Feuerbach's theorem

Remarks

Let $R$ be the circumradius, $r$ the inradius, and $N$ and $I$ the centres of the nine-point circle and of the incircle. The nine-point circle has radius $R/2$, and $N$ is the midpoint of the circumcentre $O$ and the orthocentre $H$. Write $a, b, c$ for the side lengths, and $A, B, C$ also for the angles. Two known formulas give $OH^2 = 9R^2 - (a^2 + b^2 + c^2)$ and $IH^2 = 2r^2 - 4R^2 \cos A \cos B \cos C$. With these and Euler's theorem for $OI$, Apollonius's theorem in the triangle $OIH$ gives $IN = R/2 - r$. This says the two circles are internally tangent; a similar computation gives $I_a N = R/2 + r_a$ for each excentre $I_a$ with exradius $r_a$.

In an equilateral triangle the nine-point circle and the incircle coincide, since $N = I$ and $R/2 = r$. The point of tangency with the incircle is the Feuerbach point of the triangle.

Going deeper

For a student

The theorem is a highlight of classical triangle geometry, published by Karl Wilhelm Feuerbach in 1822. It ties together the circumcircle, the incircle and the nine-point circle through the radii $R$, $r$ and $R/2$.