In $\triangle ABC$ it is given that $\angle ABC=60$.
Let the perpendicular bisector of $AC$ meet angle bisector of $\angle B$ at $X$.
Prove that $X$ lies on the circumcircle of $\triangle ABC$.
I have tried angle chasing but I am not able to conclude that $\angle XBC$ = $\angle XCB$ = $30$.
Two versions for you:
On the left, draw the circumcircle, start by forming the perpendicular bisector on $AC$, shown in gray. Then note that the angles marked by $1$ are the same since they subtend the same chord on the circle. Likewise for $2$. Finally, note that the perpendicular bisector causes $AZX \cong CZX$, so that $1=2$ and $BX$ is an angle bisector.
Or, on the right, start with the angle bisector $BX$, shown in gray. Again by inscribed angles subtending the same chords, we also have angle $1$ in two corners of $\triangle ACX$, implying it is isosceles, and hence if we draw its altitude that it will bisect $AC$ at right angles.