16. In Rt $\triangle A B C$, $\angle A=90^{\circ}, A C=3$, $A B=4, B C=5$, point $D$ is on side $B C$. If the perimeters of $\triangle A C D$ and $\triangle A B D$ are equal, then $S_{\triangle A B D}$ is ( ).
(A) $\frac{4}{3}$
(B) $\frac{3}{2}$
(C) 2
(D) $\frac{12}{5}$
(E) $\frac{5}{2}$
16. D.

Let $CD=x$.
From the fact that the perimeters of $\triangle ACD$ and $\triangle ABD$ are equal, we have
$$
\begin{array}{l}
AC + CD = AB + BD \\
\Rightarrow 3 + x = 4 + (5 - x) \\
\Rightarrow x = 3 .
\end{array}
$$

Thus, $S_{\triangle ABD} = \frac{1}{2} AB \cdot BD \sin B$
$$
= \frac{1}{2} \times 4 \times 2 \times \frac{3}{5} = \frac{12}{5} .
$$