Geometry problem proving that all lines $DE$ passes through one point

Question

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…)

Answer 1

Let $F$ be on parallel to $BC$ through $I$ so that $AF||IE$, then $\angle AFI = \angle IEC$.

Lemma: Line $EF$ goes through fixed point which is a midpoint of arc $BC$ not containing $A$.

Proof: Let line $EF$ meet $AI$ at $P$. Use Thales theorem twice. We have $${a\over b+p} = {EF\over EP} = {b\over p}\implies p ={b^2\over a-b}$$ so $P$ is fixed point on $AI$.

Now we try to identify $P$. Put $E$ in $B$. We see that $$\angle AFI = \angle IEC = \angle IBC= \angle IBA$$ so $AFE(B)I$ is concyclic. Now also see $$\angle CAP = \angle IAB = \angle BFI = \angle PBC$$ so $P$ lies also on circumcircle $ABC$ and the lemma is proven.