(14) It is known that $A$ and $B$ move on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ (with $C$ and $D$ being the left and right foci, respectively), as shown in the figure. Let the line $M A: y=k x$, and solve the system
$$
\left\{\begin{array}{l}
y=k x, \\
\frac{x^{2}}{25}+\frac{y^{2}}{16}=1,
\end{array}\right.
$$

Then
$$
\left\{\begin{array}{l}
x_{A}^{2}=\frac{25 \times 16}{16+25 k^{2}}, \\
y_{A}^{2}=k^{2} x_{A}^{2},
\end{array}\right.
$$

We can obtain
$$
|M A|^{2}=x_{A}^{2}+y_{A}^{2}=\left(1+k^{2}\right) \frac{25 \times 16}{16+25 k^{2}} .
$$

By substituting $-\frac{1}{k}$ for $k$, we get
$$
\begin{aligned}
|M B|^{2} & =\left[1+\left(-\frac{1}{k}\right)^{2}\right] \frac{25 \times 16}{16+25\left(-\frac{1}{k}\right)^{2}} \\
& =\left(1+k^{2}\right) \frac{25 \times 16}{16 k^{2}+25} .
\end{aligned}
$$

Then $S_{\triangle M B}^{2}=\frac{1}{4}|M A|^{2} \cdot|M B|^{2}$
$$
\begin{array}{l}
=\frac{1}{4} \times 25 \times 16 \times \frac{25 \times 16\left(1+k^{2}\right)^{2}}{\left(16 k^{2}+25\right)\left(16+25 k^{2}\right)} \\
=\frac{1}{4} \times 25 \times 16 \times\left[1-\frac{81}{25 \times 16 k^{2}+\frac{25 \times 16}{k^{2}}+\left(25^{2}+16^{2}\right)}\right] \\
\geqslant 100\left[1-\frac{81}{2 \sqrt{25 \times 16 k^{2} \cdot \frac{25 \times 16}{k^{2}}}+\left(25^{2}+16^{2}\right)}\right] \\
=100 \times \frac{40^{2}}{41^{2}} .
\end{array}
$$

Therefore, the minimum area of $\triangle A M B$ is $\frac{400}{41}$, which is achieved when $k= \pm 1$.
(14) Given a line segment $CD$ of length 6 with midpoint $M$, two triangles $\triangle ACD$ and $\triangle BCD$ are constructed on the same side with $CD$ as one side, and both have a perimeter of 16. It is also given that $\angle AMB=90^{\circ}$. Find the minimum area of $\triangle AMB$.