Let a hexagone with a diameter $D$ be given and let $d>\frac D 2.$ On each side of the hexagon one constructs a isosceles triangle with two equal sides of length $d$. Prove that the sum of the areas of those isoscele triangles is greater than the area of a rhombus with side lengths $d$ and a diagonal of length $D$. (The diameter of a polygon is the maximum of the lengths of all its sides and diagonals.)