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. Given that most students know more algebra than they do Euclidean geometry, the easiest way to see this is to use analytic geometry. We will find the point on the intersection of the lines CH and DJ, and then show this point is exactly distance one from B, that is, it also lies on the circle of radius 1 with center B.
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 these two equations intersect when x = -11/14, so y = -5
sqrt(3) / 14.
The point ( -11/14, -5 sqrt(3)/ 14 ) satisfies (11/14)2 + ( 5
sqrt(3) / 14 )2 = ( 121 + 75)/ 142 = 1,
thus showing it lies on the circle with center B.
Return to the 2 circle 3 line construction page
Villanova Home Page
Trisection Index of
Pages revised 26 Jan 2006