16. (12 points) As shown in Figure 1, given that $D$ is any point on side $AB$ of $\triangle ABC$ with an area of 1, $E$ is any point on side $AC$, and $F$ is any point on segment $DE$. Let $\frac{AD}{AB}=x, \frac{AE}{AC}=y, \frac{DF}{DE}=z$, and $y+z-x=\frac{1}{2}$. Try to find the maximum area of $\triangle BDF$.
16. As shown in Figure 1, connect $B E$. Then
$$
\begin{array}{l}
S_{\triangle \triangle I N}=z S_{\triangle A D E}=z(1-x) S_{\triangle A B E} \\
=z(1-x) y S_{\triangle U C C}=z(1-x) y .
\end{array}
$$

By the AM-GM inequality, we have
$$
z(1-x) y \leqslant\left[\frac{z+(1-x)+y}{3}\right]^{3}=\frac{1}{8} \text {. }
$$

Equality holds if and only if $z=1-x=y, y+z-x=\frac{1}{2}$, i.e., $x=y$ $=z=\frac{1}{2}$. 
Therefore, the maximum area of $\triangle B D F$ is $\frac{1}{8}$.