13. (15 points) In the figure, $A B C D$ is a parallelogram, $E$ is on side $A B$, $F$ is on side $D C$, $G$ is the intersection of $A F$ and $D E$, and $H$ is the intersection of $C E$ and $B F$. It is known that the area of parallelogram $A B C D$ is $1, \frac{\mathrm{AE}}{\mathrm{EB}}=\frac{1}{4}$, and the area of triangle $B H C$ is $\frac{1}{8}$. Find the area of triangle $A D G$.


【Analysis】Let out the base and height of the parallelogram, determine the position of point $F$, and then express the lengths of $C F$, $D F$, $B E$, and $A E$ using the base of the parallelogram. Then, use the relationship between the base and height of the parallelogram and the base and height of triangle $A D G$ to solve the problem.

【Solution】Solution: Let the base of parallelogram $A B C D$ be $a$, and the height be $h$, with $a h=1$. $A E=\frac{a}{5}, B E=\frac{4 a}{5}, h=\frac{1}{a}$.
1. Calculate the position of point $F$ on $C D$:
$$
\begin{array}{l}
S_{\triangle B E H}=B E \times h \div 2-S_{\triangle B C H}, \\
=\frac{4}{5} a \times \frac{1}{2 a}-\frac{1}{8}, \\
=\frac{11}{40} ;
\end{array}
$$
$h_{1}=2 \times S_{\triangle B E H} \div B E$ ( $h_{1}$ is the height from $B E$ in $\triangle B E H$ ),
$$
\begin{array}{l}
=2 \times \frac{11}{40} \div \frac{4}{5} a, \\
=\frac{55}{80 a} ;
\end{array}
$$
$S_{\triangle C F H}=C F \times\left(h-h_{1}\right) \div 2$,
$$
=C F \times h \div 2-S_{\triangle B C H} \text {, }
$$

So $C F \times\left(\frac{1}{a}-\frac{55}{80 a}\right) \div 2=C F \times \frac{1}{a} \div 2-\frac{1}{8}$,
$$
\begin{array}{c}
C F \times \frac{25}{160 a}=C F \times \frac{80}{160 a}-\frac{20}{160}, \\
C F \times \frac{55}{160 a}=\frac{20}{160}, \\
C F=\frac{4 a}{11} ; \\
D F=D C-C F=\frac{7 a}{11} ;
\end{array}
$$
2. Calculate the area of $\triangle A D G$:
$$
\begin{array}{l}
S_{\triangle A D G}=S_{\triangle A D E}-S_{\triangle A E G}, \\
=A E \times h \div 2-A E \times h_{2} \div 2,\left(h_{2} \text { is the height from } A E \text { in } \triangle A E G\right) \\
=\frac{a}{5} \times \frac{1}{a} \div 2-\frac{a}{5} \times h_{2} \div 2, \\
=\frac{1}{10}-\frac{a}{10} \times h_{2}, \text { (1) } \\
S_{\triangle A D G}=S_{\triangle A D F}-S_{\triangle D F G}, \\
=D F \times h \div 2-D F \times\left(h-h_{2}\right) \div 2, \\
=\left(D F \times h_{2}\right) \div 2, \\
=\frac{7 a}{11} \times h_{2} \div 2, \\
=\frac{7 a}{22} \times h_{2},
\end{array}
$$
Substitute (2) into (1):
$$
\begin{array}{c}
\frac{7 a}{22} \times h_{2}=\frac{1}{10}-\frac{a}{10} \times h_{2}, \\
\frac{70 a}{220} \times h_{2}=\frac{22}{220}-\frac{22 a}{220} \times h_{2}, \\
h_{2}=\frac{22}{92 a}, \\
S_{\triangle A D G}=\frac{7 a}{22} \times h_{2}, \\
=\frac{7 a}{22} \times \frac{22}{92 a}, \\
=\frac{7}{92} ;
\end{array}
$$

Answer: The area of $\triangle A D G$ is $\frac{7}{92}$.