Problem 1305

【Analysis】Let the circumradius of $\triangle B H C$ be $R^{\prime}, O$ be the circumcenter of $\triangle A B C$, the extension of $B H$ intersects $A C$ at point $S$, the extension of $C H$ intersects $B A$ at point $T$, and $O O^{\prime}$ intersects $B C$ at point $M$. Then
$\angle B H C=\pi-\angle C A B$.
By the Law of Sines, we have
$2 R \sin \angle C A B=B C=2 R^{\prime} \sin \angle B H C$
$=2 R^{\prime} \sin \angle C A B$
$\Rightarrow R=R^{\prime}$
$\Rightarrow O B=O C=R=R^{\prime}=O^{\prime} B=O^{\prime} C$
$\Rightarrow$ Quadrilateral $O B O^{\prime} C$ is a rhombus
$\Rightarrow O O^{\prime}$ is perpendicular to and bisects $B C$ at point $M$.
Taking $O$ as the origin and the direction of ray $M O$ as the positive direction of the $y$-axis, we establish a Cartesian coordinate system.
Let point $M(0,-R \sin \alpha)$. Then
$$\begin{array}{l} B(-R \cos \alpha,-R \sin \alpha), C(R \cos \alpha,-R \sin \alpha), \\ O^{\prime}(0,-2 R \sin \alpha), \end{array}$$

where $\alpha \in\left(0, \frac{\pi}{2}\right)$.
Let $A(R \cos \beta, R \sin \beta)(\beta \in(0, \pi))$.
Since $N$ is the midpoint of segment $A O^{\prime}$, we have
$$N\left(\frac{R}{2} \cos \beta, \frac{R}{2}(\sin \beta-2 \sin \alpha)\right) \text {. }$$

Then the point symmetric to $N$ with respect to $B C$ is
$$D\left(\frac{R}{2} \cos \beta,-\frac{R}{2}(\sin \beta+2 \sin \alpha)\right) \text {. }$$

Therefore, $A 、 B 、 D 、 C$ are concyclic
$\Leftrightarrow$ Point $D$ lies on the circumcircle of $\triangle A B C$
$\Leftrightarrow R^{2}=O D^{2}=\frac{R^{2}}{4}\left(1+4 \sin \alpha \cdot \sin \beta+4 \sin ^{2} \alpha\right)$
$\Leftrightarrow 4 \sin \alpha \cdot \sin \beta+4 \sin ^{2} \alpha=3$.
Notice that,
$$\begin{array}{l} b^{2}=A C^{2} \\ =R^{2}(2-2 \cos \alpha \cdot \cos \beta+2 \sin \alpha \cdot \sin \beta), \\ c^{2}=A B^{2} \\ =R^{2}(2+2 \cos \alpha \cdot \cos \beta+2 \sin \alpha \cdot \sin \beta), \\ a^{2}=B C^{2}=4 R^{2} \cos ^{2} \alpha . \end{array}$$

Then $b^{2}+c^{2}-a^{2}$
$$\begin{array}{l} =R^{2}\left(4+4 \sin \alpha \cdot \sin \beta-4 \cos ^{2} \alpha\right) \\ =R^{2}\left(4 \sin \alpha \cdot \sin \beta+4 \sin ^{2} \alpha\right) . \end{array}$$

Hence $b^{2}+c^{2}-a^{2}=3 R^{2}$
$\Leftrightarrow 4 \sin \alpha \cdot \sin \beta+4 \sin ^{2} \alpha=3$.
In summary, the original proposition holds.