A conjecture about an intrinsic similarity of non-isosceles triangles Given any non-isosceles triangle $\triangle ABC$, and denoting $AB$ its longest side, the following construction

determines the points $DFGE$. In this post is shown that the points $DFCGE$ always determine a circle. 
Let then consider the center $J$ of this circle:

By means of this new point $J$, we can draw the three circles passing by $F,C$ and $J$, by $C,G$ and $J$, and by $D,E$ and $J$. 
The two intersections of the first two circles with the third one determine two additional points $H$ and $I$.

A fourth circle passing by these two points $H$, $I$, and $C$ defines the two points $K$ and $L$ on the sides $AC$ and $BC$, respectively.


My conjecture is that the triangle $\triangle LKC$ is similar to $\triangle ABC$.


I tried to apply the techniques suggested in this post and in this other post to prove the claim, but with no success.
Thanks for your suggestions!
 A: Let us at first show that a)$H$ is the intersection of $FE$ and $DC$, b) $I$ is the intesection of $CE$ and $DG$. I will only show a), item b) is similar.
Let $H^\prime$ be intersection  of $FE$ and $DC$ (we will show that $H = H^\prime$). Denote $\angle H^\prime FD = \xi$, $\angle JFD = \mu$, $\angle JCE = \nu$. Using the fact that $\angle EFJ = \angle FEJ$, it is easy to establish that $\mu + \nu = 2\xi$ (actually $\mu = \xi + \angle JFE, \nu = \xi - \angle FEJ$). On the other hand one can calculate that $\angle CJF = \mu + \nu$, $\angle CH^\prime F = 2\xi$. Hence   $F, H^\prime, J, C$ are on a same circle. Similarly you can show this for $D, H^\prime, J, E$, and the claim follows. 
Next, note that
 $D, H, I, E$ are on a same circle. This means that $\triangle HCI$ is similar to $\triangle ECD$. Therefore:
\begin{align*}
\angle DAC + \angle ACD &= \angle CDE = \angle HIC = \pi - \angle H K C = \angle KHC + \angle ACD,  
\end{align*} 
where the third equality is due to the fact that $K, H, I, C$ are on a same circle. Hence $\angle DAC = \angle BAC =  \angle KHC$. But $\angle KHC = \angle KLC$ (because once again $H, K, C, L$ are on a same circle). Thus $\angle KLC = \angle BAC$ and we are done. 
A: Let us show first the following:

Let $\Delta ABC$ be a triangle.
  
We construct $D,E$ on $AB$ so that the triangles $\Delta CAE$, $\Delta CBC$ are isosceles.
The parallel to $CE$ through $D$ intersects $AC$ in $F$.
The parallel to $CD$ through $E$ intersects $AB$ in $G$.
In particular each triangle / trapez in the following list is isosceles: $ACE$, $AFD$, $CEDF$; $BCD$, $BGE$, $CDEG$. 
Since an isosceles trapez is inscribed in a circle, the points $C,D,E,F,G$ are on a circle. Let $J$ be its center. Then $J$ is on the perpendicular bisector of all the mentioned $3$-gons and $4$-gons w.r.t. suitable sides, in particular, $J$ it is on the angle bisectors from $A,B$ in $\Delta ABC$. So it is also on the remained angle bisector from $C$.
  We record:
  $$JC=JD=JE=JF=JG\ .$$
  Let the circles $(CJF)$ and $(CJG)$ intersect the circle $(JDE)$ in the points $H$, respectively $I$. 
Then we have the following colinearities:
  
