they do meet at the center of the triangle. to see this let we $a, b, c$ represent the points. call the points $A_1, A_2$ on $BC$ such that $BA_1 = A_1A_2 = A_2C$ the point $a_1 = 2/3 b + 1/3 c, a_2=1/3b + 2/3 c$ similarly define points $b_1=1/3 c+2/3a, b_2 = 2/3 c+ 1/3 a.$ the point where all diameters meet is mid point of $A_1B_1$ given by $(a+b+c)/3.$