Example 9. As shown in Figure 5, a circle intersects the three sides of an equilateral triangle at six points. Given that $A G=2, G F=13, F C=1, H J=7$. Find $DE$.
(33rd American High School Mathematics Examination)
Let $A H=x$. By the secant-tangent theorem, we have
$$
A H \cdot A J=A G.
$$
For $A F$,
$$
\begin{array}{l}
\because A G=2, G F \\
=13, H J=7, \\
\therefore A H(A H+7) \\
=2(2+13)=30 .
\end{array}
$$

Rearranging, we get $A H^{2}$
$$
+7 A H-30=0.
$$

Solving this, we get $A H_{1}=3, A H_{2}=-10$ (discard).
$$
\begin{array}{l}
\text { Also, } \because A C=A B, \\
\therefore B J=(2+13+1)-(3+7)=6 .
\end{array}
$$

Let $B D=a, C E=b$. By the secant-tangent theorem, we get the system of equations
$$
\left\{\begin{array}{l}
a(16-b)=6 \times(6+7), \\
b(16-a)=1 \times(1+13) .
\end{array}\right.
$$

Solving this system, we get $\left\{\begin{array}{l}a=10-\sqrt{22}, \\ b=6-\sqrt{2} .\end{array}\right.$
$$
\therefore D E=16-(a+b)=2 \sqrt{22}.
$$