Begin with the original diagram, add a few auxiliary lines, then note the similar right triangles CDG and CHD. By our construction, |CD| = 2 and |CG| = 6, and by similar triangles (or trigonometry) |CG| / |CD| = |CD| / |CH|. Thus, |CH| = |FH| = 1/3, so |AF| = 1/3.
Of course, the picture hides the hardest part of the proof: that the angle CDG is a right angle. Given that most students know more algebra than they do Euclidean geometry, an easier way to see this is to use analytic geometry. We will find the point on the intersection of the circle of radius 2 with center C and the circle of radius 3 with center D. Putting the origin at C, the equations are x2 + y2 = 22 and (x-3)2 + y2 = 32. Eliminating the y2 variable by subtracting the equations, (x-3)2 - x2 = 32 - 22, hence -6 x + 9 = 5, or x = 2/3. This is the x-coordinate of D but also of H. Since |CH| = |HF| = 2 |AF|, we see that |AF| = 1/3 as claimed.
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Pages revised 26 Jan 2006