Hubert Shutrick

The nine-point circle of a triangle is tangent to the incircle and the three excircles of the triangle.


This proof is a complement to the ctk proof where the theorem is proved for two excircles and would appear to be incomplete.

Let ABC be the triangle with angles \alpha,\beta, \gamma and let A'B'C' be the midpoint triangle so that the nine-point circle is its circumcircle. It will be sufficient to prove that it is tangent to the inside inscribed circle and the one that is outside BC.


If the inside circle touches BC at I and the outside one touches it at E, then it can be shown as in the ctk proof that A' is also the midpoint of IE.

Let FM with Fon AB and M on A'C' be the other common tangent intersecting BC at K. The foot point N of the perpendicular from A to BC is also on the nine-point circle and it can be shown as in the ctk proof that A'I is the geometric mean of A'K and A'N.

Consider inversion in the circle with centre A' and radius A'I. From the previous paragraph N goes to K and, since the nine-point circle passes through A', it must go to a line through K. We show that that line is FM by examining the angle \angle KMC'. The quadrilateral FA'MB is cyclic since \angle MA'B and \angle MFB are both \gamma and therefore \angle A'MF is \alpha + \gamma like \angle A'BF. The vector MK is then at angle \beta with A'C' which goes to itself under the inversion. However, the tangent to the nine-point circle at C' also makes that angle with A'C' because B' is in the opposite arc. The two inscribed circles being orthogonal to the inverting circle go to themselves and so the nine-point circle must be tangent to both since it is the inverse of their common tangent FM.