Prove that in an acute triangle a line drawn through $A$ and the circumcenter is perpendicular to the reflection of $BC$ on the angle bisector of $A$

Question

Here is a picture provided:

The angle shown here is what we need to prove.

This is where I got: suppose $BAC$ angle is $a$; $ABC$ is $b$; and $BCA$ is $c$. $BC$ and $B'C'$ intersect at $K$.

Then $AC'B'=c$ and $C'B'A=b$

$CKB'=b-c$

and I couldn't get any further. Any help is appreciated.

Answer 1

Let $O$ be the circumcentre. $\angle COA=2\angle B$ and $\triangle AOC$ is isosceles, so $\angle OAC=90^\circ-\angle B$.

Combined with $\angle AB'C'=\angle CBA=\angle B$ we get the result.