II. (25 points) Given $\triangle A B C$, where $B$ and $C$ are fixed points and $A$ is a moving point. Construct the diameters $A A_{1}$, $B B_{1}$, and $C C_{1}$ of the circumcircle of $\triangle A B C$. Let $M=S_{\triangle A_{1} B C}$, $N=S_{\triangle A_{2} B C}+S_{\triangle A B_{1} C}+S_{\triangle A D C_{1}}$. Determine the position of $A$ such that $M=N$. Why?
(1) If $\triangle A B C$ is an obtuse triangle, it is obvious that $M>N$;
(2) If $\triangle A B C$ is a right triangle, it is obvious that $M>N$;
(3) If $\triangle A B C$ is an acute triangle, let $O$ be the circumcenter of $\triangle A B C$, and draw $O P_{1} \perp A B, O P_{2} \perp B C, O P_{3} \perp A C$, with $P_{1}$, $P_{2}, P_{3}$ being the feet of the perpendiculars.
$\because A A_{1}$ is a diameter,
$\therefore \angle A B A_{1}=90^{\circ}, O P_{1} / / A_{1} B$.
$\therefore S_{\triangle A P_{1} o}=\frac{1}{4} S_{\triangle A B A_{1}}$.

Similarly, $S_{\triangle A P_{2} o}=\frac{1}{4} S_{\triangle A A_{1} c}$.
$\therefore \quad S_{\text {quadrilateral } A P_{1} O P_{2}}=\frac{1}{4} S_{\text {quadrilateral } A B A_{1} C}$.
$$
=\frac{1}{4}\left(M+S_{\triangle B A_{1} c}\right) .
$$

Similarly,
$S_{\text {quadrilateral } B P_{1} O P_{3}}=\frac{1}{4}\left(M+S_{\triangle A B_{1} C}\right)$;
$S_{\text {quadrilateral } C P_{1} O P_{2}}=$
$$
\frac{1}{4}\left(M+S_{\triangle A C_{1} B}\right) \text {. }
$$

Adding the three equations gives
$S_{\triangle A B C}=\frac{1}{4}(3 M+N)$.
Thus $M=N$.
$\therefore$ When point $A$ lies outside the circle with $B C$ as the diameter, and the two lines passing through $B, C$ are perpendicular to $B C$ respectively.