20.53 As shown in the figure, a circle intersects an equilateral triangle at six points. If $A G=2$, $G F=13$, $F C=1$, $H J=7$, then $D E$ equals
(A) $2 \sqrt{22}$.
(B) $7 \sqrt{3}$.
(C) 9 .
(D) 10 .
(E) 13 .
(33rd American High School Mathematics Examination, 1982)
[Solution] As shown in the figure, let $A H=y, B D=a, D E=x$, and $E C=b$.

From the given $A G=2, G F=13, \quad H J=7$ and $F C=1$, the side length of this equilateral triangle is
$$
2+13+1=16.
$$

Considering $A B$ and $A C$ as secants from point $A$ intersecting the circle at $H, J$ and $G, F$,

we have $y(y+7)=2(2+13)$, or $0=y^{2}+7 y-30=(y-3)(y+10)$,
thus $y=3$, and $B J=16-7-3=6$.
Similarly, $b(b+x)=1 \cdot(1+13)=14$,
$$
a(a+x)=6 \cdot(6+7)=78.
$$

Also,
$$
a+b+x=16,
$$

Solving the above three equations yields $x=2 \sqrt{22}$. Therefore, the answer is $(A)$.