Question 1 As shown in Figure 1, given a convex quadrilateral $ABCD$, $\angle B + \angle D < 180^{\circ}$, and $P$ is a moving point on the plane. Let
$$
\begin{array}{l} 
f(P) \\
= PA \cdot BC + \\
PD \cdot CA + \\
PC \cdot AB.
\end{array}
$$
(1) Prove:
When $f(P)$ reaches its minimum value, points $P$, $A$, $B$, and $C$ are concyclic;
(2) Let $E$ be a point on the arc $\overparen{AB}$ of the circumcircle $\odot O$ of $\triangle ABC$, satisfying
$$
\begin{array}{l}
\frac{AE}{AB} = \frac{\sqrt{3}}{2}, \frac{BC}{EC} = \sqrt{3} - 1, \\
\angle ECB = \frac{1}{2} \angle ECA,
\end{array}
$$
and $DA$, $DC$ are tangents to $\odot O$, with $AC = \sqrt{2}$. Find the minimum value of $f(P)$.
(1) Proof: As shown in Figure 1, construct \(\triangle A P B^{\prime} \backsim \triangle C P B\). Then \(\angle B P B^{\prime} = \angle C P A\),
\[
\frac{A P}{C P} = \frac{B^{\prime} P}{B P}, \quad \frac{A B^{\prime}}{B C} = \frac{P A}{P C}.
\]

Thus, \(\triangle B^{\prime} P B \backsim \triangle A P C\).
Therefore, \(\frac{B B^{\prime}}{A C} = \frac{P B}{P C}\).
Hence, \(A B^{\prime} = \frac{P A \cdot B C}{P C}, \quad B B^{\prime} = \frac{P B \cdot A C}{P C}\).
Notice that \(A B^{\prime} + A B \geqslant B B^{\prime}\), so,
\[
\frac{P A \cdot B C}{P C} + A B \geqslant \frac{P B \cdot A C}{P C},
\]

which simplifies to \(P A \cdot B C + P C \cdot A B \geqslant P B \cdot A C\).
Equality holds if and only if point \(A\) lies on segment \(B B^{\prime}\). Therefore, \(P, A, B, C\) are concyclic. Thus,
\[
\begin{array}{c}
f(P) = P A \cdot B C + P D \cdot C A + P C \cdot A B \\
\geqslant P B \cdot A C + P D \cdot C A \geqslant B D \cdot A C.
\end{array}
\]

Therefore, when and only when \(P\) is the intersection of the circumcircle of \(\triangle A B C\) and \(B D\), \(f(P)\) attains its minimum value
\[
f(P)_{\min} = A C \cdot B D.
\]

(2) Solution: As shown in Figure 1, extend \(A B\) to point \(F\) such that \(B F = B E = m\).
Let \(A E = \sqrt{3} x, A B = 2 x\),
\[
E C = y, \quad B C = (\sqrt{3} - 1) y.
\]

Since \(\angle E C B = \frac{1}{2} \angle E C A\), it is easy to prove that
\[
\angle E A B = \frac{1}{2} \angle E B A = \angle B E F = \angle F.
\]

Thus, \(A E = E F = \sqrt{3} x\), and \(\triangle F B E \backsim \triangle F E A\). Hence, \(\frac{E F}{F A} = \frac{B F}{E F}\), which simplifies to \(\frac{\sqrt{3} x}{2 x + m} = \frac{m}{\sqrt{3} x}\).
Solving this, we get \(m_1 = x, m_2 = -3 x\) (discard \(m_2\)).
By the Pythagorean theorem, \(\angle A E B = 90^\circ\). Thus, \(A B\) is the diameter of \(\odot O\), i.e., \(\angle B A D = 90^\circ\).
By Ptolemy's theorem,
\[
A C \cdot B E + A E \cdot B C = A B \cdot E C.
\]

Substituting the corresponding expressions, we get
\[
(\sqrt{3} - 1) y = \sqrt{2},
\]

which simplifies to \(B C = A C = \sqrt{2}\).
Thus, \(A B = 2\), and \(\triangle A B C\) is an isosceles right triangle. Therefore, it is easy to prove that \(\triangle A D C\) is also an isosceles right triangle.
So, \(A D = 1, B D = \sqrt{5}\).
Combining with (1), we know \(f(P)_{\min} = \sqrt{10}\).