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2.5 Feuerbach's Theorem

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2.5 Feuerbach's Theorem 2.5 Feuerbach’s theorem 117 Appendix to Chapter 2: Feuerbach’s theorem and the Apollonius problem for the excircles 2.5 Feuerbach’s theorem Theorem 2.1. The nine-point circle of a triangle is tangent internally to the incircle and externally to each of the excircles. Cb Ib Bc A Ic F Cc Fb Fc Bb I N Ac B Aa C Ab Fa Ba Ca Ia Proof. Consider the incircle (I) and the A−excircle (Ia) of triangle ABC, tangent to BC at X and X respectively. These two points are symmetric with respect to the midpoint Ma of BC. Let T be the intersection of AIa and BC. It is well known that A and T divide I and Ia harmonically. Considering the projections of these points on the line BC, we conclude 2 that Ha and T divide X and X harmonically. By Proposition ?, MaHa · MaT = MaX .It follows that inversion in the circle C := Ma(X) interchanges Ha and T . Note that the circle of inversion C is orthogonal to each of the circles (I) and (Ia). These circles are therefore invariant under the inversion. Apart from the three sidelines, the circles (I) and (Ia) have a fourth common tangent L, which is the reflection of the line BC in AIa. The inversive image of L is a circle through Ma (the center of inversion) and Ha (the image of T ). We claim that this circle is actually the nine-point circle (N) through the midpoints of the three sides. To see that, note that L is parallel to the tangent to circumcircle at A. Since this latter line is orthogonal to OA, which is parallel to the line NMa (under a homothety at the centroid), L is orthogonal to the line MaN. 118 CONTENTS A L I C B Ha X T Ma X Ia A L Mc Mb O I N T C B Ha Ma X Ia Therefore, the image of L is a circle orthogonal to the line MaN (which is invariant under the inversion). Since this circle passes through Ha and Ma, and has center on the line MaN, it must be the nine-point circle (N). From this we conclude that the nine-point circle is tangent to the incircle and the A- excircle. The point of tangency with the incircle is on the line joining Ma to the reflection of X 2.5 Feuerbach’s theorem 119 A Fe L Mc Mb I N T C B Ha X Ma X Fa Ia in the angle bisector AI. This is called the Feuerbach point of the triangle. If the incircle touches CA at Y and AB at Z, the same Feuerbach point also lies on the lines joining Mb to the reflection of Y in BI, and Mc to the reflection of Z in CI. On the other hand, the line joining Ma to the reflection of X in AI intersects the A- excircle at the point of tangency with the nine-point circle. Similar constructions give the points of tangency of the nine-point circle with the other two excircles. 120 CONTENTS 2.6 The Apollonius problem for the excircles Proposition 2.2. The radical center of the three excircles is the Spieker center, the incenter of the inferior triangle MaMbMc. Proof. The midpoint Ma of BC clearly has equal powers with respect to the B- and C- excircles. Since the line joining the excenters Ib and Ic is perpendicular to the bisector of angle A, the radical axis of the excircles is the parallel through Ma to the A-bisector. Equivalently, it is the Ma-bisector of the inferior triangle MaMbMc. Similarly, the radical axes of the C- and A- excircles, and that of the A- and B-excircles, are the bisectors of the Mb- and Mc bisectors of the same inferior triangle. The three radical axes are concurrent at the incenter of triangle MaMbMc. Since the three excircles are mutually disjoint, the radical center Sp is a point exterior to each of them. There is a unique radical circle (Sp) orthogonal to each of the excircles. Ib A Ic Mc Mb Sp B Ma C Ia Since the nine-point circle (N) is tangent externally to each of the excircles, inversion in the radical circle (Sp) gives a circle tangent internally to the excircle. If Fa is the point ( ) ( ) ( ) of tangency of N and Ia , the line SpFa intersects Ia at another point Fa. Similarly ( ) ( ) construct Fb and Fc on Ib and Ic , making use of the points of tangency Fb and Fc with ( ) N . The circle FaFbFc is tangent internally to the excircles. This is called the Apollonius circle of triangle ABC. 2.6 The Apollonius problem for the excircles 121 Fc A Fa c F Fb N Sp B Fa C Fb Each sideline is a common tangent of the three excircles. Inversion in the radical circle (Sp) yields three more circles through Sp, each tangent internally to two excircles and externally to the third. These are called the Bevan circles. Ib A Ic Sp B C Ia 122 CONTENTS Here are the eight circles each tangent to the three excircles. Ib A Ic Sp N B C Ia.
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