# Location of Feuerbach point on incircle of $\triangle ABC$

Assume we have a $\triangle ABC$ in which $b>a>c$. Then the Feuerbach point lies on arc $B'C'$ of incircle of $\triangle ABC$.
My question is why? Proofs that show the incircle and nine-point circle are tangent, do not specify the tangency point (Feuerbach point) location. • What do you mean by "existence of Feuerbach point" ? A question should be self contained. – Jean Marie Jan 24 '17 at 17:06
• @JeanMarie I edited the question. – 1Emax Jan 24 '17 at 17:12
• – Jean Marie Jan 24 '17 at 17:19
• @JeanMarie I have, but it didn't really help me with my question. – 1Emax Jan 24 '17 at 17:23
• That just depends on the positions of the incenter $I$ and the centre $N$ of the nine-point circle, i.e. the midpoint of $OH$. $I$ has the same distance from the sides of $ABC$: what can be said about $N$? – Jack D'Aurizio Jan 24 '17 at 17:30 Proving that $F$ lies on the wanted arc is the same as proving that $N$ (the midpoint of $OH$, center of the nine point circle) lies in the depicted grey region. The distance of $N$ from the $BC$-side is the average between the distance of $O$ from the $BC$-side and the distance of $H$ from the $BC$-side, hence $$d(N,BC) = \frac{R}{2}\left(\cos A+2\cos B\cos C\right)=\frac{R}{2}\cos(B-C)$$ and the trilinear or barycentric coordinates of $N$ are easy to find.
Let $A'$ and $B'$ the vertices of the grey region on the $BC$ and $CA$ sides. We just need to check that the given constraints ensure $\det(A',N,I)>0$, $\det(I,N,B')>0$ and we are done.