2. Let the incircle $\odot O$ of $\triangle A B C$ touch $B C$ at point $D$, and draw the diameter $D E$ through $D$. Connect $A E$ and extend it to intersect $B C$ at point $F$. If $B F+C D=1998$, then $B F+2 C D=$
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 $B F=p-A B=C D$.
$$
\therefore B F+2 C D=1998+\frac{1}{2} \times 1998=2997 \text {. }
$$