1. As shown in the figure, given a convex quadrilateral $ABCD$, $\angle B + \angle D < 180^\circ$, and $P$ is a moving point on the plane. Let $f(P) = PA \cdot BC + PD \cdot CA + PC \cdot AB$.
(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 $\overarc{AB}$ of the circumcircle $\odot O$ of $\triangle ABC$, such that $\frac{AE}{AB} = \frac{\sqrt{3}}{2}$, $\frac{BC}{EC} = \sqrt{3} - 1$, and $\angle ECB = \frac{1}{2} \angle ECA$. Also, $DA$ and $DC$ are tangents to $\odot O$, and $AC = \sqrt{2}$. Find the minimum value of $f(P)$.
(1) By Ptolemy's theorem, we have $P A \cdot B C + P C \cdot A B \geqslant P B \cdot A C$, with equality if and only if points $P, A, B, C$ are concyclic, and $P$ lies on the arc $\overparen{A C}$.
Therefore, $f(P) \geqslant P D \cdot C A + P B \cdot A C = A C (P D + P B)$
$\geqslant A C \cdot B D$, with equality if and only if points $B, P, D$ are collinear.
In summary, the minimum value of $f(P)$ is $A C \cdot B D$, with equality if and only if points $B, P, D$ are collinear, and $P$ lies on the circumcircle $O$ of $\triangle A B C$, i.e., points $P, A, B, C$ are concyclic.
(2) As shown in the figure, according to the problem, $\frac{A E}{A B} = \frac{\sin 2 \alpha}{\sin 3 \alpha} = \frac{2 \sin \alpha \cos \alpha}{3 \sin \alpha - 4 \sin ^{3} \alpha} = \frac{2 \cos \alpha}{4 \cos ^{2} \alpha - 1} = \frac{\sqrt{3}}{2}$
$\Rightarrow 4 \sqrt{3} \cos ^{2} \alpha - 4 \cos \alpha - \sqrt{3} = 0 \Rightarrow (2 \cos \alpha - \sqrt{3})(2 \sqrt{3} \cos \alpha + 1) = 0$
$\Rightarrow \cos \alpha = \frac{\sqrt{3}}{2}$ or $\cos \alpha = -\frac{\sqrt{3}}{6}$ (discard), thus $\alpha = 30^{\circ}$.
Similarly, $\frac{B C}{E C} = \frac{\sin \beta}{\sin (150^{\circ} - \beta)} = \sqrt{3} - 1 \Rightarrow \frac{\sin (150^{\circ} - \beta)}{\sin \beta} = \frac{1}{2} \cot \beta + \frac{\sqrt{3}}{2} = \frac{\sqrt{3} + 1}{2}$
$\Rightarrow \cot \beta = 1$, thus $\beta = 45^{\circ}$, hence $\triangle A B C$ is an isosceles right triangle.
Connecting $O C$, it is easy to see that $O A D C$ is a square, then in $\triangle A B D$, $B D^{2} = 2^{2} + 1^{2} = 5$
$\Rightarrow B D = \sqrt{5}$. By the conclusion of (1), the minimum value of $f(P)$ is $\sqrt{10}$.