Given is a triangle $ABC$ with its circumscribed circle and $| AC | <| AB |$. On the short arc $AC$, there is a variable point $D\ne A$. Let $E$ be the reflection of $A$ wrt the inner bisector of $\angle BDC$. Prove that the line $DE$ passes through a fixed point, regardless of point $D$.
Let $G,F$ are point of intersections bisector of $\angle BDC$ and $DE$ with circumcircle of $ABC$.
As $DG$ is bisector of $BDC$ then $AG$ is bisector of $BAC$. $AG=EG, \angle DEG=\angle DAG= \angle DFG$ because $E$ is reflection of $A$. So $FG=GE=AG$
So $F$ is point of intersection of circle with center $G$ ( point of intersection of angle bisector of $\angle BAC$ and circumcircle $ABC$) and radius $AG$ with circumcircle $ABC$ always lies on $DE$ not depending on $D$
Let $G$ and $M$ be the midpoints of $AE$ and arc $BC$ without $A$ respectively. Also let $F$ be a point in $(ABC)$ such that $AF||BC$. We just prove that $DEF$ are colinear,
$\angle GDE=\angle ADG=\angle AFM=\angle FAM=\angle FDM$.
So we are done.