Monge's Theorem

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The diagram below shows an example of Monge's theorem. Each pair of circles in the plane defines a pair of external common tangents. In the diagram below, we have three circles, and therefore three pairs of common tangents, coloured red, blue and green. Let \(X\) be the point of intersection of the common tangents for the circles centred at \(A\) and \(B\), let \(Y\) be the intersection of the common tangents for the circles centred at \(B\) and \(C\), and let \(Z\) be the intersection of the common tangents for the circles centred at \(C\) and \(A\).

Monge's theorem then says that the points \(X\), \(Y\) and \(Z\) are collinear.

You can experiment by moving the points \(A\), \(B\) and \(C\) to change the centres of the circles, and moving the red points to change the radii of the circles. You can see that \(X\), \(Y\) and \(Z\) remain collinear after moving the circles in this way.

The diagram below shows a proof of Monge's theorem using Desargues' theorem. The three circles centred at \(A\), \(B\) and \(C\) define a triangle \(\Delta DEF\) consisting of the common tangents outside the three circles (i.e. the line \(DE\) is the common tangent of the circles centres at \(A\) and \(B\) such that the point \(C\) and the line \(DE\) are on opposite sides of the line \(AB\)).

Also note that the line through the centres of a pair of circles also passes through the intersection point of the common tangents (and is in fact the angle bisector of the common tangents). You can see this in the diagram below, where \(AB\) passes through \(X\), \(BC\) passes through \(Y\) and \(CA\) passes through \(Z\).

Since the centre of a circle lies on the angle bisector of any pair of tangents, then the three angle bisectors of \(\Delta DEF\) (coloured in brown) pass through the points \(A\), \(B\) and \(C\). Since the angle bisectors of a triangle intersect at the incentre, then the incentre \(I\) of \(\Delta DEF\) is a point of perspective for the triangles \(\Delta ABC\) and \(\Delta DEF\). Desargues' theorem then shows that the triangles are in perspective from a line, and therefore the points \(X\), \(Y\) and \(Z\) are collinear.