# Monge via Desargues

Monge's theorem (Monge's Circle Theorem, Three Circles Theorem) claims that,

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.)

One of the proofs of the theorem has been derived from Desargues' theorem and has exploited the fact that, in a triangle, the angle bisectors concur in a point - the incenter of the triangle. As a matter of fact the relation between Desargues' and Monge's theorems is more direct as becomes clear from a demonstration below.

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As was observed on several occasions, any two circles are similar. If, in addition, their radii differ, they are also homothetic with a finite center of homothety. This means that there exists a point O - the center of homothety (dilation) - such that

1. Any two corresponding points P and P' on the two circles are collinear with O.
2. The ratio OP/OP' does not depend on P.

If such circles have external common tangents, the latter intersect at the center of homothety, and the point of tangency of a common tangent correspond to each other.

In general, a homothety is well defined by any pair of corresponding points. Since a homothety maps a segment into a parallel segment, two arbitrary segments may only define a homothety if they are parallel. Two parallel segments A1B1 and A2B2, with A1 corresponding to B1 and A2 to B2, define a unique homothety with center at the intersection of A1A2 and B1B2.

Given three circles (O1), (O2), (O3), they are pairwise homothetic. It follows that there are three homotheties HX, HY, HZ, with centers X, Y, Z, such that HX maps (O2) onto (O3), HY maps (O3) onto (O1), HZ maps (O1) onto (O2).

Choose two points A1 and B1 on (O1) and let A2 and B2 be their images under HZ: A2 = HZ(A1) and B2 = HZ(B1). As we just mentioned, A1B1||A2B2. Define A3 = HX(A2) and B3 = HX(B2). As before, A2B2||A3B3, implying A3B3||A1B1. If so, there is a unique homothety - the product of HZ and HX - that maps A1B1 onto A3B3. This is exactly HY. To see that, choose a third point C1 on (O1) and find its successive images C2 and C3 first under HZ and then under HX. The product of HZ and HX maps C1 onto C3 and hence it maps ΔA1B1C1 onto ΔA3B3C3 and also maps their circumcircles (O1) and (O3). But the latter two circles are homothetic under HY, which proves the claim.

It thus follows that that the lines joining the corresponding vertices of triangles A1A2A3 and B1B2B3 are parallel, i.e., meeting at a point at infinity. The two triangles are thus perspective from a point. By Desargues' theorem they are also perspective from a line. The latter is a paraphrase of of the fact that the points X, Y, Z are collinear.

(Since homotheties and similarities were the main instrument in the above proof, it may not be entirely out of place to mention that the triangles A1A2A3 and B1B2B3, although perspective, need not be similar, let alone homothetic.) ### Monge's Theorem

1. Three Circles and Common Tangents
2. Monge from Desargue
3. Monge via Desargue ### Desargues' Theorem 