Triangle/Nine-point circle/Fact/Proof

${\displaystyle {}U}$

${\displaystyle {}U'={\frac {1}{2}}{\left(A+B+C\right)}\,.}$

The distance between the midpoint ${\displaystyle {}A'={\frac {1}{2}}{\left(B+C\right)}}$ of the side connecting ${\displaystyle {}B}$ and ${\displaystyle {}C}$ and ${\displaystyle {}U'}$ is

${\displaystyle {}\Vert {{\frac {1}{2}}{\left(B+C\right)}-U'}\Vert =\Vert {{\frac {1}{2}}{\left(B+C\right)}-{\frac {1}{2}}{\left(A+B+C\right)}}\Vert ={\frac {1}{2}}\Vert {A}\Vert \,.}$

Since the norms of all vertices ${\displaystyle {}A,B,C}$ are equal due to the choice of ${\displaystyle {}U}$, it follows that ${\displaystyle {}U'}$ is the circumcenter of the triangle given by the midpoints of the original triangle, and that its radius is the half of the radius of the circumcircle.

${\displaystyle {}A+B+C}$

${\displaystyle {}A}$

${\displaystyle {}{\frac {1}{2}}(A+B+C)+{\frac {1}{2}}A=A+{\frac {1}{2}}(B+C)\,.}$

The distance between this point and ${\displaystyle {}U'}$ is

${\displaystyle {}\Vert {{\frac {1}{2}}(A+B+C)-{\left(A+{\frac {1}{2}}(B+C)\right)}}\Vert =\Vert {-{\frac {1}{2}}A}\Vert ={\frac {1}{2}}\Vert {A}\Vert \,.}$

${\displaystyle {}F}$

${\displaystyle {}{\frac {1}{2}}{\left({\frac {1}{2}}(B+C)\right)}+{\frac {1}{2}}{\left(A+{\frac {1}{2}}(B+C)\right)}={\frac {1}{2}}{\left(A+B+C\right)}\,,}$

which is the center of ${\displaystyle {}F}$. Hence, for each side, its midpoint, the midpoint between the opposite vertex and the orthocenter, and the foot of the corresponding altitude form a right triangle. Its circle with the hypotenuse as diameter equals ${\displaystyle {}F}$.