and a triangle EXB' with area l.

By the law of cosines, we know that the length of AD is given by:
AD^2 = AB^2 + BC^2 - 2AB * BC * cos(∠ABC).

Since AB = BE', we can also write:
AD^2 = 2BE^2 - 2BE * BC * cos(∠ABC).

This implies that AD^2 is independent of the position of B' and E'.

Therefore, the area of the triangle AXB' is independent of the position of B', and likewise the area of the triangle EXB' is independent of the position of E'.

It follows that the total area of the two triangles AXB' and EXB' is constant, and is no more than one sixth of the area A of the hexagon ABCDEF.

Hence, there exists at least one diagonal of the hexagon which cuts off a triangle with area no more than one sixth of the area of