Lemma:$A$&$B$&$C$&$D$ are distinct points and $(AC,BD)=P$.Circumcircles of $ABP$ and $CDP$ intersect at $P $and $Q$. $Q$ is the centre of spiral similarity that sends $A $to$ C$ and $B$ to$ D$.also it's the spiral similarity that sends $A $to $B$ and $C $to$ D$. the proof is trivial. Problem: Here is proof that include the main idea of problem.(but somehow not really complete) suppose that $B$ and ${B}'$ are on a ray Called $x$ and also both of these points are on one side of $A$ on ray $x$ and also suppose that $C$ and ${C}'$ are on a ray Called $y$ and also both of these points are on one side of $A$ on ray $y$ such that $AB+AC=A{B}'+A{C}'$. Circumcircles of $ABC$ and ${A}'{B}'{C}'$ intersect at $P$ and another point called $T$. $T$ is the centre of spiral similarity that sends $B$ and ${B}'$ to $C$ and ${C}'$, By our assumption we have $B{B}'=C{C}'$ so the dialation is 1 and we have $TC=TB$.Regarding to this fact that $ACTB$ is cyclic , $AD$ is the bisector of $ xAy$ so $D$ is intersection point of the bisector and circumcircle of$ ACB$. other cases can be solved like this . sincerely, Daniel23

Consider the angular bisector of $\angle BAC$ .We prove that all such varying cirles intersect on this line. Let $X$ be the point on the ray $AB$ such that $B$ lies between $A$ and $X$ and $AB = BX$ .Hence length $AX$ is a constant Let $Y$ be the point on the angular bisector of $\angle A$ such that $\angle AXY= \frac{A}{2} \Longrightarrow \Delta ACY$ congruent to $\Delta BYX \Longrightarrow BY=YC$ Since in any triangle the angular bisector and the perpendicular bisector of the opposite side intersect on the circumcircle $Y$ which is a fixed point lies on the circumcircle of $\Delta ABC$

Suppose, w.l.o.g. $AB>AC$. Let $B'\in (AB), C'\in (AC$ so that $AB'=AC'=\frac{AB+AC}{2}$. As constructed, obviously, $BB'=CC'\ (\ 1\ )$. Let $X$ from angle bisector of $\angle BAC$ so that $B'AC'X$ is cyclic. Obviously $B'X\bot AB, C'X\bot AC$ and $B'X=C'X\ (\ 2\ )$. From $(1)\wedge (2)\implies \Delta BB'X\cong\Delta CC'X$, i.e. $\angle B'BX=\angle C'CX$, hence $BACX$ is cyclic, done. Best regards, sunken rock