Let $A = (a,n), B = (0,0),$ and $C = (c,0)$ Perpendicular bisector of $BC$ is $x = \frac{c}{2}$. Perpendicular bisector of $AB$ is $y = -\frac{a}{n} x + \frac{a^2+n^2}{2n}$ $O$ is the intersection of these $2$ lines, so we get $y = -\frac{a}{n} (\frac{c}{2}) + \frac{a^2+n^2}{2n} \Rrightarrow y = \frac{a^2-ac+n^2}{2n}$ and $x = \frac{c}{2} \Rrightarrow$ $$O = (\frac{c}{2}, \frac{a^2-ac+n^2}{2n})$$Let the circumcenter of $AOB$ be $P$. As before, perpendicular bisector of $AB = -\frac{a}{n}x + \frac{a^2+n^2}{2n}$, and we can find that the perpendicular bisector of $BO$ is given by $y = \frac{-cn}{a^2-ac+n^2} x + \frac{(a^2+n^2)(a^2-2ac+c^2+n^2)}{4n(a^2-ac+n^2)}$ $P$ is the intersection of these $2$ lines, so setting them equal and solving, we get $$P = (\frac{a^2-c^2+n^2}{4(a-c)}, \frac{a^3-2a^2c+ac^2+an^2-2cn^2}{4n(a-c)}$$$X$ is the point of intersection, other than $A$, of the circle centered at $P$ and the line $AC$. We see the circle centered at $P$ is given by the equation $$(x- \frac{a^2-c^2+n^2}{4a-4c})^2 + (y - \frac{a^3-2a^2c+ac^2+an^2-2cn^2}{4n(a-c)})^2 = (\frac{a^2-c^2+n^2}{4a-4c})^2 + (\frac{a^3-2a^2c+ac^2+an^2-2cn^2}{4n(a-c)})^2$$Line $AC$ is given by the equation $y = \frac{-n}{c-a}x + \frac{cn}{c-a}$. Substituting this value of $y$ into the equation of the circle, and simplifying, we get $$x^2 + 2x(\frac{a^2-c^2+n^2}{4a-4c}) + y^2 - 2y(\frac{a^3-2a^2c+ac^2+an^2-2cn^2}{4n(a-c)}) = 0$$$$\Rrightarrow x^2 + 2x(\frac{a^2-c^2+n^2}{4a-4c}) + (\frac{-n}{c-a}x + \frac{cn}{c-a})^2 - 2(\frac{-n}{c-a}x + \frac{cn}{c-a})(\frac{a^3-2a^2c+ac^2+an^2-2cn^2}{4n(a-c)}) = 0$$Solving for $x$ in the above equation, we get $x = a, y = n$, which are the coordinates of point $A$, and $x = \frac{c}{2}, y = \frac{cn}{2(c-a)}$. Thus, $XO \perp BC$. $\blacksquare$

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