Vtyjkyuk

In an arbitrary triangle $ABC$ pick arbitrary points $Din BC$ and $Ein AC$ 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 $Hne C$

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.

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 $triangle HBFsim triangle HAG$: $angle HBC=angle HAC$ 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 $c(GFC)$.

$triangle HBFsim triangle HAG$ together with the fact that $G$ and $F$ are midpoints of $AE$ and $BD$ imply $triangle HBDsim triangle HAE$. 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.

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 $Emapsto D$ and $Gmapsto F$, so $k_1=DFover EG$ 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 $Emapsto D$ and $Amapsto B$, so $$k_2=BD over AE=2DFover 2EG=k_1$$ and rotational angle is $$angle DH_2E = angle DCE = gamma$$

So this two spiral simmetry are the same, so $H_1=H_2=H$.