[alogo] Three circles problem

Let a = (O, OB), b = (C, CB) and c = (D, DE) be three circles. (a) and (b) are tangent at B. All three have fixed radius and B moves on (a), while (b) remains tangent to (a) at B all the time. For each position of B the bisectors of angle AOB are drawn and their intersections F, G with the radical axis of the circles (b) and (c) are defined. Show that F and G move on two lines orthogonal to the common diameter AD of circles (a) and (c).

[0_0] [0_1]
[1_0] [1_1]

The proof is trivial, using coordinates with origin at O. The following geometric proof is due to Antreas Varverakis. Consider the circles d = (C',C'H), e = (C'',C''A), which are symmetric of (b) with respect to the lines OG and OF respectively. Obviously G and F are radical centers of the tripples of circles: ((b),(d),(c)) and ((b),(e),(c)) respectively. Since, in each tripple, the two last circles are fixed, the lines [LG] and [KF] are also fixed.
The line [FG] envelopes an ellipse with one focus at O. Look at ThreeCirclesProblem1.html if you are interested in.

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