16. (15 points) As shown in Figure 2, it is known that the ellipse $C$ passes through the point $M(2,1)$, with the two foci at $(-\sqrt{6}, 0)$ and $(\sqrt{6}, 0)$. $O$ is the origin, and a line $l$ parallel to $OM$ intersects the ellipse $C$ at two different points $A$ and $B$.
(1) Find the maximum value of the area of $\triangle OAB$;
(2) Prove that the lines $MA$, $MB$, and the $x$-axis form an isosceles triangle.
16. (1) Let the equation of the ellipse $C$ be
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) \text {. }
$$

From the problem, we have
$$
\left\{\begin{array} { l } 
{ a ^ { 2 } - b ^ { 2 } = 6 , } \\
{ \frac { 4 } { a ^ { 2 } } + \frac { 1 } { b ^ { 2 } } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a^{2}=8, \\
b^{2}=2 .
\end{array}\right.\right.
$$

Therefore, the equation of the ellipse is
$$
\frac{x^{2}}{8}+\frac{y^{2}}{2}=1 \text {. }
$$

Since the line $l \parallel O M$, we can set
$$
l: y=\frac{1}{2} x+m \text {. }
$$

Substituting the above equation into (1) gives
$$
x^{2}+2 m x+2 m^{2}-4=0 \text {. }
$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}+x_{2}=-2 m, x_{1} x_{2}=2 m^{2}-4 \text {. }
$$

Since the line $l$ intersects the ellipse $C$ at two distinct points $A$ and $B$, we have
$$
\Delta=(2 m)^{2}-4\left(2 m^{2}-4\right)>0 \text {. }
$$

Thus, $m \in(-2,2)$, and $m \neq 0$.
Therefore, $S_{\triangle O A B}=\frac{1}{2}|m|\left|x_{1}-x_{2}\right|$
$$
\begin{array}{l}
=\frac{1}{2}|m| \sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=|m| \sqrt{4-m^{2}}=\sqrt{m^{2}\left(4-m^{2}\right)}
\end{array}
$$
$\leqslant 4$.
The equality holds if and only if $m^{2}=4-m^{2}$, i.e., $m= \pm \sqrt{2}$.
Therefore, the maximum area of $\triangle O A B$ is 4.
(2) Let the slopes of the lines $M A$ and $M B$ be $k_{1}$ and $k_{2}$, respectively. Then
$$
k_{1}=\frac{y_{1}-1}{x_{1}-2}, k_{2}=\frac{y_{2}-1}{x_{2}-2} .
$$

We only need to prove: $k_{1}+k_{2}=0$.
In fact,
$$
\begin{array}{l}
k_{1}+k_{2}=\frac{\frac{1}{2} x_{1}+m-1}{x_{1}-2}+\frac{\frac{1}{2} x_{2}+m-1}{x_{2}-2} \\
=1+m\left(\frac{1}{x_{1}-2}+\frac{1}{x_{2}-2}\right) \\
=1+m \cdot \frac{\left(x_{1}+x_{2}\right)-4}{x_{1} x_{2}-2\left(x_{1}+x_{2}\right)+4} \\
=1+m \cdot \frac{-2 m-4}{2 m^{2}-4-2(-2 m)+4}=0 .
\end{array}
$$

Therefore, the lines $M A$ and $M B$ form an isosceles triangle with the $x$-axis.