Is the proof given below acceptable?
Claim. In any triangle $ABC$ construct isosceles triangles $\Delta ACE$ and $\Delta BDC$ 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 $ABC$ construct triangles $\Delta ACE$ and $\Delta BDC$ on sides $AC$ and $BC$ such that $\frac{AC}{AE}=\frac{BC}{BD}$ 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 $C$ .
The proof for this claim is essentially the same as the proof given above.
from Hot Weekly Questions - Mathematics Stack Exchange
Peđa Terzić
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