Trisecting the sides of a triangle. Consider the hexagon formed by the six points which trisect the sides of a triangle(two on each side).  Is is true that when we connect opposite points in this hexagon, the lines intersect at a single point ?  I think this is most likely true but I cant prove it.
 A: Yes, the diagonals meet at a common point.

Certainly, the diagonals are parallel to, and two-thirds the length of, corresponding sides of the triangle. Consider a sub-triangle (say, $\triangle AF_2 E_1$) created by one of the diagonals and an appropriate vertex. The other diagonals ($F_1D_2$ and $E_2 D_1$) are parallel to sides ($AF_2$ and $AE_1$) of that triangle at the corresponding midpoints; those diagonals, then, necessarily meet the midpoint ($M$) of the remaining side (namely, diagonal $F_2 E_1$) at their own midpoints. $\square$

Alternatively (though less-enlighteningly), you could use my Extended Ceva's Theorem from this answer, which gives this condition for concurrence:

$$\begin{align}
1 &
= \frac{|BD_1|}{|D_1C|}\;\frac{|CE_1|}{|E_1A|}\;\frac{|AF_1|}{|F_1B|} + \frac{|D_2C|}{|BD_2|}\;\frac{|E_2A|}{|CE_2|}\;\frac{|F_2B|}{|AF_2|} \\[4pt]
&+\frac{|BD_1|}{|D_1C|}\;\frac{|D_2C|}{|BD_2|}+\frac{|CE_1|}{|E_1A|}\;\frac{|E_2A|}{|CE_2|}+\frac{|AF_1|}{|F_1B|}\;\frac{|F_2B|}{|AF_2|}
\end{align}$$

In this situation, all ratios are $\frac12$, so the condition is satisfied and the lines are concurrent. $\square$
A: 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 $(a+b+c)/3.$
A: Not too hard, borrowing the diagram above.
Easy to prove triangles F1E2M and D2D1M are congruent (using similar triangles to first show F1E2||BC and F1E2=BC/3=D1D2).
Thus E2M=D1M and F1M=D2M.
M is the midpoint of F1D2 and E2D1.
Join F2 and M.
Triangles F1F2M and F1BD2 must be similar (two sides in same ratio and included angle)
Using corresponding angles, F2M||BC.
The same way we can show E1M||BC
F2ME1 must be a straight line, so concurrency is proved.
