Three Concurrent Circles: What is this about?
A Mathematical Droodle

Explanation

The applet hints at the following problem [Mathematical Quickies, #184]:

Mathematical Quickies is a collection of problems that appeared in several mathematics journals in the sections under this same caption. The problems are not necessarily trivial, but can be disposed expeditiously with a proper insight. The problem at hand is definitely of this kind.

Let A be the point of concurrency and B, C, D the common points of the given circles taken by two. No confusion will arise if we denote circles by their centers. So let O₁, O₂, O₃ be the circumcircles of ABC, ABD, and ACD, respectively.

The book takes still a round about way of getting the result. For example, the proof in the book uses the fact that triangles ACD and AO₁O₂ are similar. In addition, when comparing various angles it does not take into account all possible configurations (sometimes two angles are equal and sometimes they are supplementary.) While it is easy to dispense with comparing triangles, taking into account all possible configuration is impossible without exploiting the notion of directed angles as was done, for example, in the thirteenth proof of the Butterfly Theorem. However, the essence of the configuration can be made quite transparent with a single observation if some laxity is permitted.

I shall refer to the unmodified diagram presented by the applet on loading of the page. If you played with the applet on your own before reaching this paragraph, it's best if you reload the page before further reading.

In circle O₂, the central angle AO₂B is twice as large as the inscribed angle ADB. Since O₁O₂ is perpendicular to the chord AB and passes through its midpoint, AO₂O₁ is exactly half of AO₂B. Therefore,

AO2O1 = ADB.

Similarly, in circle O₃,

AO3O1 = ADB.

It follows that quadrilateral AO₁O₂O₃ is cyclic, i.e., points A, O₁, O₂, O₃ are indeed concyclic.

Nathan Bowler has observed that the statement just proven is the converse of the Simson-Wallace theorem. To see that, observe the midpoints of segments AB, AC, AD. The midpoints serve as the feet of the perpendiculars from A to the sides of DO₁O₂O₃. These lie on a straight line parallel to the given one. Thus in fact the statement above says that of the feet of the perpendiculars from point A to the sides of DO₁O₂O₃ are collinear, then A lies on the circumcircle of the triangle (the four points A, O₁, O₂, O₃ are concyclic.)

References

1. C. W. Trigg, Mathematical Quickies, Dover, 1985, #184