I'm trying to find a proof for this proposition, but I can't find an initial hint. Note: This proposition comes from numerical evidence (Geogebra). My first try was to use formulas from integral geometry (Crofton and Poincare Formula for convex domains), but I got stuck in the geometrical arguments about the expected intersections with lines and circles, respectively. I really think there might be an easier, clean, basic geometry way.
Let be 2 inscribed equilateral triangles in a circle. Prove that for the intersection $H$ of these triangles (an hexagon), the perimeter of $H$ multiplied by the circle diameter is 8 times the area of $H$ $$8A(H)=P(H)D$$
Solution: As Blue indicated in the comment, joining the circle center with the hexagon vertices give 6 congruent triangles. Trivially, the base of this triangle is $\frac{P(H)}{6}$ and the height is $D/4$. Therefore, $A(H)=\frac{6bh}{2}=\frac{P(H)D}{8}$
