Example 9. In $\triangle A B C$, $\angle B A C=60^{\circ}$, $\angle A C B=45^{\circ}$. (1) Find the ratio of the three sides of this triangle $A B: B C: A C$; (2) Let $P$ be a point inside $\triangle A B C$, and $P A=\sqrt{6}+\sqrt{2}, P B=3 \sqrt{2}+\sqrt{6}$, $P C=3 \sqrt{2}+2 \sqrt{6}$. Find $\angle A P B, \angle B P C$, $\angle C P A$.
(1) As shown in the figure, draw the altitude $B D$ on side $A C$, and let $A D=x$, then
$$
\begin{array}{l}
B D=\sqrt{3} x, A B=2 x \text {. } \\
\therefore D C=B D=\sqrt{3} x \text {, } \\
B C=\sqrt{6} x \text {, } \\
A C=A D+D C=(1+\sqrt{3}) x \text {. } \\
\end{array}
$$

Thus,
$$
A B: B C: A C=2: \sqrt{6}:(1+\sqrt{3}) \text {. }
$$
(2) As shown in the figure, construct $\triangle A E B \sim$ $\triangle A P C$ outside $\triangle A B C$, and make
$$
\frac{A E}{A P}=\frac{E B}{P C}=\frac{B A}{C A},
$$

i.e.,
$$
\begin{array}{l}
\frac{A E}{\sqrt{6}+\sqrt{2}}=\frac{E B}{3 \sqrt{2}+2 \sqrt{6}} \\
=\frac{2}{1+\sqrt{3}} . \\
\therefore A E=2 \sqrt{2}, B E=3 \sqrt{2}+\sqrt{6} .
\end{array}
$$

In $\triangle A P E$ and $\triangle A C B$,
$$
\begin{array}{l}
\frac{A E}{A P}=\frac{A B}{A C}=\frac{2}{1+\sqrt{3}}, \\
\angle E A P=\angle B A E+\angle P A B \\
\quad=\angle C A P+\angle P A B=\angle B A C .
\end{array}
$$
$\therefore \triangle A P E \sim \triangle A C B$.
Hence $\angle A P E=\angle A C B=45^{\circ}$,
$$
\frac{P E}{A E}=\frac{C B}{A B}=\frac{\sqrt{6}}{2} . \quad \therefore P E=2 \sqrt{3} \text {. }
$$

In $\triangle P E B$, we have
$$
B H=\sqrt{B P^{2}-\left(\frac{P E}{2}\right)^{2}}=3+2 \sqrt{3} \text {. }
$$

Thus, $P M=\frac{P E \cdot B H}{B E}=\frac{3 \sqrt{2}+\sqrt{6}}{2}$
$$
\begin{array}{c}
=\frac{P B}{2} . \\
\therefore \angle P B E=30^{\circ} .
\end{array}
$$

Therefore, $\angle B P E=\frac{1}{2}\left(180^{\circ}-\angle P B E\right)=75^{\circ}$.
$$
\begin{aligned}
\therefore & \angle A P B=\angle A P E+\angle B P E \\
& =45^{\circ}+75^{\circ}=120^{\circ} .
\end{aligned}
$$

Also, $\angle A E P=180^{\circ}-\angle A P E-\angle P A E=75^{\circ}$,
$$
\begin{aligned}
\therefore & \angle A P C=\angle A E B=\angle A E P+\angle B E P \\
& =75^{\circ}+75^{\circ}=150^{\circ} .
\end{aligned}
$$

Thus,
$$
\begin{array}{l}
\angle B P C=360^{\circ}-(\angle A P E+\angle A P C) \\
=90^{\circ} .
\end{array}
$$