Begin with the original diagram, add a few auxiliary lines, then note the similar right triangles GDA and DJA. By our construction, |AG| = 6 and |AD| = 1, and by similar triangles (or trigonometry) |AG| / |AD| = |AD| / |AJ|. Thus, |AJ| = 1 / |AG| = 1/6. Since |AF| = 2 |AJ|, we see that |AF| = 1/3.
Of course, the picture hides the hardest part of the proof: that the angle GDA is a right angle. 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 circle of radius 1 with center A and the circle of radius 3 with center C. The equations are x2 + y2 = 1 and (x-3)2 + y2 = 32. Eliminating the y2 variable by subtracting the equations, (x-3)2 - x2 = 32 - 1, hence -6 x + 9 = 8, or x = 1/6. This is the x-coordinate of D but also of J, hence showing that |AJ| = 1/6 as claimed.
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Pages revised 26 Jan 2006