Conway's circle theorem

 In my previous post on Three Youtube videos I discussed the Mathologer video which states and gives an elegant proof of John Conway's circle theorem. In this post I want to give a more pedestrian proof that requires only simple angle chasing. Some of the video commenters appear to have found similar proofs so my contribution may simply be to present a proof accompanied by diagrams to make it easy to follow along.

For completeness I'll state the theorem. It concerns an arbitrary triangle where, in this diagram, I colour the triangle sides red, green, and blue. The triangle sides are extended by red, green, and blue lines as shown; all red segments are of the same length and, similarly the blue segments and the green segments.

Conway's theorem is that the 6 endpoints lie on a circle and, moreover, the centre of this circle coincides with the in-centre of the original triangle. As shown below.

Let's label the three interior angles of the triangle as p-2a, p-2b, p-2c (p being my typographical simplification for pi). Then, of course, 

(p-2a) + (p-2b) + (p-2c) = p

and so 

a+b+c = p.

We'll draw a containing hexagon:

You'll see that I've labelled some angles around the hexagon. Their values follow directly from the fact that the diagram has 6 isosceles triangles: their bases are the edges of the hexagon.

In the quadrilateral ABCD there are two opposite angles a and b+c, which sum to p (pi). Thus ABCD is cyclic: A, B, C, D lie on a circle. Thus the unique circle through A, B, C also passes through D. Similarly this is true for the quadrilaterals BCDE and CDEF. Hence all of A, B, C, D, E, F lie on a common circle.

The centre of this circle lies on the perpendicular bisectors of the chords AB, CD, EF. Obviously (isosceles triangles again), each bisector passes through one of the triangle vertices, bisecting the angle there. Therefore these bisectors meet at the centre of the in-circle of the triangle.