TITLE: Intersection point of three circles
QUESTION [11 upvotes]: Can you provide a proof for the following proposition:

Proposition. Let $\triangle ABC$ be an arbitrary triangle with orthocenter $H$. Let $D,E,F$ be a midpoints of the $AB$,$BC$ and $AC$ , respectively. Let $A'$ be a reflection of the point $A$ with respect to the point $E$ , $B'$ reflection of the point $B$ with respect to the point $F$ and $C'$ reflection of the point $C$ with respect to the point $D$. Consider the three circles $k_1,k_2,k_3$ defined by the points $AHA'$ , $BHB'$ and $CHC$' , 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.

REPLY [11 votes]: The points $A,B,C$ are midpoints of the sides of $\triangle A'B'C'$, thus $H$ is the centre of the circumcircle $\omega$ of $\triangle A'B'C'$. Make an inversion with respect to $\omega$. The point $A$ maps to the intersection point $A_1$ of the tangents at $B',C'$ to $\omega$ (these tangents are the images of the circles $HB'A$, $HC'A$).
So, in triangle $A_1 B_1 C_1$ we join the vertices with the points of tangency of the incircle (or excircle) with the respective sides, and should prove that such three lines are concurrent. This is well known and follows from Ceva theorem, for example.