10. (20 points) Given $\angle A, \angle B, \angle C$ are the three interior angles of $\triangle ABC$, and the vector
$$
\boldsymbol{\alpha}=\left(\cos \frac{A-B}{2}, \sqrt{3} \sin \frac{A+B}{2}\right),|\boldsymbol{\alpha}|=\sqrt{2}.
$$

If when $\angle C$ is maximized, there exists a moving point $M$ such that $|\overrightarrow{M A}|$, $|\overrightarrow{A B}|$, and $|\overrightarrow{M B}|$ form an arithmetic sequence, find the maximum value of $\frac{|\overrightarrow{M C}|}{|\overrightarrow{A B}|}$.
10. Since $|\alpha|=\sqrt{2}$, we have
$$
\begin{array}{l}
\cos ^{2} \frac{A-B}{2}+3 \sin ^{2} \frac{A+B}{2} \\
=2+\frac{1}{2} \cos (A-B)-\frac{3}{2} \cos (A+B)=2 \\
\Leftrightarrow \cos (A-B)=3 \cos (A+B) \\
\Leftrightarrow 2 \sin A \cdot \sin B=\cos A \cdot \cos B \\
\Leftrightarrow \tan A \cdot \tan B=\frac{1}{2} . \\
\quad \text { Also, } \tan C=-\tan (A+B)=\frac{\tan A+\tan B}{\tan A \cdot \tan B-1} \\
\quad=-2(\tan A+\tan B) \\
\quad \leqslant-4 \sqrt{\tan A \cdot \tan B}=-2 \sqrt{2} .
\end{array}
$$

The equality holds if and only if $\tan A=\tan B=\frac{\sqrt{2}}{2}$.
$$
\text { Let }|A B|=2 c \text {. }
$$

Since $|\overrightarrow{M A}|+|\overrightarrow{M B}|=4 c$, $M$ is a moving point on the ellipse $\frac{x^{2}}{4 c^{2}}+$ $\frac{y^{2}}{3 c^{2}}=1$.
Thus, point $C\left(0, \frac{\sqrt{2}}{2} c\right)$.
Let point $M(x, y)$. Then
$$
\begin{array}{l}
|M C|^{2}=x^{2}+\left(y-\frac{\sqrt{2}}{2} c\right)^{2} \\
=4 c^{2}-\frac{4}{3} y^{2}+y^{2}-\sqrt{2} c y+\frac{c^{2}}{2} \\
=-\frac{1}{3} y^{2}-\sqrt{2} c y+\frac{9 c^{2}}{2}(|y| \leqslant \sqrt{3} c) .
\end{array}
$$

When $y=-\sqrt{3} c$,
$$
\begin{array}{l}
|M C|_{\max }^{2}=\frac{7+2 \sqrt{6}}{2} c^{2} \Rightarrow|M C|_{\max }=\frac{\sqrt{6}+1}{\sqrt{2}} c \\
\Rightarrow\left(\frac{|\overrightarrow{M C}|}{|\overrightarrow{A B}|}\right)_{\max }=\frac{2 \sqrt{3}+\sqrt{2}}{4} .
\end{array}
$$