  
*
  
*$F,H,E$
  
*$C,H,D$
  
*$A,H,J$
  
*$G,I,D$
  
*$C,I,E$
  
*$B,I,J$
  

Proof: The trapez $FDEC$ is isosceles, so the angles in $J$ of the two isosceles triangles $\Delta JDE$, $\Delta JCF$ are equal, being opposite to equal chords. In particular, the four angles (marked in green) at the bases / chords $DE$ and $CF$ are equal.
Let us show that $F,H,E$ are on a line. For this we compute the sum of he angle $\widehat{FHE}$.
$$
\begin{aligned}
\widehat{FHE}
&=
\widehat{FHJ}+\widehat{JHE}\\
&=
(180^\circ-\widehat{FCJ})+\widehat{JDE}\\
&=180^\circ\ ,
\end{aligned}   
$$
because the two "green angles" $\widehat{FCJ}$, $\widehat{JDE}$ cancel each other in measure. We have shown a first colinearity.
Let us show by copy+paste+change that $C,H,D$ are on a line. For this we compute the sum of he angle $\widehat{CHD}$.
$$
\begin{aligned}
\widehat{CHD}
&=
\widehat{CHJ}+\widehat{JHD}\\
&=
\widehat{CFJ}+(180^\circ-\widehat{JED})\\
&=180^\circ\ .
\end{aligned}   
$$
The triangles $\Delta JHC$ and $\Delta JHE$ are equal (well, congruent, but it is longer), because $JH$ is common side, $JC=JE$, and there are equal red angles in $H$. So $HC=HE$. This means, that $H$ is also on the perpendicular bisector of the side $CE$ in $\Delta ACE$, where $A,J$ live on. 
Per symmetry, the remained three colinearities are also proven.
$\square$.

To finish, we consider now also the points $K,L$. In the conditions of the previous result, we continue:

$J$ is the orthocenter of the triangle $CHI$
Let $K,L$ be the points of intersection of the circle $(CHI)$ with the lines $AC$, respectively $BC$.
  
Then we have $$
\begin{aligned}
HK &= HF = HD\ ,\\
IL &= IG = IE\ ,
\end{aligned}
$$
  and there is the congruence of quadrilaterals 
  $$ DEIH=KLIH\ .$$
Finally, $\Delta CAB\sim\Delta CLK$.

Proof:
$HJ$ is a segment on the the perpendicular bisector of the side $CI$ in $\Delta HCI$. So $J$ is on the height from $H$ of this triangle. By analogy, it is also on the height from $I$ of it, so it is the orthocenter in $\Delta CHI$. As a corollary, it is on the third height, so $CJ$ is perpendicular on $HI$. (This will soon help us to compute angles.)
Let us further show $\Delta HKF$ isosceles in $H$ by comparing the angles at the basis $KF$. Let us note in passing that $AFJE$ is inscriptible, because of the two "green angles", interior and exterior to it, in $E$ and $F$. So $\angle HFJ= \angle EFJ=\angle EAJ$ of measure $\frac 12\hat A$, $\hat A$ being the angle in $A$ of the initial triangle $\Delta ABC$. Now we start a longer computation, that joins $K$ and the needed angle to points introduced earlier and earlier:
$$
\begin{aligned}
\widehat{HKF}
&=\widehat{HIC}\qquad\text{(by construction of $K$ on $(CHI)$}\\
&=90^\circ - \widehat{JCI}\\
&=90^\circ - \widehat{JCG}+\widehat{ICG}\\
&=\frac 12(180^\circ - \hat C)+\widehat{ICB}\\
&=\frac 12(\hat A + \hat B)+\widehat{ICB}\\
&=\frac 12\hat A + (\widehat{IBC}+\widehat{ICB})\\
&=\frac 12\hat A + (180^\circ-\widehat{EIB})\\
&=\frac 12\hat A + \widehat{JDE}\\
&=\widehat{HFJ}+\widehat{JFC}\\
&=\widehat{HFC}=\widehat{HFK}\ .
\end{aligned}
$$
So $\Delta HFK$ is isosceles. We get $HK=HF$ and further $=HD$.
The other equality is fulfilled by analogy / by symmetry. 
Consider now the quadrilaterals
$$
DEIH\qquad\text{and}\qquad LKHI\ .$$
They are inscribed in circles of the same radius, $(CHI)$ and $(JDE)$. (I first saw this in the 5.th class as Problem of the 5L coin. In this second i do not have the reference in English...)
Now we start from $IH$ and construct in the one and the other circle segments  $HD=HK$. The triangles $\Delta IHD$ and $\Delta IHK$ are equal. Same for $\Delta HIE$, and $\Delta HIL$ in the other side of the picture. This shows the wanted equality of quadrilaterals.
We are now in position to compute angles in $\Delta CKL$, for instance:
$$
\begin{aligned} 
\widehat{CLK}
&=\widehat{CLI}-\widehat{KLI}\\
&=(180^\circ-\widehat{CHI})-\widehat{DEI}\\
&=\widehat{DHI}-\widehat{DEI}\\
&=180^\circ-\widehat{DEI}-\widehat{DEI}\\
&=180^\circ-\widehat{AEC}-\widehat{ACE}\\
&=\hat A\ .
\end{aligned} 
$$
$\square$
Note:
It was not my intention to get to the final the shortest path, rather i wanted to "show first everything in the given constellation"... The solution avoids intentionally inversion and metric relations.
