79. As shown in the figure, in rectangle $A B C D$, $A B=4, B C=6, E, F, G, H$ are on $A B, B C, C D, D A$ respectively, and $A E: E B=3: 1, B F: F C=2: 1, D G: G C=1: 3, A H: H D=1: 2, P$ is on $H F$, and the area of quadrilateral $A E P H$ is 5, find the area of quadrilateral $P F C G$.
Answer: 8.

Solution: As shown in the figure, connect $E H, E F, G H, G F$ and use $S_{1}, S_{2} \cdots S_{8}$ to represent the areas of these 8 small triangles.
It is easy to know that $A E=3, E B=1, B F=4, F C=2, C G=3, D G=1, H D=4, A H=2$,
$$
\begin{array}{c}
S_{1}+S_{2}=5, \quad S_{1}=\frac{1}{2} \times 2 \times 3=3, \quad S_{2}=5-3=2, \quad S_{4}=\frac{1}{2} \times 4 \times 1=2 . \\
S_{\text {trapezoid } A B F H}=\frac{1}{2} \times(2+4) \times 4=12, \\
S_{1}+S_{2}+S_{3}+S_{4}=12,
\end{array}
$$

Therefore,

So
$$
S_{3}=12-2-5=5, \quad S_{5}=\frac{1}{2} \times 4 \times 1=2, \quad S_{8}=\frac{1}{2} \times 2 \times 3=3 \text {. }
$$

Obviously
$$
S_{5}+S_{6}+S_{7}+S_{8}=12
$$

So
$$
S_{6}+S_{7}=12-2-3=7 \text {. }
$$

Let $S_{6}=x$, then $S_{7}=7-x$,

That is
$$
\frac{S_{2}}{S_{3}}=\frac{H P}{P F}=\frac{S_{6}}{S_{7}},
$$

Solving we get
$$
\begin{aligned}
\frac{2}{5} & =\frac{x}{7-x}, \\
x & =2,
\end{aligned}
$$

So
$$
S_{7}=7-x=5 .
$$

Therefore, the area of quadrilateral $P F C G$ $=S_{7}+S_{8}=5+3=8$.