Let $A N$ meet the circumcircle of $A B C$ at point $K$, the midpoint of arc $B C$ that does not contain $A$.

The orthogonal projection of $K$ onto side $B C$ is $M$. Let $R$ and $S$ be the orthogonal projections of $K$ onto lines $A B$ and $A C$, respectively. Points $R, M$, and $S$ lie in the Simson line of $K$ with respect to $A B C$. Since $K$ is in the bisector of $\angle B A C$, $A R K S$ is a kite, and the Simson line $R M S$ is perpendicular to $A N$, and therefore parallel to $P Q$.

Now consider the homothety with center $A$ that takes $O$ to $K$. Since $O P \perp A B$ and $K R \perp A B$, $O P$ and $K R$ are parallel, which means that $P$ is taken to $R$. Finally, line $P Q$ is parallel to line $R S$, so line $P Q$ is taken to line $R S$ by the homothety. Then $Q$ is taken to $M$, and since $O$ is taken to $K$, line $O Q$ is taken to line $M K$. We are done now: this means that $O Q$ is parallel to $M K$, which is perpendicular to $B C$ (it is its perpendicular bisector, as $M B=M C$ and $K B=K C$.)

Consider a cartesian plane with $A=(0,0)$ as the origin and the bisector $A N$ as $x$-axis. Thus $A B$ has equation $y=m x$ and $A C$ has equation $y=-m x$. Let $B=(b, m b)$ and $C=(c,-m c)$. By symmetry, the problem is immediate if $A B=A C$, that is, if $b=c$. Suppose that $b \neq c$ from now on. Line $B C$ has slope $\frac{m b-(-m c)}{b-c}=\frac{m(b+c)}{b-c}$. Let $N=(n, 0)$.

Point $M$ is the midpoint $\left(\frac{b+c}{2}, \frac{m b-m c}{2}\right)$ of $B C$, so $A M$ has slope $\frac{m(b-c)}{b+c}$.

The line through $N$ that is perpendicular to the $x$-axis $A N$ is $x=n$. Therefore
$$P=(n, m n) \quad \text{ and } \quad Q=\left(n, \frac{m(b-c) n}{b+c}\right) .$$
In the right triangle $A P O$, with altitude $A N$, $A N \cdot A O=A P^{2}$. Thus
$$n \cdot A O=(0-n)^{2}+(0-m n)^{2} \Longleftrightarrow A O=n\left(m^{2}+1\right) \Longrightarrow O=\left(n\left(m^{2}+1\right), 0\right) .$$
Finally, the slope of $O Q$ is
$$\frac{\frac{m(b-c) n}{b+c}-0}{n-n\left(m^{2}+1\right)}=-\frac{b-c}{(b+c) m} .$$
Since the product of the slopes of $O Q$ and $B C$ is
$$-\frac{b-c}{(b+c) m} \cdot \frac{m(b+c)}{b-c}=-1,$$
$O Q$ and $B C$ are perpendicular, and we are done.