Another beauty hidden in a simple triangle (3) In an arbitrary triangle $ABC$ pick arbitrary points $D\in BC$ and $E\in AC$ such that $DE \nparallel AB$. Denote midpoint of segment $BD$ with $F$ and midpoint of segment $AE$ with $G$. 
Now draw circumscribed circles of triangles $ABC$, $FGC$ and $DEC$. Prove that all circles intersect at some point $H\ne C$ 
I have tried to work with circle centers. If I were able to prove that centers of all three circles lie on the same line, the rest of the proof would be trivial. However, after several unsuccessful attempts I decided to post this problem here.  

 A: I don't know if I'm missing something. 
Let circles $(CDE)$ and $(CFG)$ meet second time at $H_1$, then spiral similarity with center at $H_1$ takes $E\mapsto D$ and $G\mapsto F$, so $k_1={DF\over EG}$ and rotational angle is $$\angle DH_1E = \angle DCE =: \gamma$$
Let circles $(CDE)$ and $(ABC)$ meet second time at $H_2$, then spiral similarity with center at $H_2$ takes $E\mapsto D$ and $A\mapsto B$, so $$k_2={BD \over AE}={2DF\over 2EG}=k_1$$ and rotational angle is $$\angle DH_2E = \angle DCE = \gamma$$
So this two spiral simmetry are the same, so $H_1=H_2=H$.
A: Let $H$ be the intersection of $c(ABC)$ and $c(GFC)$. We will show that $H$ belongs to $c(EDC)$, for it suffices to prove $\angle EHD=\angle C$.
Note that $\triangle HBF\sim \triangle HAG$: $\angle HBC=\angle HAC$ as inscribed angles over $HC$ in $c(ABC)$, and $\angle BHF=\angle BHA-\angle FHA=\angle C-\angle FHA$ and also $\angle AHG=\angle FHG-\angle FHA=\angle C-\angle FHA$, where we again used inscribed angles in $c(ABC)$ and $c(GFC)$. 
$\triangle HBF\sim \triangle HAG$ together with the fact that $G$ and $F$ are midpoints of $AE$ and $BD$ imply $\triangle HBD\sim \triangle HAE$. So $\angle EHA=\angle DHB$, which implies $\angle EHG=\angle DHF$, so $\angle EHD= \angle EHF-\angle DHF= \angle EHF-\angle EHG= \angle GHF=\angle C$.
Hopefully I didn't make typos in these angles.   
