Picturing the Proof of the 2 Circles 3 Lines construction

Begin with the original diagram, add several auxiliary circles and lines, then note the similar triangles ACF and GCH.
Since |AC| = 1, |GC| = 3, and |GH| =1, we see that |AF| = 1/3.

Of course, the picture hides the hardest part of the proof: showing that the circle with center B and the two lines CH and DJ go through the point E.  Possibly the easiest way to see this is to use analytic geometry.  We will find the point E2 on the intersection of the lines CH and DJ, and then show this point is exactly distance one from B, that is, the point E2 is the same as the point E.   For convenience, we will make B the origin, then one can easily verify the coordinates of these points:
A  (-1,0)
B  (0,0)
C  $(-1/2, \sqrt{3}/2)$
D  $(-3/2, -\sqrt{3}/2)$
H  $(-1, -\sqrt{3} )$
J  (1,0).
Then the slope of the line DJ is $\sqrt{3}/5$ and the equation is  $y = (\sqrt{3}/5) x - (\sqrt{3}/5)$.
The slope of the line CH is  $3 \sqrt{3}$ and the equation is  $y = 3 \sqrt{3} x + 2 \sqrt{3}$.
Some algebra shows that the point E2 where these two equations intersect has $x = -11/14$, so $y = -5 \sqrt{3} / 14$.
The point  E2 $( -11/14, -5 \sqrt{3}/ 14 )$ satisfies  $(11/14)^2 + ( 5 \sqrt{3} / 14 )^2 = ( 121 + 75)/ 14^2 = 1$,
thus showing it is the point E from our construction.

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