Can you provide a proof for the following proposition:
Proposition. Let $\triangle ABC$ be an arbitrary triangle with nine-point center $N$ and circumcenter $O$. Let $A',B',C'$ be a reflection points of the points $A,B,C$ with respect to the point $N$. Consider the three circles $k_1,k_2,k_3$ defined by the points $AOA'$ , $BOB'$ and $COC'$ , respectively. I claim that $k_1$,$k_2$ and $k_3$ meet at a common point $𝑃$.
GeoGebra applet that demonstrates this proposition can be found here.
Best Answer
Such things are quick in complex numbers. Let $O=0$ be the origin, $ABC$ be the unit circle. The centroid of $ABC$ is $G=(A+B+C)/3$, the Euler circle is the image of the circle $ABC$ under homothety $f\colon X\to (3G-X)/2$ centered in $G$ with coefficient $-1/2$, thus $N=f(O)=(A+B+C)/2$. Next, $A'=2N-A=B+C$. The circle through $O,A,B+C$ has equation $g_A(z):=(A^2-BC)z\bar{z}+(B+C-A)z+A(BC-AB-AC)\bar{z}=0$. Such three circles have a common point different from $O$ if their equations are linearly dependent. They are: $(B-C)g_A+(C-A)g_B+(A-B)g_C=0$.