Trisecting the sides of a triangle.

Question

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.

Answer 1

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 $(a+b+c)/3.$

Answer 2

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. $\square$

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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. $\square$