Monge from Desargues
given three circles of distinct radii situated entirely in each other's exterior, then the three points of intersection of the pairs of external common tangents are collinear.
(The theorem is valid even if the circles intersect and has a sensible interpretation when two or three circles have the same radius.)
Following is a proof derived from Desargues' theorem.
Denote the centers of the three circles as A, B, C. Let A', B', C' be the intersections of the common tangents external to ΔABC, see the applet. In ΔA'B'C' the lines AA', BB', CC' serve as bisectors of angles A', B', and C', respectively. Therefore, the three lines concur at the incenter (I') of ΔA'B'C'. This shows that the two triangles ABC and A'B'C' are perspective from a point. By Desargues' theorem, they are also perspective from a line.
- J. McCleary, An Application of Desargues' Theorem, Mathematics Magazine, Vol. 55, No. 4 (Spet., 1982), 233-235.
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