This is essentially the same proof but it is a bit simplified and provides a more precise statement. We suppose that $A=0$, $B=1$, $C=z$. Then $E=\rho e^{i\alpha} z$ and $D=1+\rho e^{i\beta}(z-1)$ (so we are making no a priori assumptions about the swing angles). Then $F=\lambda \rho z e^{i\alpha}$ and $G=\lambda \rho ze^{-i\beta}$ plus a term which is constant (i.e., independent of $z$) and so irrelevant in this context.

From this we immediately get the following refined version. The midpoint $H$ is independent of $z$ if and only if the swing angles are related as in the OP. One can also extend to the case where $H$ is a suitable point on the perpendicular bisector of $FG$ and this version is sharp (i.e., it only holds for such points on the bisector).

This is essentially the same proof but it is a bit simplified and provides a more precise statement. We suppose that $A=0$, $B=1$, $C=z$. Then $E=\rho e^{i\alpha} z$ and $D=1+\rho e^{i\beta}(z-1)$ (so we are making no a priori assumptions about the swing angles). Then $F=\lambda \rho z e^{i\alpha}$ and $G=\lambda \rho ze^{-i\beta}$ plus a term which is constant (i.e., independent of $z$) and so irrelevant in this context. By the way, replacing $E$ and $D$ by $F$ and $G$ is a red herring.

From this we immediately get the following refined version. The midpoint $H$ is independent of $z$ if and only if the swing angles are related as in the OP. One can also extend to the case where $H$ is a suitable point on the perpendicular bisector of $FG$ and this version is sharp (i.e., it only holds for such points on the bisector).