5. C.

Let $S_{\triangle C E G}=x$.
$$
\begin{array}{l}
\text { By } \frac{S_{\triangle A B G}}{S_{\triangle A B G}}=\frac{S_{\triangle B C G}}{S_{\triangle C E G}} \Rightarrow \frac{S_{\triangle A B G}}{14}=\frac{8+6}{x} \\
\Rightarrow S_{\triangle A B G}=\frac{196}{x} . \\
\text { By } \frac{S_{\triangle A B G}}{S_{\triangle A C G}}=\frac{S_{\triangle B D G}}{S_{\triangle C D G}} \Rightarrow \frac{\frac{196}{x}}{14+x}=\frac{8}{6} \\
\Rightarrow x^{2}+14 x-147=0 \Rightarrow x=7 \\
\Rightarrow S_{\triangle A C D}=14+7+6=27 \\
\Rightarrow S_{\triangle A B C}=\frac{8+6}{6} \times 27=63 .
\end{array}
$$
5. As shown in Figure $2, D, E, F$ are points on the sides $BC, CA, AB$ of $\triangle ABC$, respectively, and $AD, BE, CF$ intersect at point $G$. If $S_{\triangle BDG}=8, S_{\triangle CDG}=6, S_{\triangle AEG}=14$, then $S_{\triangle ABC}$ equals $(\quad)$.
(A) 56
(B) 62
(C) 63
(D) Uncertain