Intersection point of three circles

Question

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.

Answer 1

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.

Answer 2

Switching the roles of $ABC$ and $A'B'C'$, consider the circles passing through the vertices of a triangle $A,B,C$, midpoints of the opposite sides $A',B',C'$ and the circumcenter $O$.

It is straightforward to observe that the circle through $A,O,A'$ also passes through the point $A_1$ of intersection of $BC$ with the tangent in $A$ to the circumcircle of $\Delta ABC$. Moreover, $A_1O$ is a diameter of this circle.

Constructing in the same way $B_1,C_1$ the intersection points of the tangents in $B,C$ to the circumcircle of $\Delta ABC$ it is well known that $A_1,B_1,C_1$ are colinear (this can be proved using Menelaus's theorem or Pascal's theorem). Since the centers of the circles $(AOA',BOB',COC')$ are the midpoints of $OA_1,OB_1,OC_1$, they are also colinear. Thus we have three circles passing through a point $O$ having colinear centers. These circles meet again.