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: that the line CE goes through the point H defined by the two circles. Given that most students know more algebra than they do Euclidean geometry, the easiest way to see this is to use analytic geometry.
For convenience, we will make A the origin, then
one can easily verify the coordinates of these points:
A (0,0)
B (1,0)
C (-1/2, sqrt(3)/2)
D (-1/2, -sqrt(3)/2)
E = midpoint between D and B = (1/4, -sqrt(3)/4 )
Then the slope of the line CE is 3 sqrt(3) and the equation is y = 3
sqrt(3) x - sqrt(3).
This shows that the y-intercept is -sqrt(3) which is precisely the point H
defined by the two circles.
Alternatively, one can easily see that the x-intercept is 1/3 which is precisely
the point F.
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Pages revised 14 Feb 2006