$A'$ is a moving point of side $BC$ of $\triangle ABC$. The perpendicular bisector of $A'B$ and $A'C$ cuts side $AB$ and $AC$ respectively at $B'$ and $C'$. Line $d$ passes through $A'$ and is perpendicular to $B'C'$. Prove that $d$ passes through a fixed point.
I have predicted that $d$ would pass through point $A''$ in which $AA'' \perp BC$ and $A''$ lies on the circumcircle of $\triangle ABC$. But I haven't found out a way to prove that yet.