In plane geometry, the Conway circle theorem states that when the sides meeting at each vertex of a triangle are extended by the length of the opposite side, the six endpoints of the three resulting line segments lie on a circle whose centre is the incentre of the triangle. The circle on which these six points lie is called the Conway circle of the triangle.[1] [2] The theorem and circle are named after mathematician John Horton Conway.
Let I be the center of the incircle of triangle ABC, r its radius and Fa, Fb and Fc the three points where the incircle touches the triangle sides a, b and c. Since the (extended) triangle sides are tangents of the incircle it follows that IFa, IFb and IFc are perpendicular to a, b and c. Furthermore the following equalities for line segments hold. |AFc|=|AFb|, |BFc|=|Ba|, |CFa|=|Cb|. With that the six triangles IFcPa, IFcQb, IFaPb, IFaQc, IFbQa and IFbPc all have a side of length |AFc|+|BFc|+|CFa| and a side of length r with a right angle between them. This means that due SAS congruence theorem for triangles all six triangles are congruent, which yields |IPa|=|IQa|=|IPb|=|IQb|=|IPc|=|IQc|. So the six points Pa, Qa, Pb, Qb, Pc and Qc have all the same distance from the triangle incenter I, that is they lie on a common circle with center I.
The radius of the Conway circle is
\sqrt{r2+
| ||||
| s |
r
s
Conway's circle is a special case of a more general circle for a triangle that can be obtained as follows: Given any △ABC with an arbitrary point P on line AB. Construct BQ = BP, CR = CQ, AS = AR, BT = BS, CU = CT. Then AU = AP, and PQRSTU is cyclic.[3]
If you you place P on the extended triangle side AB such that BP=b and BP being completely outside the triangle then the above constructions yield Conway's circle theorem.