In the triangle $\triangle ABC$ $O,H$ are the center of the circle $\odot(ABC)$ and the orthocenter, respectively. Let the Macbeath Inellipse $\Gamma$'s touchpoints $BC,AC,AB$ at points $D,E,F$, respectively, and draw the contact triangle $DEF$.
Extending side $EF$ intersects the circle $\odot (ABC)$ at points $A_1$ and $A_2$. Similarly define the points $B_1,B_2,C_1,C_2$.
Prove that an ellipse $W$ can be inscribed into a hexagon formed by straight lines $DA_1,DA_2,FC_1,FC_2,EB_1,EB_2$ in which one of the foci coincides with the point $O$.
A homothetic argument would work well here; however this seems to not get me any place.
Best Answer
This is a partial answer. The OP asks for a proof that i) an ellipse $W$ can be inscribed and that ii) $W$ has a focus at $O$. I'll show that i) is a special case of a general theorem, but I don't have much to say about ii).
A corollary of Carnot's Theorem for conics states that for a triangle $ABC$ the points $A_1,A_2,B_1,\dots,C_2$ lie on a conic if and only if the lines $AA_1,AA_2,BB_1,\dots,CC_2$ touch a conic. See the figure above, screencapped from Szilasi, Two applications of the theorem of Carnot. As far as I can tell this corollary has been folklore for over a century, but lately it has been referred to as Bradley's Theorem.
Carnot's Theorem, its dual, and their converses are described and proven in Hatton, Projective Geometry (1913), pgs 186-188. Bradley's Theorem follows from these, and can be visualized by lining up and merging the two figures on pg 187 (shown below, screencapped from Hatton).
You can find more papers on the topic by using search terms like "Carnot's Theorem Conics Bradley". (Giving the corollary a name has made it easier to search the topic.)
The theorem relates to your question directly. Triangle $DEF$ in the question becomes triangle $ABC$ in the theorem. The circumcircle in your question becomes the conic that cuts the triangle (the dashed ellipse in the first figure above). In your question it cuts the sides outside the triangle, whereas in the figure it cuts inside, but this makes no difference.
As for ii) hopefully somebody else will answer. I tried one other inellipse and the conic $W$ in that case had foci distinct from the inellipse foci. So there is something special about starting from the Macbeath inellipse in this construction.