I like this problem! Invert and reflect everything across the internal angle bisector of $\angle BAC$. Then $A'$ becomes $\triangle {ABC}$'s circumcenter ${O}$, $E, F$ is on the external angle bisector of $\angle BAC$ such that $AB=BE$ and $AC=CF$. Let $BE$ and $CF$ intersect at $G$ and $BF$, $CE$ at $K$. We only need $OK \parallel EF$. Main claim: $K$ is the symmedian of $\triangle {ABC}$, thus ${OK}$ is $\triangle {ABC}$'s Brocard axis. Let $CE$ cut $AB$ at ${P}$ and $BF$ cut $AC$ at ${Q}$. Obviously $ABGC$ is a parallelogram and by $AD=BC$, $AB \cdot AC=BC^2$. $\frac{AP}{PB}=\frac{FC}{CG}=\frac{AC}{AB}=\frac{AC^2}{BC^2}$ and similarly $\frac{AQ}{QC}=\frac{BA^2}{BC^2}$, so the conclusion follows. Let $O_B, O_C$ be the circumcenters of the ${B}-$ and ${C}-$ Apolonius circles respectively. $$\frac{ \overline {AO_B}}{ \overline {AO_C}}=\frac{ \frac{AB^2 \cdot AC}{AB^2-BC^2}}{ \frac{AC^2 \cdot AB}{AC^2-BC^2}}=\frac{ \frac{AB^2 \cdot AC}{AB^2-AB \cdot AC}}{ \frac{AC^2 \cdot AB}{AC^2-AB \cdot AC}}=-1$$So $O_BO_C \perp EF$, which is $OK \parallel EF$. Done!

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a beautiful solution,I like it!

In $\triangle {ABC}$, ${D}$ is on the internal angle bisector of $\angle BAC$ and $\angle ADB=\angle ACD$. $E, F$ is on the external angle bisector of $\angle BAC$, such that $AE=BE$ and $AF=CF$. The circumcircles of $\triangle ACE$ and $\triangle ABF$ intersects at ${A}$ and ${K}$ and $A'$ is the reflection of ${A}$ with respect to $BC$. Prove that: if $AD=BC$, then the circumcenter of $\triangle AKA'$ is on line $AD$.