Is the proof given below acceptable?
Claim. In any triangle $\triangle ABC$ construct isosceles triangles $\triangle ACE$ and $\triangle BDC$ on sides $AC$ and $BC$ , with apices at points $A$ and $B$ , such that $\angle EAC+\angle CBD=180^{\circ}$ holds true. Let points $F$ and $G$ divide legs $AE$ 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 $C$ .
GeoGebra applet that demonstrates this claim can be found here.
The following proof is inspired by this answer to my previous question.
Proof. Consider $A$, $B$ , $C$ as complex numbers and choose a $\lambda \in \mathbb{R}$. Denote $\angle EAC=\alpha$ and $\angle CBD=\beta$ . Then, $$F=A+\lambda(E-A)=A+\lambda(\cos \alpha +i \sin \alpha)(C-A)$$ $$G=B+\lambda(D-B)=B+\lambda(\cos (-\beta) +i \sin (-\beta))(C-B)$$ $$H=\frac{1}{2}(F+G)=$$ $$\frac{1}{2}(A+\lambda(\cos \alpha +i \sin \alpha)(C-A)+B+\lambda(-\cos \alpha -i \sin \alpha)(C-B))=$$ $$\frac{1}{2}(A+\lambda(\cos \alpha + i\sin \alpha)C-\lambda(\cos \alpha + i\sin \alpha)A+ $$$$B-\lambda(\cos \alpha + i\sin \alpha)C+\lambda(\cos \alpha + i\sin \alpha)B)=$$ $$\frac{1}{2}(A(1-\lambda(\cos \alpha + i\sin \alpha))+B(1+\lambda(\cos \alpha + i\sin \alpha)))$$
This shows that $H$ is independent of the location of $C$.
Q.E.D.
EDIT
It is possible to generalize this claim even further.
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 $C$ .
GeoGebra applet that demonstrates this claim can be found here.
Proof. 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.
Another way to see this is with spiral similarities. We’ll use the fact that spiral similarities come in pairs.
Reflect $C$ over $A$ to $I$. Since $IAF \sim CBG$, there’s a spiral similarity $OIAF \sim OCBG$ with some center $O$. $OIF \sim OCG$ pairs with $OIC \sim OFG$, which extends to $OIAC \sim OFHG$ using the midpoints of $IC$ and $FG$. Then $OIA \sim OFH$ pairs with $OIF \sim OAH$, while $OIA \sim OCB$ pairs with $OIC \sim OAB$. These combine to $OICF \sim OABH$. Thus the angles of $\triangle ABH$ equal the angles of $\triangle ICF$, which are independent of the position of $C$ because the ratios $AI : AC : AF$ and angles $∠IAC, ∠CAF$ are fixed.