In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If $AG=2$, $GF=13$, $FC=1$, and $HJ=7$, then $DE$ equals 

[asy]
size(200);
defaultpen(fontsize(10));
real r=sqrt(22);
pair B=origin, A=16*dir(60), C=(16,0), D=(10-r,0), E=(10+r,0), F=C+1*dir(120), G=C+14*dir(120), H=13*dir(60), J=6*dir(60), O=circumcenter(G,H,J);
dot(A^^B^^C^^D^^E^^F^^G^^H^^J);
draw(Circle(O, abs(O-D))^^A--B--C--cycle, linewidth(0.7));
label("$A$", A, N);
label("$B$", B, dir(210));
label("$C$", C, dir(330));
label("$D$", D, SW);
label("$E$", E, SE);
label("$F$", F, dir(170));
label("$G$", G, dir(250));
label("$H$", H, SE);
label("$J$", J, dir(0));
label("2", A--G, dir(30));
label("13", F--G, dir(180+30));
label("1", F--C, dir(30));
label("7", H--J, dir(-30));[/asy]

$\textbf {(A) } 2\sqrt{22} \qquad \textbf {(B) } 7\sqrt{3} \qquad \textbf {(C) } 9 \qquad \textbf {(D) } 10 \qquad \textbf {(E) } 13$
1. We start with the given equations:
   \[
   x + y + z = 16
   \]
   \[
   x(x + y) = 6 \cdot 13 = 78
   \]
   \[
   z(y + z) = 14
   \]

2. From the first equation, we can express \(x\) and \(z\) in terms of \(y\):
   \[
   x = 16 - y - z
   \]

3. Substitute \(x = 16 - y - z\) into the second equation:
   \[
   (16 - y - z)(16 - z) = 78
   \]
   Expanding and simplifying:
   \[
   256 - 16z - 16y + yz + z^2 = 78
   \]
   \[
   256 - 16z - 16y + yz + z^2 = 78
   \]
   \[
   z^2 + yz - 16z - 16y + 178 = 0
   \]

4. Similarly, substitute \(z = 16 - x - y\) into the third equation:
   \[
   (16 - x - y)(16 - y) = 14
   \]
   Expanding and simplifying:
   \[
   256 - 16y - 16x + xy + y^2 = 14
   \]
   \[
   256 - 16y - 16x + xy + y^2 = 14
   \]
   \[
   y^2 + xy - 16y - 16x + 242 = 0
   \]

5. To solve these equations, we can use the substitution method. From the first equation, we have:
   \[
   x + y + z = 16 \implies x = 16 - y - z
   \]

6. Substitute \(x = 16 - y - z\) into the second equation:
   \[
   (16 - y - z)(16 - z) = 78
   \]
   \[
   256 - 16z - 16y + yz + z^2 = 78
   \]
   \[
   z^2 + yz - 16z - 16y + 178 = 0
   \]

7. Similarly, substitute \(z = 16 - x - y\) into the third equation:
   \[
   (16 - x - y)(16 - y) = 14
   \]
   \[
   256 - 16y - 16x + xy + y^2 = 14
   \]
   \[
   y^2 + xy - 16y - 16x + 242 = 0
   \]

8. Solving these equations, we get:
   \[
   x = 10 - \frac{y}{2}
   \]
   \[
   z = 6 - \frac{y}{2}
   \]

9. Substitute these into the second equation:
   \[
   (10 - \frac{y}{2})(10 + \frac{y}{2}) = 78
   \]
   \[
   100 - \left(\frac{y}{2}\right)^2 = 78
   \]
   \[
   \left(\frac{y}{2}\right)^2 = 22
   \]
   \[
   y = 2\sqrt{22}
   \]

10. Verify by substituting \(y = 2\sqrt{22}\) into the third equation:
    \[
    (6 - \frac{y}{2})(6 + \frac{y}{2}) = 14
    \]
    \[
    (6 - \sqrt{22})(6 + \sqrt{22}) = 14
    \]
    \[
    36 - 22 = 14
    \]
    This confirms that \(y = 2\sqrt{22}\) is correct.

The final answer is \(\boxed{2\sqrt{22}}\)