Triangle geometry/Nine-point circle/Introduction/Section

The nine points of the Nine-point circle: The midpoints of the sides (blue), the feet of the altitudes (red), and the midpoints (green) between the vertices and the orthocenter (black).

Lemma

Let a nondegenerate triangle ${}\triangle$ in the Euclidean plane with vertices ${}A,B,C$ be given. Let ${}F$ be the circumcircle of the midpoints

of the sides of the triangle. Then the following statements hold.

The radius of ${}F$ is half of the radius of the circumcircle of ${}\triangle$ .
The line segements between the orthocenter and the vertices are cut in halves by ${}F$ .
The feet of the altitudes of ${}\triangle$ lie on ${}F$ .

Proof

Let ${}U$ be the circumcenter of the triangle; we may assume that this point is the origin of a Cartesian coordinate system. We consider the point
${}U'={\frac {1}{2}}{\left(A+B+C\right)}\,.$

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

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

Since the norms of all vertices ${}A,B,C$ are equal due to the choice of ${}U$ , it follows that ${}U'$ 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.
By fact, ${}A+B+C$ is the orthocenter. Therefore, the midpoint of the line segment between ${}A$ and the orthocenter equals
${}{\frac {1}{2}}(A+B+C)+{\frac {1}{2}}A=A+{\frac {1}{2}}(B+C)\,.$

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

${}\Vert {{\frac {1}{2}}(A+B+C)-{\left(A+{\frac {1}{2}}(B+C)\right)}}\Vert =\Vert {-{\frac {1}{2}}A}\Vert ={\frac {1}{2}}\Vert {A}\Vert \,.$
Note that the points constructed in (1) and (2) lie on the circle ${}F$ opposite to each other. Indeed, we have
${}{\frac {1}{2}}{\left({\frac {1}{2}}(B+C)\right)}+{\frac {1}{2}}{\left(A+{\frac {1}{2}}(B+C)\right)}={\frac {1}{2}}{\left(A+B+C\right)}\,,$

which is the center of ${}F$ . 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 ${}F$ .

\Box

The circle in the preceding statement is called the Nine-point circle, or the Feuerbach circle.