16. As shown in Figure 6, given that $B D$ and $C E$ are the angle bisectors of $\triangle A B C$, $F$ is the midpoint of side $D E$, and the distances from $F$ to $A C$, $A B$, and $B C$ are $F G = a$, $F H = b$, and $F M = c$ respectively. If $c^{2}-c-2 a b+\frac{1}{2} m^{2}-2 m+\frac{5}{2}=0$,
(1) Find the values of $a$, $b$, $c$, and $m$;
(2) Prove: $D G=\frac{1}{4}(B C-C D)$.
16. (1) As shown in Figure 9, draw $E Q \perp A C$ at point $Q$, $E N \perp B C$ at point $N$, and $D K \perp B C$ at point $K$.

Since $C E$ is the angle bisector of $\angle A C B$, we have
$$
E Q=E N .
$$

In $\triangle D E Q$, because $F$ is the midpoint of $D E$ and $F G \parallel E Q$, we have
$$
E N=E Q=2 F G=2 a \text {. }
$$

Similarly, $D K=2 F H=2 b$.
In quadrilateral $E N K D$, since
$$
\begin{array}{l}
E N \parallel F M \parallel D K \Rightarrow E N+D K=2 F M \\
\Rightarrow 2 a+2 b=2 c \Rightarrow a+b=c .
\end{array}
$$

From the given condition,
$$
a b=\frac{1}{2} c^{2}-\frac{1}{2} c+\frac{1}{4} m^{2}-m+\frac{5}{4} \text {. }
$$

Thus, $a$ and $b$ are the two real roots of the quadratic equation
$$
x^{2}-c x+\frac{1}{2} c^{2}-\frac{1}{2} c+\frac{1}{4} m^{2}-m+\frac{5}{4}=0
$$

Notice that,
$$
\begin{array}{l}
\Delta=c^{2}-4\left(\frac{1}{2} c^{2}-\frac{1}{2} c+\frac{1}{4} m^{2}-m+\frac{5}{4}\right) \\
=-(c-1)^{2}-(m-2)^{2} \leqslant 0 .
\end{array}
$$

Therefore, $\Delta=0$.
Thus, $c=1, m=2$.
Since $a+b=c$ and $a=b$, we have $a=b=\frac{1}{2} c=\frac{1}{2}$.
(2) From (1), we know that $E N=D K$. Therefore, $E D \parallel B C$.

Since $E C$ bisects $\angle E B C$, we have
$$
\angle E C D=\angle E C B=\angle C E D \Rightarrow E D=D C \text {. }
$$

Similarly, $E B=E D$.
Thus, $\triangle E Q D \cong \triangle D K C$ and $\triangle E Q D \cong \triangle E N B$.
Therefore, $B N=D Q=C K$.
Then $B C-C D=B N+C K=2 D Q=4 D G$
$\Rightarrow D G=\frac{B C-C D}{4}$. 
(Ro Jinhai provided)