Geometry problem proving that all lines $DE$ passes through one point 
Let $I$ is the incenter of $\triangle ABC$. Let $K$ be the circumcircle of $ABC$. Let $D$ be a variable point on arc $AB$ on $K$ not containing $C$. Let $E$ be a point on line segment $BC$ such that $\angle ADI = \angle IEC$. Prove that as $D$ varies on arc $AB$, the line $DE$ passes through one point.

First after testing a couple cases i found the point is the midpoint of arc $BC$. I've tried direct proof and reverse reconstructing the problem, but nothing worked. Can anyone help?
(edit: I reflected $E'$ in the bisector of $C$ and created a cyclic quadrilateral $ADIE'$. Maybe it is a useful property...)
(edit 2: I suspect inversion might be involved...)

Here is a detailed proof, also using "the point $F$" as in the post of
Aqua, it is the point that best "geometrically interpolates" between $D$ and $E$. The definition of $F$ below will be the one, that lets us give a direct (instead of indirect) proof.

We start by constructing the parallel $(f)$ to $BC$ through $I$. Then let us draw the circle $(\Gamma)$ through $A,D,I$, it intersects $(f)$ first time $I$, and for a second time (counting multiplicities) in a point $F$. This is the definition of $F$. By construction, we have the same measure $x$ of angles...
$$
x
=\widehat{IEC}
=\widehat{ADI}
=\widehat{AFI}
\ .
$$
The points $A'$, and $S$ are the intersections of the angle bisector $AI$ (in $A$ of $\Delta ABC$) with the side $BC$, and respectively with the circumcenter $(K)=\odot(ABC)$.
Claim:

*

*$(1)$ The points $S,D,F$ are colinear.

*$(2)$ The points $S,E,F$ are colinear.




It is clear, that the OP follows from the above, showing that $DE$ passes through the fixed point $S$, the mid point of the arc $\overset\frown{BC}$ opposite to the vertex $A$.

Proof of the claim:
$(1)$
Using the two circles, the given $(K)$ and the constructed $(\Gamma)$ we have
the equality of measures of angles:
$$
\begin{aligned}
\widehat{ADS}
&\overset{(K)}=
\widehat{ABS}
=
\widehat{ABC}+\widehat{CBS}
=
\hat B+\widehat{CAS}
=
\hat B+\frac 12\hat A\ ,
\\
\widehat{ADF}
&\overset{(\Gamma)}=
\widehat{FIS}
=
\widehat{BA'S}
=
\frac12(\overset\frown{BS}+\overset\frown{AC})
=
\frac12\overset\frown{BS}+\frac12\overset\frown{AC}
=
\widehat{BAS}+\widehat{ABC}
=
\frac 12\hat A+\hat B\ ,
\\
&\qquad\text{ giving }
\\
\widehat{ADS}
&=
\widehat{ADF}
\ ,
\end{aligned}
$$
so the points $D,F,S$ are colinear. This concludes $(1)$.

$(2)$ We will show that $F,E,S$ are colinear by the reciprocal of the theorem of Thales, showing that proportions of corresponding sides in the a posteriori valid similarity $\Delta SA'E\sim\Delta SI$ are equal. Here is the computational path:
$$
\begin{aligned}
\frac {FI}{EA'}
&=
\frac {FI}{EA'} &&\text{ using }\Delta AIF\sim\Delta IA'E
\\
&=
\frac {AB}{BA'} &&\text{ using the angle bisector $BI$ in}\Delta ABA'
\\
&=
\frac {SC}{SA'} &&\text{ using }\Delta SA'C\sim\Delta BA'A
\\
&=
\frac {SI}{SA'} &&\text{ using $\Delta SIC$ isosceles, $SC=SI$}
\ .
\end{aligned}
$$
The needed proportions are thus equal, so $F,E,S$ are colinear. This concludes $(2)$, and the claim.
