Let's say I have a triangle $ABC$, the middle of the sides are called $A'$, $B'$ and $C'$.
I have proved that $\Omega$, the orthocenter of $ABC$, is the barycentre of $A'B'C'$ with masses $\tan \alpha$, $\tan \beta$ and $\tan \gamma$ on $A'$, $B'$ and $C'$. Now I have to deduce from this that $\Omega$ is also the barycentre of $A$, $B$ and $C$ with masses $a$, $b$ and $c$. I want to find $a, b, c$ so that $\Omega$ is the barycentre of $ABC$.
Thank you in advance!
It's not true that the orthocentre of $ABC$ is the barycentre of $A'B'C'$ with masses $\tan\alpha$, $\tan\beta$ and $\tan\gamma$. In a right triangle, the orthocentre is at the corner $C$ with the right angle, whereas that barycentre would be at $C'$, since $\tan\gamma$ goes to infinity. More generally, as long as all angles are acute, that barycentre would be within $A'B'C'$, wheras the orthocentre of $ABC$ doesn't have to be.
If your result were correct, it would be straightforward to obtain a corresponding mass distribution on $A$, $B$ and $C$: Move half the mass on the midpoints to each of the adjacent corners; that moves all the mass to the corners without moving the barycentre. So the result would be masses of $(\tan\alpha+\tan\beta)/2$ on $C$ etc.
For the correct barycentric coordinates of the orthocentre, see this Wikipedia section.