# 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.

What if applet does not run? |

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

- Any two corresponding points P and P' on the two circles are collinear with O.
- 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 A_{1}B_{1} and A_{2}B_{2}, with A_{1} corresponding to B_{1} and A_{2} to B_{2}, define a unique homothety with center at the intersection of A_{1}A_{2} and B_{1}B_{2}.

Given three circles (O_{1}), (O_{2}), (O_{3}), they are pairwise homothetic. It follows that there are three homotheties H_{X}, H_{Y}, H_{Z}, with centers X, Y, Z, such that H_{X} maps (O_{2}) onto (O_{3}), H_{Y} maps (O_{3}) onto (O_{1}), H_{Z} maps (O_{1}) onto (O_{2}).

Choose two points A_{1} and B_{1} on (O_{1}) and let A_{2} and B_{2} be their images under H_{Z}: _{2} = H_{Z}(A_{1})_{2} = H_{Z}(B_{1})._{1}B_{1}||A_{2}B_{2}. Define _{3} = H_{X}(A_{2})_{3} = H_{X}(B_{2})._{2}B_{2}||A_{3}B_{3}, implying A_{3}B_{3}||A_{1}B_{1}. If so, there is a unique homothety - the product of H_{Z} and H_{X} - that maps A_{1}B_{1} onto A_{3}B_{3}. This is exactly H_{Y}. To see that, choose a third point C_{1} on (O_{1}) and find its successive images C_{2} and C_{3} first under H_{Z} and then under H_{X}. The product of H_{Z} and H_{X} maps C_{1} onto C_{3} and hence it maps ΔA_{1}B_{1}C_{1} onto ΔA_{3}B_{3}C_{3} and also maps their circumcircles (O_{1}) and (O_{3}). But the latter two circles are homothetic under H_{Y}, which proves the claim.

It thus follows that that the lines joining the corresponding vertices of triangles A_{1}A_{2}A_{3} and B_{1}B_{2}B_{3} 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 A_{1}A_{2}A_{3} and B_{1}B_{2}B_{3}, although perspective, need not be similar, let alone homothetic.)

### Monge's Theorem

- Three Circles and Common Tangents
- Monge from Desargue
- Monge via Desargue

### Desargues' Theorem

- Desargues' Theorem
- 2N-Wing Butterfly Problem
- Cevian Triangle
- Do You Speak Mathematics?
- Desargues in the Bride's Chair (with Pythagoras)
- Menelaus From Ceva
- Monge from Desargues
- Monge via Desargue
- Nobbs' Points, Gergonne Line
- Soddy Circles and David Eppstein's Centers
- Pascal Lines: Steiner and Kirkman Theorems II
- Pole and Polar with Respect to a Triangle
- Desargues' Hexagon
- The Lepidoptera of the Circles

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

71924671