Generalizing Bottema's theorem

Can you provide another proof for the claim given below?

Claim. In any triangle $$\triangle ABC$$ construct triangles $$\triangle ACE$$ and $$\triangle BDC$$ on sides $$AC$$ and $$BC$$ such that $$\frac{AE}{AC}=\frac{BD}{BC}$$ and $$\angle EAC+\angle CBD=180^{\circ}$$ hold true. Let points $$F$$ and $$G$$ divide sides $$AE$$ and $$BD$$ respectively in the same arbitrary ratio . The midpoint $$H$$ of the line segment that connects points $$F$$ and $$G$$ is independent of the location of $$C$$ .

Proof.(Complex numbers) Consider $$A$$, $$B$$ , $$C$$ as complex numbers and choose a $$\lambda \in \mathbb{R}$$. Denote $$\angle EAC=\alpha$$ , $$\angle CBD=\beta$$ and $$\frac{AE}{AC}=\frac{BD}{BC}=k$$ . Then, $$F=A+\lambda(E-A)=A+\lambda \cdot k(\cos \alpha +i \sin \alpha)(C-A)$$ $$G=B+\lambda(D-B)=B+\lambda \cdot k(\cos (-\beta) +i \sin (-\beta))(C-B)$$ $$H=\frac{1}{2}(F+G)=\frac{1}{2}(A+\lambda \cdot k(\cos \alpha +i \sin \alpha)(C-A)+$$$$B+\lambda \cdot k(-\cos \alpha -i \sin \alpha)(C-B))=$$ $$\frac{1}{2}(A+\lambda \cdot k(\cos \alpha + i\sin \alpha)C-\lambda \cdot k(\cos \alpha + i\sin \alpha)A+$$$$B-\lambda \cdot k(\cos \alpha + i\sin \alpha)C+\lambda \cdot k(\cos \alpha + i\sin \alpha)B)=$$ $$\frac{1}{2}(A(1-\lambda \cdot k(\cos \alpha + i\sin \alpha))+B(1+\lambda \cdot k(\cos \alpha + i\sin \alpha)))$$ This shows that $$H$$ is independent of the location of $$C$$.

Q.E.D.

The GeoGebra applet that demonstrates this claim can be found here.

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).