14. In $\triangle A B C$, points $A^{\prime}, B^{\prime}, C^{\prime}$ are on sides $B C, A C, A B$ respectively, $A A^{\prime}, B B^{\prime}, C C^{\prime}$ intersect at point $O$, and $\frac{A O}{O A^{\prime}}+\frac{B O}{O B^{\prime}}+\frac{C O}{O C^{\prime}}=92$. Find the value of $\frac{A O}{O A^{\prime}} \cdot \frac{B O}{O B^{\prime}} - \frac{C O}{O C^{\prime}}$.
14. As shown in the figure, since $\triangle A O B$ and $\triangle A^{\prime} O B$ have the same height, therefore,
$$
\begin{aligned}
\frac{A O}{O A^{\prime}} & =\frac{S_{\triangle A O B}}{S_{\triangle A^{\prime} O B}} \\
& =\frac{S_{\triangle C O A}}{S_{\triangle C O A^{\prime}}}+S_{\triangle C O A} \\
& =\frac{S_{\triangle A O B}+S}{S_{\triangle A^{\prime} O B}+S_{\triangle C O A^{\prime}}} \\
& =\frac{S_{\triangle B O C}+S}{S_{\triangle B O C}}.
\end{aligned}
$$

Let $x=S_{\triangle B O C}, y=S_{\triangle C O A}$,
$z=S_{\triangle A O B}$, then
$$
\frac{A O}{O A^{\prime}}=\frac{z+y}{x}, \frac{B O}{O B^{\prime}}=\frac{x+z}{y}, \frac{C O}{O C^{\prime}}=\frac{y+x}{z} \text {. }
$$

Therefore,
$$
\begin{array}{l}
\frac{A O}{O A^{\prime}} \cdot \frac{B O}{O B^{\prime}} \cdot \frac{C O}{O C^{\prime}} \\
=\frac{(z+y)(x+z)(y+x)}{x y z} \\
=\frac{z+y}{x}+\frac{x+z}{y}+\frac{y+x}{z}+2 \\
=\left(\frac{A O}{O A^{\prime}}+\frac{B O}{O B^{\prime}}+\frac{C O}{O C^{\prime}}\right)+2 \\
=92+2=94 \text {. } \\
\end{array}
$$