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Explanation

Three rays emanating from a point O form three angles. Assuming none of these exceeds 180°, we may inscribe into each angle a circle. In the applet the circles have centers at P, Q, R. The rays serve as common internal tangents of a pair of circles. For each pair, draw the second internal tangent - the reflection of the first one in the line joining the centers of the circles. Then the second tangents are concurrent.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

This is a part of a theorem proved by Professor Mannheim (1864) of l'École Polytechnique [F. G.-M., Exercices de Géométrie, Éditions Jacques Gabay, sixiéme édition, 1991, p. 326 (Th. 164)]:

 Given three (either tangent or non intersecting) circles, if three common internal tangents of the circles taken two by two are concurrent, so are the other three internal tangents.

### Proof

Denote the cross points of the two internal tangents of the circles taken two by two as A, B, C. (If the circles touch each other, their two internal tangents coalesce, and then our notation applies to the common point of tangency of the circles.) Let three of the tangents, which are in our setting OA, OB, and OC, meet in a point O, naturally. Let X stand for the point common to some other pair of tangents, say AX and BX. We want to show that CX is one of the remaining tangents. (For convenience, we shall identify the circles by their centers: (P), (Q), (R).)

Assuming CX is not tangent to, say (Q), draw DX, with D on OC tangent to (Q). By construction then, the quadrilaterals AOBX and AODX are exscriptible. The sides of the former are tangent to the circle (R), those of the latter are tangent to the circle (Q). By a theorem of J. Steiner,

 AO + AX = BO + BX and AO + DX = DO + DX,

which, by transitivity, implies

 BO + BX = DO + DX.

By converse of Steiner's theorem, the quadrilateral BODX is exscriptible and its sides are tangent to the circle (P). In particular, DX is tangent to (P) and is, therefore, a common internal tangent of (P) and (Q). But there are just two such tangents, and they cross in C. Wherefrom, C = D.

### Remark

The same method proves a similar statement concerning external tangents. External tangents should be considered for two pairs of circles, while the third pair should provide internal tangents.