Prove that in an arbitrary convex hexagon there is a diagonal that cuts off from it a triangle whose area does not exceed 61 of the area of the hexagon. What are the properties of a convex hexagon, each diagonal of which is cut off from it is a triangle whose area is not less than 61 the area of the hexagon? geometrygeometric inequalityhexagonconvexdiagonalareaUkrainian TYM