(14) Let $O$ be a point inside $\triangle A B C$, and $\overrightarrow{A O}=\frac{1}{3} \overrightarrow{A B}+\frac{1}{4} \overrightarrow{A C}$, then the ratio of the area of $\triangle O A B$ to the area of $\triangle O B C$ is $\qquad$ .
(14) $\frac{3}{5}$ Hint: From $\overrightarrow{A O}=\frac{1}{3} \overrightarrow{A B}+\frac{1}{4} \overrightarrow{A C}$, we can get
$$
5 \overrightarrow{O A}+4 \overrightarrow{O B}+3 \overrightarrow{O C}=\overrightarrow{0} \text {. }
$$

As shown in the figure, extend $O A$ to $A^{\prime}$ such that $\left|O A^{\prime}\right|=$ $5|O A|$, similarly extend $O B$ to $B^{\prime}$ such that $\left|O B^{\prime}\right|=$ $4|O B|$, and extend $O C$ to $C^{\prime}$ such that $\left|O C^{\prime}\right|=$ $3|O C|$. Then
$$
\overrightarrow{O A^{\prime}}+\overrightarrow{O B^{\prime}}+\overrightarrow{O C^{\prime}}=\overrightarrow{0} \text {. }
$$

Therefore, $O$ is the centroid of $\triangle A^{\prime} B^{\prime} C^{\prime}$. Thus
$$
S_{\triangle O A^{\prime} B^{\prime}}=S_{\triangle O B^{\prime} C^{\prime}}=S_{\triangle O C^{\prime} A^{\prime}} \text {. }
$$

Also,
$$
\begin{array}{l}
\frac{S_{\triangle O A B}}{S_{\triangle O A^{\prime} B^{\prime}}}=\frac{O A}{O A^{\prime}} \cdot \frac{O B}{O B^{\prime}}=\frac{1}{20} \\
\frac{S_{\triangle O B C}}{S_{\triangle O B^{\prime} C^{\prime}}}=\frac{O B}{O B^{\prime}} \cdot \frac{O C}{O C^{\prime}}=\frac{1}{12}
\end{array}
$$

Therefore, $\frac{S_{\triangle O A B}}{S_{\triangle O B C}}=\frac{3}{5}$.