Example 4 As shown in Figure 2, given a convex quadrilateral $ABCD, \angle B + \angle D < 180^{\circ}, P$ is a moving point on the plane. Let
$$
\begin{aligned}
f(P)= & PA \cdot BC + \\
& PD \cdot CA + \\
& PC \cdot AB.
\end{aligned}
$$
(1) Prove that when $f(P)$ reaches its minimum value, points $P, A, B, C$ are concyclic;
(2) Let $E$ be a point on the arc $\overparen{AB}$ of the circumcircle $\odot O$ of $\triangle ABC$, satisfying $\frac{AE}{AB}=\frac{\sqrt{3}}{2}, \frac{BC}{EC}=\sqrt{3}-1, \angle ECB = \frac{1}{2} \angle ECA$, and $DA, DC$ are tangents to $\odot O$, $AC = \sqrt{2}$. Find the minimum value of $f(P)$.
(1) Proof: In quadrilateral $PABC$ (including concave and degenerate quadrilaterals), by the generalized Ptolemy's theorem, we have
$PA \cdot BC + PC \cdot AB \geqslant PB \cdot AC$.
Thus, $f(P) \geqslant PB \cdot AC + PD \cdot CA$
$= AC(BP + PD) \geqslant AC \cdot BD$.
Equality holds if and only if $P$ is the intersection of the circumcircle of $\triangle ABC$ and $BD$. Therefore, the minimum value of $f(P)$ is $AC \cdot BD$.
(2) Solution: As shown in Figure 3, let $AE = \sqrt{3} x$, $AB = 2x$, $BC = (\sqrt{3} - 1) y$, $EC = y$, and the midpoint of $\overparen{AE}$ be $F$. Then $BF = AE = \sqrt{3} x$, $BE = EF = FA$.
In quadrilateral $ABEF$, by Ptolemy's theorem, we have
$AF \cdot BE + EF \cdot AB = AE \cdot BF$.
Thus, $BE^2 + BE \cdot 2x = (\sqrt{3} x)^2$,
$(BE - x)(BE + 3x) = 0$.
Therefore, $BE = x$.
Since $AE^2 + BE^2 = 4x^2 = AB^2$, we have $\angle AEB = 90^\circ$, and $AB$ is the diameter.
In quadrilateral $ACBE$, by Ptolemy's theorem, we have
$$
AE \cdot BC + BE \cdot AC = EC \cdot AB,
$$
which simplifies to $\sqrt{3} x (\sqrt{3} - 1) y + x \sqrt{2} = y \cdot 2x$.
Thus, $y = \frac{\sqrt{2} + \sqrt{6}}{2}$.
Since $BC = (\sqrt{3} - 1) y = \sqrt{2} = AC$ and $\angle ACB = 90^\circ$, $\triangle ABC$ is an isosceles right triangle with $AB = \sqrt{2} AC = 2$.

Also, $\angle DCA = \angle DAC = \angle ABC = 45^\circ$, so $\triangle ACD$ is also an isosceles right triangle.
Thus, $AD = \frac{AC}{\sqrt{2}} = 1$.
Since $DA$ is a tangent and $AB$ is the diameter, $\angle DAB = 90^\circ$.
Therefore, $BD = \sqrt{AD^2 + AB^2} = \sqrt{5}$.
Hence, the minimum value of $f(P)$ is $AC \cdot BD = \sqrt{2} \times \sqrt{5} = \sqrt{10}$.