Question 405635
Our diagram looks something like this:


{{{drawing(400, 400, -10, 10, -10, 10,


triangle(-3, -3, 3, -3, -1, 4),
triangle(-3, -3, 3, -3, 0, -8.196),
triangle(-3, -3, -1, 4, -7, 2),
triangle(3, -3, -1, 4, 7, 4),
locate(-3, -3, A),
locate(3, -3, B),
locate(-1, 4, C),
locate(0, -8.196, D),
locate(7, 4, E),
locate(-7, 2, F),
line(-3, -3, 7, 4),
line(-1, 4, 0, -8.196),
line(3, -3, -7, 2)


)
}}}
(It may appear that these three lines intersect at a point, but that might not necessarily be the case!)


This may seem like an unusual solution, but I believe it is by far the easiest method, and more efficient than trying to brute-force the problem. I will begin by introducing complex numbers into the problem. We can assume that points A, B, ..., F are points on the complex plane.


Suppose that A is the origin of the complex plane, i.e. A = 0, and that B and C are arbitrary complex numbers. It follows that:


{{{F = C*omega}}} and {{{B = D*omega}}} where {{{omega = e^(i*pi/3) = cos(pi/3) + i*sin(pi/3) = 1/2 + (sqrt(3)/2)i}}} (The Greek letter "omega" denotes a root of unity, in this case, a 6th root of unity. These equations basically state that F is equal to point C rotated about {{{pi/6}}}, and B is the same as D rotated about {{{pi/6}}}.) We want to show that CD and BF have equal magnitudes. Since CD = D - C and BF = F - B (this is similar to subtracting vectors), then we want to show that

 
|D - C| = |F - B| where |z| denotes the magnitude of the complex number z.


Note that {{{F = C*omega}}} and {{{B = D*omega}}}. We can substitute these into our equation to obtain


|D - C| = |(C - D)*omega|. Here, we note that the magnitude of D-C is the same as C-D (i.e. a complex number in the form a+bi has the same magnitude as the number -a-bi). Also, the magnitude of {{{omega}}} is equal to {{{sqrt((1/2)^2 + (sqrt(3)/2)^2)}}} which is 1. Hence, the magnitude remains unchanged, and we conclude that |D - C| = |F - B| and that CD = BF. By a simple symmetry argument (e.g. assigning B as the origin to show that BF = AE), we conclude that CD = BF = AE, and we are done.