So, let’s call the intersection of BF and DE, ‘G’. You probably know that bisects both segments.
Let’s call where AG intersect BC, ‘H.’
Let’s call where CD intersects EF, ‘I’.
Let’s call where AI intersects BC, ‘J’.
You want to know if BH=HJ=JC.
No.
Assume the length of the sides of the triangle are 1 unit =AB=BC=CA.
The midpoint of AC, which is F, creates the length, .5=AF.
So, for AFB, AB, the hypoteneuse = 1; AF = .5, therefore, BF = .866.
GF = .433 (half of BF)
AGF is a right triangle. The sine of GAF = .433/.5 = .866 = 40 degrees = HAC (also)
If BAF is 60 degrees and GAF is 40 degrees, this means BAH is 20 degrees.
If BAH is 20 degrees, by symmetry, JAC is 20 degrees.
If HAC is 40 degrees and JAC is 20 degrees, then HAJ is 20 degrees.
Congratulations, you’ve discovered how to trisect an angle (only works if the angle is 60 degrees).
Anyway, in triangle AJC, we know that JAC is 20 degrees and ACJ is 60 degrees, that means angle AJC is 100.
If we make the length of AC =1, then CJ should be .33…
However if we use our law of sines, the sine of AJC (100 degrees) divided by the length of AC (1 unit) should equal the sine of JAC (20 degrees) divided by CJ. Solving for CJ, we get ~.347
And so, CJ is not perfectly one third the length of BC, thus the three segments can not all be equal to each other.