https://www.cut-the-knot.org/Curriculum/Geometry/CircleOfSimilitude1.shtml

At some point, the book says, "It follows at once from the definition ..." What follows is rephrased as Statement 4 in the above page. I've been lookng back and forth for a few hours now, but do not see anything immediate. Does any body see what I am missing.

Thank you,
Alex

I've been looking at this on and off, and I finally think I see what to do. Referring to the diagram in your applet, since line TL1 is tangent to circle 1 and L1K1 is secant to circle 1, it follows that the angle K1O1L1 is twice angle TL1L2. Let's call these angles x and 2x. Therefore |K1L1| = 2*R1*sin(x). However, the perpendicular distance h of T from line L1L2 is h = |TL1|*sin(x). Similarly, there are angles y and 2y at circle 2, and |K2L2| = 2*R2*sin(y), while h = |TL2|*sin(y).

Now since |TL1|/|TL2| = R1/R2, we get the result that |K1L1| = 2*R1/|TL1|*h = 2*R2/|TL2|*h = |K2L2|. This is obviously reversible, so you get the converse too. I got this idea after playing with your applet for a long time and noticing that the distance from T to L1L2 seemed visually to vary in proportion to the lengths of the secants. From there, it was just a matter of trying to build a path of angle and line relationships between the secants and the perpendicular from T to L1L2.

I must say, though, that this amount of fiddling is not my idea of "follows immediately from the definition." Also, now that I think of it, you could avoid reference to trig functions by noting that the triangle L1O1B1 is similar to TL1F, where B1 is the bisector of L1K1 and F is the foot of T on L1L2. Then you have a more "old school" type of proof.

--Stuart Anderson