Problem 546

2. 2997.

As shown in Figure 6, let
$\odot O$ be tangent to $AB$ and
$AC$ at points $M$ and $N$, respectively. Draw
a line $GH \parallel BC$ through
point $E$, intersecting
$AB$ and $AC$ at
points $G$ and $H$, respectively. Then $GH$ is tangent
to $\odot O$ at point $E$, and
$\triangle A G E \backsim \triangle A B F, \triangle A G H \backsim \triangle A B C$.
Let the perimeters of $\triangle A G H$ and $\triangle A B C$ be $2 p^{\prime}$ and $2 p$, respectively. Then
$$\begin{aligned} & A G+G E=A G+G M=A M=A N=A H+H N \\ = & A H+H E=p^{\prime} . \end{aligned}$$

Thus, $\frac{p^{\prime}}{p}=\frac{2 p^{\prime}}{2 p}=\frac{A G}{A B}=\frac{G E}{B F}=\frac{A G+G E}{A B+B F}$
$$=\frac{p^{\prime}}{A B+B F} \text {. }$$
$\therefore p=A B+B F$. Hence $B F=p-A B=C D$.
$$\therefore B F+2 C D=1998+\frac{1}{2} \times 1998=2997 \text {. }$$