Triangle/Nine-point circle/Fact/Proof

Proof

1. 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

   $U^{\prime }=\frac{1}{2}(A+B+C).$

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

   $\Vert \frac{1}{2}(B+C)-U^{\prime }\Vert =\Vert \frac{1}{2}(B+C)-\frac{1}{2}(A+B+C)\Vert =\frac{1}{2}\Vert A\Vert .$

   Since the norms of all vertices $A,B,C$ are equal due to the choice of $U$, it follows that $U^{\prime }$ 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.

2. 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 $U^{\prime }$ is

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

3. Note that the points constructed in (1) and (2) lie on the circle $F$ opposite to each other. Indeed, we have

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

   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 $F$.

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