The nine-point circle

Idea. All nine points are at distance $\frac12$ from $n=\frac{a+b+c}2$, the midpoint of $OH$.

Midpoints of the sides. For $BC$: $\frac{b+c}2-n=-\frac a2$, distance $\frac12$. Same for the others.

Midpoints of $AH$, $BH$, $CH$. Midpoint of $AH$ is $\frac{a+h}2=\frac{2a+b+c}2$; minus $n$ gives $\frac a2$, distance $\frac12$. Same for $B$ and $C$.

Feet of the altitudes. The foot from $A$ is $d=\frac{a+b+c-bc\bar a}2$, so
$$d-n=-\frac{bc\bar a}2,\qquad|d-n|=\frac{|b||c||\bar a|}2=\frac12.$$
Same for the feet from $B$ and $C$.

Conclusion. The nine points lie on the circle of centre $n$ and radius $\frac12$; for a general $R$, radius $\frac R2$. $\blacksquare$

Remark. The nine-point circle is the image of the circumcircle under the homothety of centre $G$ and ratio $-\frac12$ (and also of centre $H$ and ratio $\frac12$).

Lesson. Complex numbers on the unit circle turn "nine points on a circle" into nine one-line modulus computations.