By the properties of inscribed angles $\angle CXO = \angle ABO$ (Fig. 8) independent of whether the point $X$ lies on the side $CA$ or on the extension of $CA$ over $A$. Since $O$ is the circumcentre of $ABC$, we have $\angle ABO = \angle BAO$. Let the intersection of lines $XO$ and $BC$ be $Y$ and let the point on the circumcircle of $ABC$ lying diametrically opposite $A$ be $Z$. Since $\angle CXY = \angle ZAB$, and $\angle YCX = \angle BZA$ as inscribed angles subtending the same arc $AB$, the triangles $CXY$ and $ZAB$ are similar. Hence $\angle CYX = \angle ZBA = 90^\circ$, as $\angle ZBA$ is subtended by a diameter.

Fig. 8

Obviously $\angle OBC = \angle OCB$. If $X$ lies on segment $AC$ (Fig. 9), then $\angle OBX = \angle OAX = \angle OAC = \angle OCA$. If $X$ lies on the extension of $AC$ over $A$, then $\angle OBX = 180^\circ - \angle OAX = \angle OAC = \angle OCA$. In both cases we have $\angle OBX = \angle OCA$ and $\angle XBC = \angle OBX + \angle OBC = \angle OCA + \angle OCB = \angle ACB = \angle XCB$. Therefore $XBC$ is an isosceles triangle with the apex $X$, hence $X$ lies on the perpendicular bisector of $BC$. Since $O$ lies on the same line, the line $XO$ is the perpendicular bisector of the side $BC$.

Fig. 9