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.
Return to the 2 circle 3 line construction
Villanova Home Page Trisection Index of Pages revised 26 Jan 2006