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...)


Comment:
As can be seen AD is always tangent to a circle center on I. Due to assumption the ratio of radius of inscribed circle (IG=r) and radii of these circles is constant.  Consider position D' . $\angle D'AI=90^o$, E' is intersection of diameter D'P and BC. Let the foot of altitude from I on BC be G. Segments AI and IG are opposite to equal angles AD'I and IEC. The measures of AI and IG are constant . In this position  DE(D'E') is coincident on diameter D'P and  crosses midpoint P of arc BC. In position where D and E are coincident on B clearly $\angle ADI(=ABI) = \angle IEC(=IBC)$   because IB is bisector of ABC. In position where D is between A and B , ID  and IE are parallel with BC  and AD respectively, So $\angle ADE=\angle IEF$. The segments opposite to angles ADI and IEF are AI and IF. The measure of these two segments are constant.Now for three positions along arc AB the statement is true. This deduces that $\angle IEC=\angle ADI$  for all points along arc AB, because it is true for  three positions. This is competent with our assumption.
