1. As shown in Figure 3, $O$ is a point inside $\triangle A B C$, extending $A O$ intersects $B C$ at point $D$, extending $B O$ intersects $A C$ at point $E$, the areas of $\triangle B O D$, $\triangle A O E$, and $\triangle A O B$ are 7, $14$, and $21$ respectively. Then $\frac{B D}{C D}=$ $\qquad$ .
$$
=1 \cdot \frac{7}{8} \text {. }
$$

As shown in Figure 8, connect $C O$, and let the areas of $\triangle O C D$ and $\triangle O C E$ be $x$ and $y$ respectively. From the given conditions, we have
$$
\left\{\begin{array}{l}
\frac{14 + y}{21} = \frac{x}{7}, \\
\frac{7 + x}{21} = \frac{y}{14}.
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
x=8, \\
y=10.
\end{array}\right.\right.
$$

Therefore, $\frac{B D}{C D}=\frac{7}{8}$.