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Conway circle theorem
Geometrical construction based on extending the sides of a triangle From Wikipedia, the free encyclopedia
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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][3] The theorem and circle are named after mathematician John Horton Conway.

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Proof
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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|=|BFa|, |CFa|=|CFb|. 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.
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Additional properties
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The radius of the Conway circle is
where and are the inradius and semiperimeter of the triangle.[3]
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Generalisation

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.[4]
If 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.
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