What is the simplest way to find $\frac{n}{7}th$ of a line with mathematical proof? I'd like to know if it is possible to find out the simplest way to get $\frac{1}{7}$, $\frac{2}{7}$, $\frac{3}{7}$, $\frac{4}{7}$, $\frac{5}{7}$ and $\frac{6}{7}$ of a line in 2-dimensional geometry. Straight edge and compass preferred as tools.
I have drawn one example which shows $\frac{1}{7}$, $\frac{6}{7}$, $\frac{3}{7}$ and $\frac{4}{7}$ parts of the line below. Dotted lines are half points of the square. Blue ones are aid lines. The last line from B to G intersects point H, which is giving hoped fractions of seven.

In addition I'm forward to see mathematical proof of the geometry.
 A: From first chapter of "Four pillars of geometry", John Stillwell

A: Draw a line at an angle to the segment you want to divide, from one end of that segment. Mark off seven equal intervals on the line. Connect the endpoint to the other end of your segment. Then draw six parallels.
No time for a figure now.
A: You can divide a line into any number, m, of equal points by:
At one endpoint of the line draw another line at an angle and mark off a unit measurement.  Extend the line are mark of m marks of measure.  
Connect the m-th mark of the second line to the end-point of the original line.  We'll call this line Fred.
At each of the m - 1 measure marks construct a line through the measure mark parallel to Fred.  These lines will all intersect the original line
By Thale's Theorem of intercepts the corresponding segments of the original line will be proportional to the segments of the second line.  In other words, you've divided the lines into m equal segments.
We can take this further.  If you have a line measuring x (any x; maybe not rational; maybe not even constructable) and a line measuring y.  And any third line.  You can divide the third line it two segments with  x to y ratio. (One side x/(x+y) and the other y/(x+y) times the original).
A: Edited after the OP clarified (in comments) what the claim was:
Here's how you can use coordinates to analyze your construction.  Let's put coordinates on the plane so that (in your diagram) point $E$ is the origin $(0,0)$, and the major grid lines correspond to a spacing of $2$.  Then $D=(0,7)$, $C=(14,7)$, $A=(14,-7)$, $F=(7,-7)$, and $B=(0,-7)$.
With those coordinates, line $EC$ has slope $1/2$ and is given by $y=\frac{1}{2}x$, and line $DF$ has slope $-2$ and is given by $y=7-2x$. They intersect at $G$, which has coordinates $(\frac{14}{5},\frac{7}{5})$.
Now we know that line $BG$ has slope $3$ and is given by the equation $y=3x-7$. Meanwhile, line $EA$ has slope $-1/2$ and is given by $y=-\frac{1}{2}x$.  These intersect each other at $H$, which has coordinates $(2,-1)$.
Now if you construct a line parallel to $BD$ through $H$, that line will have the equation $x=2$.  That line cuts the side of the square $CD$ exactly 1/7 of the way across its length.  So this construction does, in fact, work.
Now, whether this is the simplest construction possible is another matter. As the other answerers have indicated, there is a general algorithm for partitioning any segment into $n$ equal parts, where $n$ is any positive integer.  Whether that method is "simpler" than yours for the case $n=7$ depends, I suppose, on how you measure "simplicity".  The other method certainly wins in terms of generalizability: it's not at all clear to me how (or even if) your method could be adapted for other values of $n$.
A: You refer to your picture and indicate that you can isolate the 1/7, 3/7, 4/7 and 6/7 elements, but not 2/7 and 5/7 portions.
You seem savvy enough to gather that once you have the 'unit' length segment that you can simply use the compass set to that length and easily segment the line piecemeal. Also, other learned respondents have already alluded to the traditional techniques for equal-parts division of a segment, yet neither seem to have been the method you seek.
The greater part of the solution I believe you are seeking has already been provided by yourself. I have included a twin picture, and the first rendering reflects the above-mentioned traditional solutions, adapted to accommodate your own workings re your supplied picture, with a few extra touches.
The second rendering is a compass-free (straight lines only) creation, again accommodating your supplied output.
Consider the manner in which you produced 'H' (red point). If you proceed further using the same technique and create similar points for each quadrant, the framework is essentially all there. You just need to add in the green auxiliary lines.
(Where the green lines intersect the blue lines, two alternatives would also have sufficed:  by either bisecting the vertices of or adding in the diagonals of the square).
A further embellishment I opted to include were all the horizontal auxiliary lines (non-black) and the respective pink points. Although unnecessary, they serve to demonstrate the inherent symmetry of the framework, such that the equal sevenths segmentation can readily be produced both horizontally and vertically.
https://i.stack.imgur.com/F2YFL.png
